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CHAPTER NO.1 REAL NUMBERS (Weightage 10 – 11 Marks)
BASIC CONCEPTS:
1. Rational numbers are numbers which can be written in the form of ,
where both ‘p’ and ‘q’ are integers and q≠ 0.
2. Irrational numbers are numbers which can’t be express in the form of
rational numbers.
3. The set of rational and irrational numbers together are called Real
numbers.
4. The fundamental theorem of Arithmetic: “ Every composite number can be
expressed as a product of primes and this factorization is unique ( apart
from the order in which the prime factors occur).
5. H.C.F : Product of the smallest powers of each common prime factors in
the numbers.
6. L.C.M : Product of the greatest powers of each prime factors in the
numbers.
7. The product of two positive integers is equal to the product of their L.C.M
and H.C.F.
8. Let x = be a rational numbers, such that the prime factorization of “q” is
of the form 2n.5m where n,m are non negative integers. Then ‘x’ has a
decimal expansion which terminates
9. If ‘q’ is not in the form of 2n.5m then that ‘x’ has a decimal expansion
which is non – terminating repeating ( recurring )
10. (i) a , x are two positive integers and a≠1 and x = an then = n
(ii) Logarithm of any number to a given base is the value of index to which
the base must be raised to get the given number.
11. LAWS OF LOGARITHMS:
i. = +
ii. = -
iii. = m.
iv. = 1
v. = 0
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12.
N W Z Q R
R = Q QI
SHORT ANSWER QUESTIONS (S.A.Q) (2 MARKS)
1. Find the L.C.M and H.C.F of 220 and 284 by applying the prime factorization
method ?
2. Without actually performing long division express the following in the
decimal expression
i. ii.
3. Expand the following logarithms
i.
ii.
iii.
iv.
4. Simplify each of the following expression as log N.
i. Log 2 + log 5
ii. Log 10 + 2 log3 – log 2
iii. 3 log4
iv. 2 log 3 – 3 log
VERY SHORT ANSWER QUESTIONS ( V.S.A.Q ) (1 MARK)
5. Express 156 as a product of the prime factors.
6. Show on the number line.
7. Express 0.375 in form
8. = x express this in the powers
Real Numbers ( R )
Irrational
Numbers
( QI )
Rational Numbers ( Q )
Integers (Z)
Whole Numbers (W)
Natural Numbers ( N)
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9. Determine the value of the following .
i. ii. )
ESSAY QUESTIONS ( E.Q) ( 4 MARKS)
10. Prove that is an irrational number by the method of contradiction.
11. Prove that 5 - is an irrational number.
12. Prove that 3 - 2 is an irrational number.
MULTIPLE CHOICE QUESTIONS ( ½ MARK )
13. Which of the following is rational number between and 1 ( )
a.
b.
c.
d. 0
14. 7.7 ia a ( )
a. Rational
b. Irrational
c. Both
d. Neither rational nor
irrational
15. Which of the following is irrational ( )
a.
b.
c.
d.
16. form of 0.875 is ( )
a.
b.
c.
d.
17. = ( )
a. 2
b. 3
c. -2
d. -3
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FILL UP THE BLANKS
18. Decimal expansion of is . . . . . . . . . . . . . .
19. = 2 express in the form of logarithm . . . . . . . . . . . . . .
20. = 1024 express in the form of logarithm . . . . . . . . . . . . . .
21. Logarithmic expansion of log 10000 is . . . . . . . . . . . . . .
22. Simplify log 16 – 2log 2 = . . . . . . . . . . ..
23. Value of = . . . . . . . . . . .
24. Logarithmic expansion of log is . . . . . . . .. . . . . . .
25. Value of = . . . . . . . . . . . . . . . . .
MATCHING
GROUP.A GROUP . B
26. = ( ) a) 1
27. Log m.x = ( ) b) 0
28. = ( ) c) -1
29. = ( ) d) -2
30. = ( ) e) 2
f) m. log x
g) log m + log x
ANSWERS:
1. 220 = x 5 x 11 ( mark )
284 = x 71 ( mark )
H.C.F. = = 4 ( mark )
L.C.M = x 5 x 11 x 71 = 15,620 ( mark )
2. (i) = ( mark )
= ( mark )
= ( mark )
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= 0.52 ( mark )
3. (i) = + - ( 1 mark)
= 2 log p + 3 log q - logr ( 1 mark)
4. (i) log 2 + log 5 = log ( 2x 5 ) ( 1 mark )
= log 10 ( 1 mark )
(ii) log 10 + 2 log3 – log 2
= log 10 + - log 2 ( mark )
= log ( ) ( mark )
5
= log ( ) ( mark )
= log 45 ( mark )
5. 156 = 2 x 78
= 2 x 2 x 39
= 2 x 2 x 3 x 13 ( mark )
= x 3I x 13I ( mark )
6.
( 1 mark)
( Divide 1 unit into 4 equal parts )
7. 0.375 =
= ( mark )
=
= ( mark )
0
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8. = x
9 = 3x ( = n x = mn ) 1 mark
9. = x
243 = 3x ( mark )
35 = 3x
5 = x ( mark )
= x = 5
10. Let us assume, to the contrary that is rational
= form
Let us divide both ‘r’ and ‘s’ with their common factor
then = ( where a and b are co-primes)
b = a
3b2 = a2
3 divides a2
Hence 3 divides a (1) ( 1 mark )
let a = 3c ( where ‘c’ is a positive integer)
3b2 = a2
3b2 = (3c)2
3b2 = 9c2
b2 = 3c2
3 divides b2
3 divides b ( 2 ) ( 1 mark )
from (1) and (2) 3 divides both ‘a’ and b’ ½ mark
But ‘a’ and ‘b’ are co-primes ½ mark
Our assumption is not correct ½ mark
is irrational ½ mark
11. Let us assume that 5 - is rational number ½ mark
5 - = form (where a, b are relative primes ) ½ mark
5 - = ½ mark
As is rational 5 - is also rational ½ mark
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But, 5 - = ½ mark
is also a rational number
It is false
So our assumption is not correct ½ mark
5 - is an irrational number ½ mark
12. Let us assume that 3 - 2 is rational number ½ mark
3 - 2 = form (where a, b are relative primes ) ½ mark
3 - =2 ½ mark
as is rational 3 - is also rational ½ mark
But 3 - =2 ½ mark
is also a rational number
It is false
So our assumption is not correct ½ mark
3 - 2 is an irrational number ½ mark
13. a
14. a
15. c
16. c
17. c
18. = = = 0.115
19. 8x = 2 x =
20. 210 = 1024 10 =
21. Log 10000 = log 24 x 54
= log 24 + log 54
= 4 log 2 + 4 log5
= 4 ( log2 + log5 )
22. Log 16 – 2 log2 = log16 – log 22
= log 16 – log4
= log
= log 4
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23. 0
24. = log 343 – log125
= log 73 – log 53
= 3log7 – 3 log5
= 3(log7 – log 5)
25. = = = 8(1) = 8
26. f
27. g
28. b
29. a
30. c
CHAPTER NO.2 SETS (weightage 8 – 10 marks )
BASIC CONCEPTS :
1. A Set is a well defined collection of objects
2. An object belong to a set is known as an element of the Set
3. A Set can be represented in two ways (i) Roaster (or) tabular form (ii) Set
builder form In roaster form all the elements of a set are listed, the
elements being separated by commas and enclosed within braces, { }.
4. Elements in a Set are expressed by means of property or rules possessed by
the elements of a Set, this form is called as Set builder form
5. A Set which does not contain any elements is called the empty Set or the
null Set
6. A Set is called a finite set if it is possible to count the number of elements
in it.
7. A Set is called an infinite set if the number of elements in it is not finite
8. The number of elements in a set is called the cardinal number of the set.
9. Universal Set is denoted by ‘ ’ . Universal set is usually drawn as rectangle.
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10. If every element in the set A is in set B , then A is called a subset to B. It is
denoted by A B
11. Two sets A and B are said to be equal, if every element of A is also an
element of B and every element of B is also an element of A.
12. Union of sets A and B is denoted as A B.
A B = { x/x A or x B }
13. Intersection of Sets A and B is denoted by A B.
A B = { x/ x A and x B }
14. Difference of Sets A and B is denoted by A – B ( or ) B – A
A – B = { x/x A and x B }
B – A = { x/x B and x A }
A – B ≠ B – A
SHORT ANSWER QUESTIONS ( S.A.Q) 2 MARKS
1. A = { 1, 2, 5}, b = {2, 3, 4, 5 } then A B, A B
2. Write the following sets in the roaster form
i. A = { x/x N, x<7 }
ii. B = { x/ x is a letter in the word “school”}
3. A = {x:x2 = 25 and 6x=15 } verify whether it is empty set. Justify your
answer.
4. A = { 1, 3, 5, 7 }, B = { 1, 2, 3, 4, 6} then find A – B , B – A .
5. Give two examples from your daily life about disjoint sets.
6. List all the subsets of the following sets
i. { x, y, z }
ii. { a, b, c, d}
7. A = { 1, 2, 3, 4 }, B= { 1, 2, 3, 4, 5, 6, 7, 8 } then find A B, and A B.
What do you notice about the result
8. A = { 1, 2, 3, 4, 5 }, B= { 4, 5, 6, 7 } then find A – B , B – A, are they equal?
VERY SHORT ANSWER QUESTIONS (V.S.A.Q) 1 MARK
9. Write the following sets in roaster form and set builder form
i. The set of all Natural numbers which divide 42.
ii. The set of Natural numbers which are less than 10
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10. Let ‘A’ be the set of prime numbers less than 6 and P the set of prime
factors of 30. Check is A and P are equal?
11. { 2, 4, 6, 8, 10 } ≠ { x : x=2n+1 and x N} state the reason for this.
12. A = { 1, 2, 3 } and B = { 3, 4, 5 } then illustrate A B in venn – diagram
13. State the principle to find the cardinal number of n(A B)
14. A = { 0, 2, 4 } then find A and A A, and comment the result.
ESSAY EQUESTIONS (E.Q) 4 marks
15. If A = {x: x is an even natural number }
B= { x : x is an odd natural number }
C= { x : x is a prime number }
D = { x: x is a multiple of 3 } then find
a. A
B
b. A
B
c. C –
D
d. A
C
16. A = { 3, 6, 9, 12, 15, 18, 21 }, B = { 4, 8, 12, 16, 20 } C = {2, 4, 6, 8, 10, 12,
14, 16 } D={ 5, 10, 15, 20} then find the following .
a. A –
B
b. D –
B
c. A
C
d. B
D
MULTIPLE CHOICE QUESTIONS ( ½ MARK)
17. A = { x : x is a boy }, B = { x: x is a girl } ( )
a) A B =
b) A B =
c) A – B = B
d) A – B = 0
18. Shaded area represents in the following diagram ( )
a) A – B
b) B – A
c) A B
d) A B
19. Number of all the subsets of a Set A = { 1, 2, 3 } is ( )
a) 3
b) 8
c) 9
d) 6
20. V = { a, e , I, o, u }, B = { a, I, k, u }, the V – B = ( )
a) { a, e, I, o, u }
b) { a , e, o, u }
c) {e, o }
d) {a, e, o }
A
B
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21. Which of the following are equal sets ( )
a) A = { 1, 0 }; B= { a, b }
b) A = { a, o }; B = {b, o }
c) A = { 3, 6, 9 } ; B= { 9,
3, 6 }
d) A = { 1, 3, 5 } ; B = { 3,
5, 7 }
22. Which of the following is an infinite set ( )
a) A set of all the months in
a year
b) { 1, 2, 3, . . . . . . . . . . . 99,
100 }
c) Set of primes less than 99
d) Set of all odd prime
23. If sets A and B are disjoint sets then ( )
a) A B =
b) A B =
c) A – B =
d) B – A =
24. A = . . . . . ( )
a) b) A c) 0 d) 1
25. Set builder form of A – B is . . . ( )
a) { x/ x A and x B }
b) { x/x A or x B }
c) { x/x A and x B }
d) { x/x B and x A }
FILL UP THE BLANKS ( ½ MARK )
26. C= { x : x is a two digit natural number such that the sum of its digits is 8}
roaster form of this set = . . . . . . . . . . . . . . . .
27. B = { 5, 25, 125, 625 } set builder form of this set = . . . . . . . . . . . . . . . . . .
28. A = { x : x is a natural number greater than 50 and less than 100 } roaster
from of this set is = . . . . . . . . . . . . . . . . . . . . . .
29. If A and B are disjoint sets then A B = . . . . . . . .
30. The set theory was developed by . . . . . . . . . . . . . . .
31. Every set is a . . . . . . . . . . . . to itself
32. Null set is a . . . . . . . . . . . . . . . . to every set
33. Number of all the sub sets to a set which contain ‘n’ elements in it is . . .
. . . . . . . .
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34. A – B = A and B – A = B then sets A and B are said to be . . . . . . . . . . . . . . .
sets
35. Set builder form of A B = . . . . . . . . . . . . . . .
MATCHING :
GROUP – A GROUP – B
36. { x/x A or x B } ( ) a) A – B
37. { x/ x A and x B } ( ) b) A B
38. { x/x A and x B } ( ) c) B – A
39. { x/x B and x A } ( ) d) A B
40. A B and B A then ( ) e) A ≠ B f)A = B
ANSWERS :
1. A = { 1, 2, 5 } ; B = { 2, 3, 4, 5 }
A B = { 1, 2, 5 } { 2, 3, 4, 5 }
= { 1, 2, 3, 4, 5 } ( 1 mark)
A B = { 1, 2, 5 } { 2, 3, 4, 5 }
= { 2, 5 } ( 1 mark)
2. (i) A = { x/x N, x<7 }
Roaster form of A = { 1, 2, 3, 4, 5, 6 } ( 1 mark )
(ii) B = { x/ x is a letter in the word “school”}
Roaster form of B = { c, h, l, o, s } ( 1 mark )
3. A = {x:x2 = 25 and 6x=15 }
x2 = 25 x= x = +5 or -5
Solution set = { 5 , -5 } (1) ( 1 mark )
6x = 15 x = x =
Solution set = { } (2) ( ½ mark )
Common solution of (1) and (2) = { } =
Hence A is a null set ( ½ mark )
4. A = { 1, 3, 5, 7 } ; B = { 1, 2, 3, 4, 6 }
A – B = {1, 3, 5, 7 } - { 1, 2, 3, 4, 6 } = { 5, 7 } ( 1 mark)
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B – A ={ 1, 2, 3, 4, 6 } - {1, 3, 5, 7 } = { 2, 4, 6 } ( 1 mark)
5. (i) Set of all the students in our school
Set of all the teachers in our school
(ii) set of all the employees in our village
Set of all the un employees in our village
6. (i) { x, y, z }
Number of all the subsets = 2n = 23 = 8
List of all the sub sets { x }, { y } , { z } , { x, y } , { x, z } , { y, z } , { x, y, z }
and ( 1 mark)
(ii) { a, b, c, d }
Number of all the subsets = 2n = 24 = 16
List of the sub sets { a } , { b } , { c }, { d }, { a, b }, { a, c }, { a , d }, { b, c }, {
b, d }, { c , d } , { a , b, c }, { a, b, d }, { a , c, d }, { b, c, d }, { a, b, c,
d } and ( 1 mark )
7. A = { 1, 2, 3, 4 } B = { 1, 2, 3, 4, 5, 6, 7, 8 }
A B = { 1, 2, 3, 4 } { 1, 2, 3, 4, 5, 6, 7, 8 }
= { 1, 2, 3, 4, 5, 6, 7, 8 } (1) ( ½ mark )
A B= { 1, 2, 3, 4 } { 1, 2, 3, 4, 5, 6, 7, 8 }
= { 1, 2, 3, 4} (2) ( ½ mark )
From (1) and (2) I observe that A B = B ( ½ mark )
A B = A ( ½ mark )
8. A = { 1, 2, 3, 4, 5 } ; B = { 4, 5, 6, 7 } ( ½ mark )
A – B = { 1, 2, 3, 4, 5 } - { 4, 5, 6, 7 } = { 1, 2, 3} ( ½ mark )
B – A = { 4, 5, 6, 7 } - { 1, 2, 3, 4, 5 } = { 6, 7 } ( ½ mark )
A – B ≠ B – A
9. ( i ) The set of all natural numbers which divide 42.
Roaster form = { 1, 2, 3, 6, 7, 14, 21, 42 } ( ½ mark )
Set builder form = { x/x is a number which divide 42 } ( ½ mark )
10. ‘A’ be the set of prime numbers less than 6
A = { 2, 3, 5 }
‘P’ the set of prime factors of 30
P = { 2, 3, 5 }
A = P ( 1 mark )
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11. { 2, 4, 6, 8, 10 } is a finite set contain even natural numbers less than 11
{ x : s=2n+1 and x N } = { 1, 3, 5, 7, 9, . . . . . . . . . } is a infinite set ,
containing all odd natural numbers. Both are not equal.
{ 2, 4, 6, 8, 10 } ≠ { { x : x = 2n + 1 and x N }
12. A = { 1, 2, 3 } ; B = { 3, 4, 5 }
1 mark
A B = { 1, 2, 3 } { 3, 4, 5 } = { 3 }
A B = = shaded area = { 3 }
13. n(A B ) = n(A) + n(B) – n(A B) 1 mark
14. A = { 0, 2, 4 }
A = { 0, 2, 4 } { }
= { }
= ( ½ mark )
A A = { 0, 2, 4 } { 0, 2, 4 }
= { 0, 2, 4 }
= A ( ½ mark )
I observe that A = and A A = A
15. A = {x: x is an even natural number } = { 2, 4, 6, 8, 10 . . . . . . . } ( ½ mark )
B= { x : x is an odd natural number } = { 1, 3, 5, 7, 9, . . . . . . . . }
( ½ mark )
C= { x : x is a prime number } = { 2, 3, 5, 7, 11, 13, . . . . . . } ( ½ mark )
D = { x: x is a multiple of 3 } = { 3, 6, 9, 12, 15, . . . . . . . . } ( ½ mark )
(i) A B = { 2, 4, 6, 8, 10 . . . . . . . } { 1, 3, 5, 7, 9, . . . . . . . . }
= { 1,2,3,4,5,6,7,8,9,10 . . . . . . . . . } ( ½ mark )
(ii) A B = { 2, 4, 6, 8, 10 . . . . . . . } { 1, 3, 5, 7, 9, . . . . . . . . }
= { } = ( ½ mark )
(iii) C – D = { 2, 3, 5, 7, 11, 13, . . . . . . } - { 2, 3, 5, 7, 11, 13, . . . . . . }
= { 2, 5, 7, 11, 13 . . . . } ( ½ mark )
A
1 3
2
4
5
B
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(iv) A C = { 2, 4, 6, 8, 10 . . . . . . . } { 2, 3, 5, 7, 11, 13, . . . . . . }
= { 2 } ( ½ mark )
16. A – B = { 3, 6, 9, 12, 15, 18, 21 } - { 4, 8, 12, 16, 20 } = { 3,6,,9,15,18,21 } ( 1
mark )
D – B = { 5, 10, 15, 20} - { 4, 8, 12, 16, 20 } = { 5, 10, 15 } ( 1 mark )
A C = { 3, 6, 9, 12, 15, 18, 21 } {2, 4, 6, 8, 10, 12, 14, 16 }
= { 2,3,4,6,8,9, 10,12,14,15,16,18,21 } ( 1 mark )
B D = { 4, 8, 12, 16, 20 } { 5, 10, 15, 20} = { 20 } ( 1 mark )
17. b
18. b
19. b
20. c
21. c
22. d
23. a
24. a
25. c
26. { 17, 26, 35, 44, 53, 62, 71 }
27. { x/x = 5n, x N , n < 5 }
28. { 51, 52, 53, . . . . . . . . . . . 99 }
29. ( null set )
30. George Cantor
31. Sub set
32. Sub set
33. 2n
34. Disjoint Sets
35. A B = { x / x A and x B }
36. b
37. d
38. a
39. c
40. f
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CHAPTER 3
POLYNOMIALS Important formulae
1. Polynomials: Polynomials are algebraic expressions constructed using
constants and variables.
Coefficients operate on variables, which can be raised to various
powers of non – negative integer exponents.
Ex: - 2x2+5x, 3x
2-5x+6, -6y, x
2 etc are polynomials.
etc are polynomials.
2. Degree of a polynomials: If P(x) is a polynomial in x, the highest power of x
in P(x) is called the degree of the polynomial P(x)
S.No Polynomial Degree Name of the
Polynomial
1 3x, + 5 1 Linear polynomial
2 x2+5x+4
2x2 – 3x -
2 Quadratic polynomial
3 3x2-
5x2+3x+1
9x2-5x+
3
Cubic polynomial
3. General form of a polynomial
P(x) = a0xn + a1x
n-1+ a2x
n-2+ …… + an-1x+an is a polynomial of n
th degree,
where a0,a1,a2,……,an-1,an, are real coefficient and a0 0
4. Zero of a polynomial: A real number F is said to be a Zero of a polynomial
P(x) if P(x) =0
Note: 1. The zero of the linear polynomial ax + b is- in general for a linear
polynomial ax+b, (a 0) the graph of y = ax+b is a straight line which
intersects the x – axis at exactly one point, namely (- ,0) and zero of this
polynomial is the x-coordinate -
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2. The zeroes of a quadratic polynomial ax2 + bx + c are precisely the x-
coordinates of the points where the curve representing y = ax2 + bx + c
intersects the x-axis.
3. If α and are the zeroes of the polynomial ax2 + bx + c then x =
Here = is called discriminant.
S.No Discriminant Nature of the
roots
No of
Zeroes
Figure
1 > 0 Real and un equal 2
2 = 0 Real and equal 1
3 < 0 Imaginary of
complex
0
6. Maximum no of zeroes of quadratic polynomial = 2
7. Maximum no of zeroes of cubic polynomial = 3
8. Maximum no of zeroes of polynomial of degree n = n
9. If α and are the zeroes of the quadratic polynomial P(x) = ax2 + bx
+ c where a 0 then general form of P(x) = k[x2 – (α + ) x + α ]
where k is a constant.
10. (i) Product of the zeroes α + = - =-
(ii) Product of the zeroes α = =-
11. If α are zeroes of a cubic polynomial P(x) = ax3 + bx
2 + cx +
d then general form is P(x) = x3 - x
2 (α + + )+ x (α + -
α
(i) Sum of the zeroes α = - =
(ii) Product of the zeroes α = –
(iii) α + =
12 Division Algorithm: For any polynomial P(x) any non-zero
polynomial g(x) there are polynomials and r(x) such that
p (x) = g(x) q(x) + r(x)
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Here p(x) is the dividend
g (x) is the divisor
q(x) is the quotient and
r (x) is the remainder.
2 Marks problems
1. If p(t) = t3 – 1 find the values of p(1), p(-1) p(0), p(2) and p(-2)
Sol: p(t) = t3 – 1
p(1) = 13-1 = 0
p(-1) = (-1)3-1 = -1-1=-2
p(0) = (0)3-1 = 0-1=-1
p(2) = (2)3-1 = 8-1=7
p(-2) = (-2)3-1 = -8-1 = -9
2. Find a quadratic polynomial if the zeroes of it are 2 and -1/3 respectively?
Sol: Let α and are the zeroes of the polynomial ax2+bx+c
Here α
α =2+ ( ) = 2
α = 2( ) =
ax2+bx+c = k[x
2 – (α + ) x + α ]
= k[x2 – x + ( )]
= k[x2 – x ]
= k[–
]
For k = 3, quadratic polynomial = –
3. On dividing x3-3x
2+x+2 by a polynomial g(x), the quotient and remainder
were x-2 and -2x+4 respectively find g(x)
Sol: p(x) = x3-3x
2+x+2, g(x) = x- 2, r(x) = -2x+4
g(x) = ?
According division rule,
Dividend = Divisor x quotient + remainder
p (x) = g(x) q(x) + r(x)
x3-3x
2+x+2, g(x) ( x- 2)+( -2x+4)
x3-3x
2+x+2+2x-4 = g(x) ( x- 2)
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x3-3x
2+3x-1= g(x) ( x- 2)
g(x) = (x3-3x
2+3x-2) (x-2)
x-2) x3-3x
2+3x-2 (x
2-x+1
x3-2x
2
(-) (+)
-x2+3x
-x2+2x
x-2
x-2
0
g(x) = x2-x+1
4. Find the zeroes of the quadratic polynomial x2+7x+10 and verify the
relationship between the zeroes and the coefficient?
Sol: x2+7x+10 = x
2+2x+5x+10
= x(x+2) + 5(x+2)
= (x+2)(x+5)
So the value of x2+7x+10 is zero when
x+2=0 or x+5 = 0
i.e: x=-2 or x = -5
Sum of the zeroes = -2+(-5) = -2-5 = = -
Product of the zeroes = (-2) (-5) = =-
5. If p(x) = 5x7-6x
5+7x-6 find (i) Coefficient of x
5 (ii) degree of p(x) (iii)
constant term
Sol: p(x) = 5x7-6x
5+7x-6
(i) Coefficient of x5 =-6
(ii) Degree of p(x) =7
(iii) Constant term = -6
1 Mark Problems
1. Find the zeroes of the polynomial p(y) = y2-1?
Sol: Let p(y) = 0
y2-1 = 0
(y-1) (y+1) = 0
y-1 = 0 or y+1=0
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y=1 or y=-1
Zeroes of p(y)= y2-1 is 1 and -1
2. If the sum of the zeroes of the quadratic polynomial kx2-3x+1 is 1 then find
the value of k?
Sol: Sum of the zeroes of the polynomial ax2+bx+c =
Sum of the zeroes of kx2-3x+1=1
=1
=1
k = 3
3. Write any three quadratic polynomials in different terms?
Sol: 3x2-2x+5, x
2-5, 6x
2+5x-1
4. Check whether -3 and 3 are the zeroes of the polynomial x2-9?
Sol: Let p(x) = x2-9
p (-3) = (-3)2-9 = 9-0 = 0
p(3) = (3)2-9 = 9-9 = 0
-3 and 3 are the zeroes of the polynomial x2-9
5. Find the zeroes of the cubic polynomial x2-x
3
Sol: x2-x
3 = x
2(1-x)
So the value of x2-x
3 is zero when
x2 = 0 or 1-x = 0
x = 0 or x = 1
Zeroes if the cubic polynomial x2-x
3 = 0,1
6. If are the zeroes of the polynomial 4x3+8x
2-6x-2 then find the value of
α + ?
Sol: If α are the zeroes of ax3+bx
2+cx+d then α + =
Here a = 4, b=8, c=-6, d=-2
α + = = =
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4 Marks Problems
1. Verify that 1, -1 and -3 are the zeroes of the cubic polynomial x3+3x
2-x-3
and check the relationship between zeroes and the coefficient?
Sol: p(x) = x3+3x
2-x-3
Here a= 1, b=3, c=-1, d=-3
p(1) = (1)3+3(1)
2-1-3 =1+3-1-3=0
p(-1) = (-1)3-3(-1)
2-1-3 =1+3-1-3=0
p(-3) = (-3)3-3(-3)
2-(-3)-3 =-27+27+3-3=0
1, -1 and -3 are the zeroes p(x)
Sum of the zeroes α+ = 1+(-1)+(-3)
= 1-1-3 = = = =
Product of the zeroes α = (1)(-1)(-3)
= = = -
α + =(1)(-1)+(-1)(-3)+(-3)(1)
= -1+3-3
= =
2. Divide 3x2-x
3-3x+5 by x-1-x
2 and verify the division algorithm?
Sol: Dividend 3x2-x
3-3x+5 = -x
3+3x
2-3x+5
Divisor = x-1-x2
= - x2+x-1
- x2+x-1) x
3+3x
2-3x+5 (x-2
-x3+x
2-x
2x2-2x+5
2x2-2x+2
3
Quotient = x-2
Remainder = 3
According to division rule,
Dividend = Divisor quotient + remainder
= (- x2+x-1) (x-2) +3
= -x3+x
2-x+2x
2-2x+2+3
= -x3+3x
2-3x+5
= Dividenal
Division rule is verified
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3. Find all the zeroes of 2x4-3x
3-3x
2+6x-2 if you know that two of its zeroes are
and ?
Sol: Let p(x) = 2x4-3x
3-3x
2+6x-2
Since and are two zeroes of p(x)
Now p(x) can be divided by (x- ) (x+ )= x2-2
x2-2) 2x
4-3x
3-3x
2+6x-2 (2x
2-3x+1
2x4
-4x2
-3x2+x
2+6x
-3x2
+6x
x2-2
x2-2
0
Quotient = 2x2-3x+1
= 2x2-2x-x+1
= 2x(x-1)-1(x-1)
= (x-1)(2x-1)
So the value of 2x2-3x+1 is zero when
x-1 = 0 or 2x-1=0
x = 1 or 2x = 1
x = ½
The remaining two zeroes are 1 and ½
All the zeroes p(x) are , , 1 and ½
5 Marks Problems
1. Draw the graph of the polynomial p(x) = x2
– x – 12 and find the zeroes.
Justify the answers?
Sol: p(x) = x2
– x – 12
Let y = x2
– x – 12
x -4 -3 -2 -1 0 1 2 3 4 5
y 8 0 -6 -10 -12 -12 -10 -6 0 8
(x,y) (-
4,8)
(-
3,0)
(-2,-
6)
(-1,-
10)
(0,-
12)
(1,-
12)
(2,-
10)
(3,-
6)
(4,0) (5,8)
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Scale: on x-axis 1cm = 1unit
on y-axis 1cm = 2units
The graph of y = x2 – x – 12 intersects the x – axis at (-3,0) and (4,0)
The zeroes of x2
– x – 12 are 4 and -3
Proof:
p(x) = x2 – x – 12
= x2-4x+3x-12
= x(x-4)+3(x-4)
= (x-4)(x+3)
To find the zeroes of x2
– x – 12 either (x-4) or (x+3) should be equal to zero
x – 4 = 0 or x+3=0
x = 4 or x=-3
The zeroes of x2
– x – 12 are 4 and 3
2. Draw the graph of the polynomial p(x) = x2
– 6x +9 and find the zeroes.
Justify the answers?
Sol: p(x) = x2
– 6x + 9
Let y = x2
– 6x + 9
x -3 -2 -1 0 1 2 3 4
y 36 25 16 9 4 1 0 1
(x,y) (-3,36) (-2,25) (-1,16) (0,9) (1,4) (2,1) (3,0) (4,1)
Scale: on x-axis 1cm = 1unit
on y-axis 1cm = 2units
The graph of y = x2 – 6x + 9 intersects the x – axis at (3,0)
The zeroes of x2
– 6x + 9 are 3, 3
Proof:
p(x) = x2 – 6x + 9
= x2-3x-3x+9
= x(x-3)-3(x-3)
= (x-3)(x-3)
To get the zeroes of x2 – 6x + 9 either (x-3) or (x-3) should be equal to zero
x – 3 = 0 or x - 3=0
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x = 3 or x=3
The zeroes of x2
– 6x + 9 are 3, 3.
PART –B
Multiple Choice Questions
1. The remainder when 2x2+3x+1 is divided by x+2 is _________ [ ]
a) -3 b) 3 c) 15 d)12
2. If x = 1 is a zero of the polynomial f(x) = x3-2x
2+4x+k, then the value of k is
a) 9 b) 6 c) -3 d) 2
3. If are the zeros of the polynomial f(x) = x2-x-1 then + = ___ [ ]
a) 0 b) -1 c) 1 d) 2
4. The product of the zeroes of the polynomial x3-4x
2+x-6 is ______[ ]
a) 4 b) -4 c) 1 d) 6
5. A point on the graph of y = x2 – 3x -4 is [ ]
a) (-1, 0) b) (2,0) c) (3,0) d) (-2, -6)
6. Which of the following is not a polynomial [ ]
a) 5x2+7x-3 b) +3 c) x
4- d) x
2/3+6x-1/2
7. The zeroes of p(x) = ax+b is ________ [ ]
a) b) c) d)
8. The sum of the zeroes of 2x3-5x
2-14x+8 is ________ [ ]
a) b) -7 c) 4 d)
9. If the graph of a polynomial does not intersect the x-axis then number of
zeroes of the polynomial is _________ [ ]
a) 0 b) 1 c) 2 d) 3
10. If p(m) = m2-3m+1then the value of p(-1) is ______ [ ]
a) 4 b) -1 c) 5 d) 3
11. The graph of y = ax + b intersects the x –axis [ ]
a) (0, ) b) (- ,0) c) ( d) (0, )
12. Zeroes of the cubic polynomial x3-4x are _____ [ ]
a) 0,2,4 b) -2,0,2 c) 0,3,-2 d) 1,0,2
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25
Fill in the blanks
1. Zeroes of the quadratic polynomial 4u2+8u are ____________
2. If one zero of the polynomial x2-4x+3 is 1 the other zero is _________
3. If are the zeroes of x3-5x
2-2x+24, then α + =___________
4. The x- coordinates of the points where the graph of y = x2
– 6x + 8 intersects
the x-axis are ________
5. The value of the polynomial px2+qx+r at x=-1 is _______
III Match the following
A B
1 Cubic polynomial [ ] a) 2x4-2/3x
3+x-
2
2 Quadratic polynomial [ ] b) -2/7
3 Linear polynomial [ ] c)
4 Constant polynomial [ ] d) 2x3-3x+5
5 Bi quadratic
polynomial
[ ] e) x2-5x+6
f) x3/2
-5x2
g)
I 1.b 2.c 3.b 4.d 5.a 6.d
7.b 8.a 9.a 10.c 11.c 12.v
II. 1.0,-2 2.3 3.-2 4.-2,4 5.p-q+r
III. 1.d 2.e 3.c 4.b 5.a
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CHAPTER 4 PAIR OF LINIOR EQUATIONS IN TWO VARIABLES
Key Points:
Solution for pair of linear equations using graph model:
Pair of linear equations are a1x+b1y+c1 = 0, a2x+b2y+c2 = 0
1. If the ratios are the two lines having only one common point. The
pair of linear equations has unique solution. The equations are known as
“Consistent pair of linear equations:
2. If the ratios are = the two line have no common point and the lines
are parallel. The pair of linear equations have no solution. The equations are
known as consistent pairs of linear equation.
3. If the ratios are = = the two lines have infinite numbers of common
point and the lines are coincident lines (dependent lines). The pair of linear
equations have infinite solution. The equations are known as dependent pair
of linear equation.
5 Marks Questions
1. Check whether the following equations are consistent or inconsistent solve
the graphically?
Sol: 2x + y – 6 = 0 4x – 2y – 4 = 0
a1 = 2, b1 = 1, c1 = 6 a2 = 4, b2 = -2, c2 = -4
= = , = , = =
Here the ratios are then the ratios are not equal.
The equations are consistent, they have only one solution
2x + y – 6 = 0 4x – 2y – 4 = 0
2x + y = 6 4x – 2y = 4
y = 6 – 2x 2 (2x – y) =4
2x – y = 4/2 = 2
y = 2x – 2
y = 6 – 2x y = 2x – 2
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x = 0 y = 6 – 2 0 =
6
(0,6) x = 0 y = 2 0 – 2 = -
2
(0,-2)
x = 1 y = 6 – 2 1 =
4
(1,4) x = 1 y = 2 1 – 2 =
0
(1,0)
x = 2 y = 6 – 2 2 =
2
(2,2) x = 2 y = 2 2 – 2 =
2
(2,2)
x = 3 y = 6 – 2 3 =
0
(3,0) x = 3 y = 2 3 – 2 =
4
(3,4)
x = 4 y = 6 – 2 4 =
-2
(4,-2) x = 4 y = 2 4 – 2 =
6
(4,6)
2. 2x -3 y = 8 4x – 6y = 9
Graphically
Sol: 2x -3 y = 8
4x – 6y = 9
= = , = = , =
=
The equation are inconsistent the lines are parallel and no solution
2x -3 y = 8 4x – 6y = 9
3y = 2x – 8 6y = 4x – 9
y = y =
x = 0 y = = =-
2.6
(0,-2.6) x = 0 y = = =-
1.5
(0, -
1.5)
x = 1 y = = =-
2
(1,-2) x = 1 y = = =
-0.8
(1,-0.8)
x = 3 y = = =-
0.6
(3,-0.6) x = 3 y = = =
0.5
(3,0.5)
x = 4 y = = = 0 (4,0) x = 6 y = = =
2.5
(6,2.5)
x = 5 y = = = (5,0.6)
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0.6
Solution: Lines are parallel having no solutions
3. 9x +3 y +12 = 0 18x + 6y +2 4 = 0
Graphically
9x +3 y +12 = 0
18x + 6y +2 4 = 0
= = , = = , = =
=
So the equations are dependent they have infinite solutions.
9x +3 y +12 = 0 18x + 6y +2 4 = 0
3y = -9x – 12 6y = -18x – 24
y = y =
x = 0 y = = = -4 (0,-4)
x = -1 y = = =
=-1
(-1,-1)
x = -2 y = = =
=2
(-2,2)
x = -3 y = = =
=5
(-3,5)
x = 0 y = = = -4 (0,-4)
x = -1 y = = =
=-1
(-1,-1)
x = -2 y = = =
=2
(-2,2)
x = -3 y = = =
=5
(-3,5)
Solution: The lines are coincident they have infinite solution.
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4. 10 Students of class x took part in a mathematics quiz. If the number of girls
is 4 more than the number of boys, then find the number of boys and the
number of girls who took part in the quiz?
Sol: Number of boys = x
Number of girls = y
Total number of students participated in the quiz is x + y = 10 ……..(1)
The number of girls is 4 more then the number of boys
y = x + 4
-x + y = 4 ……… (2)
x + y = 10
-x + y = 4
=
The above equations are consistent they have only one solution
x + y = 10 -x + y = 4
y = -x + 10 y = x + 4
y = -x + 10 y = x + 4
x = 1 y = -1 +10 = 9 (1,9) x = 1 y = 1 + 4 = 5 (1,5)
x = 2 y = -2 +10 = 8 (2,8) x = 2 y = 2 + 4 = 6 (2,6)
x = 3 y = -3 +10 = 7 (3,7) x = 3 y = 3 + 4 = 7 (3,7)
x = 4 y = -4 +10 = 6 (4, 6) x = 4 y = 4 + 4 = 8 (4,8)
Solution: The common point of the lines is (3,7) number of boys = 3, no of
girls is 7
Try These
1. 4x + 2y – 10 = 0
3x – 2y – 4 = 0
2. 6x + 3y – 15 = 0
2x + y – 5 = 0
3. 6x + 8y – 4 = 0
3x + 4y – 2 = 0
4. 2x – 3y – 5 = 0
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4x – 6y – 15 = 0
5. 3x – y = 7
2x + 3y = 1
6. In a garden there are some bees and flowers if one bee sits on each flower
then one bee will be left. If two bees sit on each flower one flower will be
left. Find the number of bees and number of flowers?
7. 5 pencils and 7 pens together cost 50Rs. Where as 7 pencils and 5 pens
together cost 46Rs. Find the cost of one pencil and that of one pen?
4 Marks Questions
Key Points
Salvation method for the pair of linear equations
1. Model method
2. Graphical method
3. Substitution method
4. Elimination method
Solve the equation using substitution method
1) x + 2y = -1 …… (1)
2x – 3y = 12 ……… (2)
Sol: From equation (1) x + 2y = -1
2y = -x – 1
y =
Substituting the y value in equation (2)
2x – 3 ( )= 12
2x + = 12
=12
= 12
9x + 3 = 12 2 = 24
7x = 24 – 3 = 21
x = = 3
Substituting the x value in y
y = = = -2
x = 3, y = -2
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2. x + =6
3x - = 5
x + =6 3x - = 5
=6 = 5
xy + 6 = 6y 3xy – 8 = 5y
xy – 6y + 6 = 0 3xy – 5y – 8= 0
xy – 6y = -6 ….. (1) 3xy – 5y = 8 ………. (2)
y(x – 6) = -6
y =
Substituting y value in equation …. (2)
3x )-5 ( ) = 8
+ = 8
= 8
-18x + 30 = 8(x – 6)
-18x + 30 = 8x – 48
-18x – 8x = -30 – 48
-26x = -78
x = =3
Put x = 3 value in y
y = = = 2
x= 3, y = 2
3. The sum of a two digit number and the number obtained by reversing the
digits is 66. If the digits of the number differ by 2 find the number?
Sol: Ones place of a two digit number is = x
In tenth digit place is = y
Then the two digit number is = 10y + x
If revers the digit number is = 10x + y
Sum of the two numbers = 66
(10y + x) + (10x + y) = 66
11x + 11y = 66
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11 (x + y) = 66
x + y = = 6 ….. (1)
Difference of the two digits is = 2
x – y = 2 ……. (2)
x + y = 6
y = 6 – x
Substituting the y value in equation (2)
x – (6 – x) = 2
x – 6 + x = 2
2x = 2 +6 = 8
x = =4
Substituting x = 4 in y value
y = 6 – 4 = 2
x = 4, y = 2 the number is = 10y + x = 10 2+4 = 24
Reverse the number is = 10x + y = 10 4+2 = 42
By using elimination method to solve the problems
1. 3x + 5y =9 ….. (1)
3x + 2y = 4 ……….. (2)
Sol: Equating the variables in the above equation
(1) 2 ……. 16x + 10y = 18
(2) 5 ……. 10x + 10y = 20
x = -2
Substituting the x value in (2)
3 (-2) + 2y = 4
-6 + 2y = 4
2y = 4 + 6 = 10
y = = 5
x = -2, y = 5
2. The taxi charges in Hyderabad are fixed, along with the charge for the
distance covered. For a distance of 10k.m, the charge paid is 220Rs for a
journey of 15km the charge paid is 310Rs.
Sol: The fixed charge is = x
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Charge for 1km is = y
For a distance of 10km, paid by the charge is = 220Rs
x + 10y = 220 ….. (1)
Paid the charge for the journey of 15km is = 310Rs
x + 15y = 310 ….. (2)
i) (2) – (1) x + 15y = 310
x + 10y = 220
5y = 90
y = = 18
Substituting the y = 18 value in 1
x + 10(18) = 220
x + 180 = 220
x = 220 – 180 = 40
The fixed charge is x = 40Rs
Charge for the journey is = 18Rs
ii) To pay for travelling a distance of 25 km is
= Fixed charge + 25 charge for 1km
= 40 + (25 18)
= 40 + 450 = 490Rs
Equations reducible to a pair of linear equations in two variables
Solve each of the following points of equations by reducing then to a pair of
linear equations.
1. + = 2
- = -1
Sol: 2 ( )+ 3( )= 2
4 ( )- 9 ( )= -1
Say = p. = d
2p + 3d = 2 ……. (1)
4p – 9d = -1 ……. (2)
Equal the variables of d
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1 3 ……… 6p + 9d = 6
……. 4p – 9d = -1
10p = 5
p = =
Substituting p = ½ in ….. 1
2 ½ + 3d = 2
1 + 3d = 2
3d = 2-1 = 1
d = 1/3
p = . then
= . =
= 2 , = 3
x = 22 = 4, y = 3
2 = 9
x = 4, y = 9
2. + =
- =
Sol: Say = p = d
p + d = p – d =
4p + 4d = 3 …. (1) =
p – d = 2 =
4 (p – d) = -1
4p – 4d = -1 ……. (2)
1 + 2 4p + 4d = 3
4p – 4d = -1
8p = 2
p = =
Substituting the p = ¼ values in 1
4 (1/4) + 4d = 3
1 + 4d = 3
4d = 3 – 1 = 2
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d = =
but p = , d = then
= = , =
3x + y = 4…(3) 3x – y = 2 …… (4)
3 + 4 3x + y = 4
3x – y = 2
6x = 6
x = = 1
Substituting the x = 1 in 3
3(1) + y =4
y = 4 – 3 = 1
x = 1, y = 1
Try These
Solve each of the following pairs of equations by reducing them to a pair of linear
equations
1. 6x + 3y = 6xy
2x + 4y = 5xy
2. = 2
= 6
3 A man travels 370km partly, by train and partly by car. If he covers 250km
by train and the rest by car, it takes him 4 hours, but if he travels 130km by
train and the rest by car, it takes 18 minutes more find the speed of the train
and that of the car?
Elimination method
1. An algebraic text book has a total of 1382 pages, it is broken up into two
parts. The second part of the book has 64 pages more than the first part. How
many pages are in each part of the book?
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36
2. The ratio of incomes of two persons is 9:7 and the ratio of their expenditures
is 4 : 3 if each of them manages to save 2000 Rs per month, find their
monthly income.
3. In a competitive exam, 3 marks are to be awarded for every correct answer
and for every wrong answer, 1 mark will be deducted, madhu scored 40
marks in this exam had 4 marks been awarded for each correct answer and 2
marks deducted for each in correct answer, madhu would have scored 50
marks. How many questions were there in the test?
Substitution Method
Solve the given pair of equations using substitution method
1. 2x + 3y = 9
3x + 4y = 5
2. 0.2x + 0.3y = 13
0.4x + 0.5y = 2.3
3. 3x – 5y = -1
x – y = -1
2Marks and 1 Mark Questions
1. Give an example and explain to consistent pairs of linear equations?
Sol: Example: 2x + y – 5 = 0
3x – 2y – 4 = 0
To observe the ratios = = , = the ratio are
so the linear equations are represents two lines intersecting. They are
consistent pairs of linear equations.
2. Give an example and explain to dependent pair of linear equations?
Sol: x – y – 2 = 0
2x + 2y – 4 = 0
To observe the ratios of the linear equations = = , =
= they are = The linear equations represents two lines are
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coincident lines. They have infinite number of solutions. They are dependent
pair of linear equations.
3. Give an example and explain to inconsistent pair of linear equations?
Sol: 4x – 6y – 15 = 0
2x – 3y – 5 = 0
To observe the ratios of the linear equations = =
= = they are The linear equations represents two
lines are parallel lines. They have no solution they are inconsistent pair of
linear equations.
4. By comparing of ratios , , find out what ever the lines represented
by the following pairs of linear equations intersect at a point are parallel or
are coincident.
a) 5x – 4y + 8 = 0
7x + 6y – 9 = 0
= = , =
So, The linear equations represents intersecting lines.
b) 9x + 3y + 12 = 0
18x + 6y + 24 = 0
= = = =
So, = the linear equations represents the coincident lines.
c) 6x – 3y + 10 =0
2x – y + 9 = 0
= = =
So, the linear equation represents the parallel lines.
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5. Solve the equations by elimination method
x + y = 5
x – y = 1
Sol: x + y = 5 (1)
x – y = 1 (2)
2x = 6
x = = 3
Substituting the x = 3 in (1)
3 + y = 5
y = 5 -3 = 2
x = 3, y = 2
6. Reducing than to a pair of linear equations
= 2 = 6
+ = 2 = 6
+ = 2 - = 6
Say = p = d then
d + p = 2, d – p =6
Choose the correct answers
1. If y = 1 which x value satisfy the linear equation - = -2
[ ]
a) 1 b) 2 c) 3 d) 4
2. 6x – 3y = 11, -10x +6y = -22 pair of equations are [ ]
a) consistent b) inconsistent c) dependentd) quadratic
3. The point which corresponds to how much money you have to earn through
sales in order to equal the money you spent in production is called[ ]
a) break even point b) consistent point
c) inconsistent point d) quadratic point
4. Relation between distance and speed to find the time [ ]
a) b) c) d)
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5. In which quadratic the point (-5, -5) is [ ]
a) Q1 b)Q2 c)Q3 d)Q4
6. Which ratio satisfies dependent pair of linear equation [ ]
a) b) c) d)
7. Number of solutions for x + y – 9 = 0, x – y – 2 = 0 is [ ]
a) 1 b) 0 c) Infinite d) 9
8. General form of the linear equation is [ ]
a) ax + by = 0 b) ax + by + c =0
c) ax = by d) x/y
9. If 2x + y = 7, x + y = 5 then y= [ ]
a) 2 b) 3 c) 5 d) 7
10. If x = 0 they y value in 3x + 2y = 4 [ ]
a) 3 b) 2 c) 4 d) 1
CHAPTER 5
QUADRATIC EQUATIONS
Key concepts
a,b,c are real numbers and a 0 then ax2+bx+c=0 is called quadratic
equation. ax2+bx+c=0 (a 0)is the general form of quadratic equation. y=
ax2+bx+c=0 is called quadratic function.
Solution of a quadratic equation by factorization and completing the square
Formula to solve quadratic equation is x=
is called the discriminant of quadratic equation.
By using discriminant we determine the roots of the quadratic equation.
i) If > 0 the roots of the quadratic equation is two distinct real roots.
ii) If = 0 the roots of the quadratic equation its two equal real roots.
iii) If < 0 the quadratic equation is no real roots.
4 Marks Questions
1. Find the roots of the following quadratic equations by factorization
i)
Sol:
100x2-20x+1=0
100x2-10x-10x+1=0
ii)
Sol:
x2+7x+5 =0
x2+5x+2x+5 =0
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10x(10x-1)-1(10x-1)=0
(10x-1) (10x-1)=0
10x-1=0 10 x-1=0
10x=1 or 10x=1
x = or x =
x( x+5)+ ( +5)=0
( x+5)(x+ =0
x+5=0 or x+ =0
x= or x=-
x= ,
So the roots are real and equal
iii) 3(x-4)2 – 5(x-4)=12
Sol: 3(x-4)2 – 5(x-4)-12=0
Say x-4=a
3a2-5a-12=0
3a2-9a+4a-12=0
3a(a-3)+4(a-3)=0
(a-3)(3a+4)=0
a-3=0 or 3a+4=0
Put a=x-4
(x-4)-3=0 or 3(x-4)+4=0
x-7=0 3x-12+4=0
x=7 3x-8=0
x=
x=7,
2. Find the determinations of a rectangle whose perimeter is 28 meters and
whose area is 40 square meters?
Sol: Length of the rectangle = l
Breadth = b
Perimeter is =28mt
2(l + b) =28
l + b= =14
l + b = 14
b=14-l …….(1)
Area is = 40 sq. mts
lb = 40 ……(2)
Substituting the b value in (2)
l (14-l) = 40
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14l – 12 = 40
l 2 - 14l+40=0
l 2 -10l -4l+40=0
l (l-10)-4 (l-10)=0
(l - 10)(l - 4)=0
l - 10 = 0 or l - 4=0
l=10 or l = 4
Substituting the l value in b
l = 10 l = 4
b = 14 – 10 b=14-4 = 10
l = 10, b = 4 or l = 4, b = 10
3. Solution of quadratic equation by completing the square?
(i) x2
- 10x + 9 = 0
Sol: x2
- 10x + 9 = 0
x2
- 10x = -9
x2
– 2.x.5 = -9
Add both sides the square value of 5
x2
– 2.x.5+52 = -9 +5
2
x2
– 2.x.5+52 = -9 +25
(x - 5)2 = 16
x – 5 =
x – 5 = 4
x = 5 + 4 or 5 – 4
x= 9 or 1
(ii) 4x2 + 4 x + 3 = 0
Sol: 4x2 + 4 x = -3
(2x)2 + 4 x = - 3
(2x)2 + 2.2x. = -3
Add both sides the value ( )2
(2x)2 + 2.2x. + ( )
2 = -3 + ( )
2
(2x+ )2
= -3 + 3
(2x+ )2
= 0
2x+ = 0
2x = -
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x =-
4. - = , x -4,7
Sol: =
=
=
11 (x + 4) (x – 7) = 30x – 11
(x + 4) (x – 7) =
(x + 4) (x – 7) = -30
x2 – 7x + 4x – 28 = -30
x2 – 3x – 28 = -30
x2 – 3x – 28 + 30 = 0
x2 – 3x + 2 = 0
x2 – 3x = -2
x2 – .x.3 = -2
x2 – 2.x. = -2
x2 – 2.x. + ( )
2 = -2+( )
2
(x - )2 = -2+
(x - )2 =
(x - )2 =
x - =
x - =
x = or
or
or
x = 2 or 1
4. The difference of squares of two numbers is 180. The square of the smaller
number is 8 times the longer number. Find the two numbers?
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Sol: The large number is = x
Smaller number is = y say
Difference of squares of two numbers is = 180
x2 – y
2 = 180 ……. (1)
The square of the smaller number is 8 times the larger number
y2 = 8x
Substituting the value in 1
x2
– 8x = 180
x2 – 2.x.4 = 180
x2 – 2. x. 4
2 = 180 + 4
2
(x – 4)2 = 180+16
(x – 4)2
= 196
x – 4 = = 14
x – 4 = 14 or -14
x = 14 + 4 or -14 + 4
= 18 or -10
Then the value of y
y2 = 8 18 y
2 = 8 -10
y2 = 144 y
2 =-80
y2
= 12
The large no is = 18
Smaller no is = 12
5. A train travels 360 km at a uniform speed. If the speed had been 5km/h
more, it would have taken 1 hour less for the same journey. Find the speed
of the train.
Sol: The train travels at a uniform speed is = 360k.m
Speed of the train = x k.m
Time taken to complete the journey =
=
If the speed has been more 5km/h then the total speed is = x + 5
Time taken to complete the journey with more speed is =
For the same journey it would take 1 hour less if the speed is x + 5
- = 1
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360 - ) = 1
- =
=
=
x (x + 5) = 360 5
x2 + 5x = 1800
x2
+ 5x – 1800 = 0
x2 + 45x – 40x – 1800 = 0
x (x + 45) – 40 (x + 45) = 0
(x + 45) (x – 40) = 0
x + 45 = 0 x – 40 = 0
x = -45 x = 40
Speed of the train is = 40 km/h
Solution of A quadratic equation with quadratic formula
6. Find the nature of the roots of the following quadratic equation. if real roots
exist, find them
Sol: 2x2 – 6x + 3 = 0
a = 2, b = -6, c = 3
b2 – 4ac = (-6)
2 - 4 2 3
= 36 – 24 = 12 > 0
The roots of the quadratic equation is real different roots
x =
=
=
=
=
=
=
The roots x = ,
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7. 2x2
+ kx + 3 = 0 find the value of k the quadratic equation have two equal
roots
Sol: If 2x2 + kx + 3 = 0 have two equal roots then
b2 – 4ac = 0
a = 2, b = k, c = 3 then
k2 – 4 2 = 0
k2 – 24 = 0
k2 = 24
k = = = 2
k = 2
8. The sum of the ages of two friends is 20 years four years ago the product of
their ages is years was 48. Is the situation is possible? If so, determine their
present age?
Sol: Age of the one person = x day
Age of the second person = 20-x
4 years ago age of the first person = x – 4
Age of the second person = 20 – x – 4
= 16 – x
4 years ago the product of their ages in years was = 48
(x – 4) (16 – x) = 48
16x – x2 – 64 + 4x = 48
-x2 + 20x – 64 = 48
-x2 + 20x – 64 – 48 =0
-x2 + 20x-112 = 0
x2 – 20x+ 112 = 0
Observe the nature of the roots of the quadratic equation is
a = 1, b = -20, c = 112
b2 – 4ac = (-20)
2 – 4 1 112
= 400 – 448 = -48 < 0
So the quadratic equation has no real roots, so the situation is not possible.
Try These
Solve by Factorization
1) 2x2
– x + = 0
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2) In a class of 60 students, each boy contributed rupees equal to the number of
girls and each girl contributed rupees equal to the number of boys. If the
total money then collected was 1600 Rs. How many boys are there in the
class?
3) The attitude of a right triangle is 7cm less than its base. If the hypotenuse is
13cm, find the other two sides?
Solve by completing the square
4) Find two consecutive odd positive integers, sum of whose squares is 290?
5) Solve x2 + 5 = -6x
6) In a class test the sum of Moulikis marks in mathematics and English is 30.
If she got 2 marks more in mathematics and 3 marks less in English, the
product of her marks would have been 210. Find her marks in the two
subjects.
Solve by formula
7) Find the nature of the roots of the equation 3x2 - 4 =0 if real roots,
exist find them?
8) kx (x – 2) + 6 = 0 having two equal roots, find the k value?
9) Is it possible to design a regular park of perimeter 80m and area 400m2?If
so, find its length and breadth?
2 Marks and 1 Mark Questions
Check whether the following equation are quadratic are not?
1. (x – 3)(2x + 1) = x (x + 5)
Sol: 2x2 + x – 6x – 3= x
2 + 5x
2x2 – 5x – 3 = x
2 + 5x
2x2 – 5x – 3 – x
2 – 5x = 0
x2 – 10x – 3 = 0
It is quadratic equation.
2. (x + 2)3 = 2x (x
2 – 1)
x3 + 3.x
2. 2 + 3.x.2
2 + 2
3 = 2x
3 – 2x
x3 + 6x
2 + 12x + 8 = 2x
3 – 2x
2x3 – 2x – x
3 – 6x
2 – 12x – 8 =0
x3- 6x
2 – 14x – 8 = 0
It is not quadratic equation
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3. The product of two consecutive positive integers is 306. We need to find the
integers form the quadratic equation?
Sol: Say two consecutive positive integers = a, a+1
Its product is = 306
a (a + 1) = 306
a2
+ a = 306
a2 + a – 306 = 0
4. Verify that 1 and are the roots of the equation 2x2 – 5x + 3 = 0
Sol: If 1 and are the roots x = 1 or x =
Supplement the value in the quadratic equation it satisfies it becomes to
zero.
If x = 1 = 2(1)2 – 5(1) + 3
= 2 – 5 + 3 = 5 – 5 = 0
If x = = 2( )2 – 5( ) + 3
= 2( ) - + 3
= - + 3
=
=
= = 0
So, 1 and are roots of the quadratic equation.
5. Solve x2 – 3x – 10 = 0 by factorization?
Sol: x2 – 3x – 10 = 0
x2 – 5x + 2x – 10 = 0
x (x – 5) + 2 (x – 5) = 0
(x – 5) (x + 2)= 0
x – 5 = 0 or x + 2 = 0
x = 5 or x = -2
6. Find two numbers whose sum is 27 and product is 182?
Sol: Two numbers = a, b say
Sum of two numbers = a + b = 27 …….. (1)
Product of two numbers = ab = 182 …….(2)
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a + b = 27
b = 27 – a
Substituting the b value in (2)
a (27 – a) = 182
27a – a2 = 182
a2 – 27a + 182 = 0
a2 – 14a – 13a + 182 = 0
a (a – 14) – 13 (a – 14) = 0
(a – 14) (a – 13) = 0
a – 14 = 0 or a – 13 = 0
a = 14 or a = 13
b = 27 – 14 b = 27 – 13
= 13 = 14
Two numbers are = 13 and 14
7. Define discriminant, why discriminant used?
Sol: b2 – 4ac is called the discriminant of the quadratic equation
Discriminant can determine the nature of the roots of the quadratic equation.
8. Determine the roots of the equation 2x2 - 2 x + 1 = 0
Sol: a = 2, b = - 2 x, c = 1
b2 – 4ac = (- 2 )
2 – 4 (2) (1)
= (4 2) – 8
= 8 – 8 = 0
The nature of the roots of the quadratic equation is real and equal.
Choose the Correct Answers
1. If the root of the quadratic equation is not real then the discriminant is [ ]
a) b2 – 4ac = 0 b) b
2 – 4ac = 0 c) b
2 – 4ac < 0 d) b
2
– 4ac
2. Roots of 4x2 + 4x + 1 = 0 [ ]
a) 4,1 b) , c) 0.5, 2.5 d) ,
3. Sum of the roots of ax2 + bx + c = 0 [ ]
a) b) c) d)
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4. Product of the roots of 5x2 + 6x + 1 = 0 [ ]
a) b) c) d)
5. Number of roots for x2 – 9 = 0 [ ]
a) 1 b) 2 c) 3 d) 9
6. Quadratic formula is
7. Draw a graph with quadratic equation, its b2 – 4ac >0 it cuts the x – axis
_________ times
8. Draw a graph with quadratic equation, if the curve not cut the x – axis then
b2 – 4ac ___________
9. b2 – 4ac is called _______
10. In general form of the quadratic equation a = 0, then the equation is
__________
CHAPTER 6
PROGRESSIONS
Arithmetic Progression (A.P)
An Arithmetic Progression is a list of numbers in which each term is
obtained by adding a fixed number „d‟ to the preceding term, except the first term.
The fixed term„d‟ is called the „common difference‟ if a1, a2, a3,….. an are
terms of an A.P then d = a2 – a1 = a3 – a2 = ……… =an – an-1
General form of A.P
a, a + d, a + 2d, a + 3d, ……..
The nth
term of general term or last term of an AP is given by an = l = a + (n-
1) d
Sum of n terms of the AP where a, a + d, a + 2d, a + 3d,…..is Sn = [2a + (n
– 1) d]
When last term l is given then Sn = [a + l]
If Sn is the sum of first n terms of an AP then the nth
term of the AP is an = sn
– sn-1
Sum of first n natural numbers =
1 + 2 + 3 + 4 +………+ n =
Sum of the squares of first n natural numbers
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12 + 2
2 + 3
2 + 4
2 +………+ n
2 =
Sum of the cubes of first n natural numbers =
13 + 2
3+ 3
3 + 4
3 +………+ n
3 =
=
Let sum of first n natural numbers = S1
Sum of the squares of first n natural numbers = S2
Sum of the cubes of first n natural numbers = S3
9 = S3 (1+8S1) and
S3=
If a, b, c are in AP then b = and „b‟ is called the arithmetic mean of „a‟
and „c‟
Geometric Progression (GP)
A list of numbers obtained by multiplying the proceeding term by a fixed
number is said to be Geometric Progression (GP)
The fixed number is called the common ratio (r)
a, ar, ar2, …… is called the general form of a G.P
Common ratio r = = = ……… =
nth
term of G.P is an = arn-1
Sum of n terms of a G.P = if r > 1
Sn = if r <1
Sum of infinite terms of a GP is Sa =
If a, b, c are in GP then b = and „b‟ is called the Geometric mean of „a‟
and „c‟.
2Marks Problems
1.Find the common difference d and written three more terms of A.P 2, , 3, , ….?
Sol: 2, , 3, , …….
a = a1 = 2, a2 =
Common difference d = a2 – a1
= – 2
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=
d =
General term an = a + (n – 1) d
a5 = a + 4d
= 2 + 4 ( )
= 2 + 2
= 4
a6 = a5 + d
= 4 +
=
a7 = a6 + d
= +
= 5
Next three more terms are 4, 5
2. Which term of the A.P: 21, 18, 15, …….is -81?
Sol: 21, 18, 15, …….
Here a = 21, d = a2 – a1 = 18 – 21 = -3, an = -81
an = a + (n – 1)d
-81 = 21 + (n – 1) (-3)
-81 = 21 – 3n + 3
-81 = 24 – 3n
-81 – 24 = -3n
-105 = -3n
n = 35
35th
term in the given AP becomes -81
3. Find the 31st term of an AP whose 11
th term is 38 and 16
th term is 73?
Sol: nth
term of an AP is an = a + (n – 1) d
11th
term a11 = 38
a + 10d = 38 (1)
16th
term a16 = 73
a + 15d = 73 (2)
Solving (1) and (2)
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a + 15d = 73
a + 10d = 38
5d = 35
d = 7
(1) a + 10(7) = 38
a + 70 = 38
a = 38 – 70
a = -32
nth
term of AP is an = a + (n – 1) d
31st term a31 = a + 30d
= -32 + 30(7)
= -32 + 210
= 178
a31 = 178
31st term of AP = 178
4.How many terms of the AP. 24, 21,18, ….. Must be taken so that their sum is 78?
Sol: 24, 21,18, …….
a = 24, d = a2 – a1 = 21 – 24 = -3
Sn = 78, n =?
Sum of n terms of an AP is Sn = [2a + (n – 1) d]
78 = [2 (24) + (n – 1) (-3)]
78 = [48 – 3n + 3]
78 = [51 – 3n]
156 = 3n (17 – n)
52 = n (17 – n)
52 = 17n – n2
n2 – 17n + 52 = 0
n2 – 13n – 4n + 52 = 0
n (n – 13) – 4 (n – 13) = 0
(n – 13) (n – 4) = 0
n – 13 =0 or n – 4 = 0
n = 13 n = 4
Required no of terms = 13 or 4
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5. Find the 20th
and nth
term of the GP
Sol:
Here a = , r = = = =
n th
term of GP is an = arn-1
20th
term of GP is a20 = ar19
= ( )
19
a20 =
n th
term is an = arn-1
= ( )n-1
an =
6. Which term of the GP is
Sol:
Here a = , r = = = = an =
n th
term of GP is an = arn-1
= ( )n-1
=
2187 = 3n
37 = 3
n
n = 7
7th
term of the given GP is
7. The 4th
term of a geometric progression is and the seventh term is find
the geometric series?
Sol: n th
term of GP is an = arn-1
4th
term of GP is a4 =
ar
3 = (1)
7th
term of GP is a7 =
ar
3 = (2)
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=
r3 =
r3 =
r3 =
3
r =
(1) ar3 =
a 3 =
a =
a =
a =
a1 = a =
a2 = ar = ( ) =
a3 = ar2 = ( )
2 = = 1
a4 = ar3 = ( )3 = =
The required GP is , , 1, ,…..
1 Mark Problems
1. How many two digit numbers are divisible by 3?
Sol: The list of two digit numbers divisible by 3 is
12, 15, 18, …., 99
This is an A.P with
a = 12, d = a2 – a1 = 15 – 12 = 3, an = 99
an = a + (n – 1) d
99 = 12 + (n – 1) (3)
99 = 12 + 3n – 3
99 = 9 + 3n
99 – 9 = 3n
90 = 3n
n = 30
There are 30 two digit numbers divisible by 3.
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2. Find the 20th
term from the end of the AP 3, 8, 13, ….. 253?
Sol: Given AP is 3, 8, 13, ….. 253
Let 253, ….. 13, 8, 3 be the AP
Here a = 253 d = 3-8 = -5
nth
term of AP is an = a + (n – 1) d
20th
term of AP is a20 = 253 + (20 – 1) (-5)
= 253 – 95
= 158
a20 = 158
20th
term from the end of the AP 3, 8, 13, ….. 253 is 158
3. Find the sum of first 1000 positive integers?
Sol: S = 1 + 2 + 3 + 4 + ……. + 1000
Sn = [a + l]
= [1 + 1000]
= 500 1001
= 500500
Or
Sum of first n natural numbers =
Sum of first 1000 natural numbers =
=
= 500500
4. Write GP if a = 256, r = -
Sol: General form of GP = a, ar, ar2, ar
3 ……
= 256, 256 (- ), 256 (- )2, 256 (- )
3
= 256, -128, 64, -32, ……..
5. Find x so that x, x+2, x+6 are consecutive terms of a geometric progression.
Sol: Three consecutive terms of GP are x, x+2, x+6
In a GP =
=
(x+2)2 = x(x + 6)
x2 + 4x + 4 = x
2 + 6x
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4x + 4 = 6x
4 = 6x – 4x
4 = 2x
x = 2
6. Which term of the GP 2, 2 , 4, ….. is 128?
Sol: 2, 2 , 4, …..
Here a = 2 , r = = = , an = 128, n = ?
General term of AP is an = arn-1
128 = n-1
64 = n-1
64 = (21/2
)n-1
26 =
6 =
12 = n -1
n = 12 + 1
n = 13
13th
term of the GP is 128
7. Find the common ratio of the GP 25, -5, 1, -
Sol: GP 25, -5, 1, -
Common ratio r = = = -
4 Marks Problems
1. A manufacturer of TV sets produced 600 sets in the third year and 700 sets
in the seventh year assuming that the production increases uniformly?
(i) The production in the 1st year
(ii) The production in the 10th
year
(iii) The total production in first 7 year
Sol: Since the production increases uniformly by a fixed number every year, the
number of TV sets manufactured in 1st, 2
nd, 3
rd, year will form an AP
Let us denote the number of TV sets manufactured in the nth
year by an
an = a + (n – 1) d
a3 = 600
a + 2d = 600 (1)
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a7 = 700
a + 6d = 700 (2)
Solving (1) and (2)
a + 6d = 700
a + 2d = 600
4d = 100
d = 100/4
d = 25
(1) a + 2 (25) = 600
a + 50 = 600
a = 600 – 50
a = 550
(i) The production in the 1st year = 550
(ii) The production in the 10th
year a10 = a + 9d
= 550 + 9(25)
= 550 + 225
The production in the 10th
year a10 = 775
(iii) The total production in first 7 years
Sn = [2a + (n – 1) d]
S7 = [2(550) + (7 – 1) 25]
= [1100 + 150]
= 1250
= 7 625
= 4375
Total production of TV sets in first 7 years = 4375
2. If the geometric progressions 162, 54, 18,….. and , , ,… have their nth
term equal find the value of n?
Sol: In the GP 162, 54, 18, ……
a = 162, r = = =
nth
term of the GP is an = arn-1
= 162 (n-1
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In the GP , , ,…
a = , r = = = = 3
nth term of the GP is an = arn-1
= (n-1
According to problem nth
term of two GPs are equal
162 (n-1
= (n-1
162 = (
n-1
162 = 3n-1
3n-1
81 81 = 32n-2
38 = 3
2n-2
8 = 2n – 2
8 + 2 = 2n
10 = 2n
n = 5
Value of n = 5
3. The number of bacteria in a certain culture triples every hour. If there were
30 bacteria present in the culture originally. Then what would be number of
bacteria in fourth hour?
What would be number of bacteria in 10th
hour?
What would be number of bacteria in 20th
hour?
What would be number of bacteria in nth
hour?
Sol: No of bacteria in first hour = 30
Since for every hour no of bacteria triples
No of bacteria in second hour = 3 30 = 90
No of bacteria in third hour = 3 90 = 270
No of bacteria in fourth hour = 3 270 = 810
30, 90, 270, 810, …….. forms a GP with
a = 30, r = = = 3
nth term of the GP is an = arn-1
Number of bacteria in 10th
hour is a10 = 30 (3)9
Number of bacteria in 20th
hour is a20 = 30 (3)19
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Number of bacteria in nth
hour is an = 30 (3)n-1
4. A sum of Rs 700 is to be used to give seven cash prizes to students of a
school for their overall academic performance. If each prize is Rs. 20 less
than its preceding prize. Find the value of each prize?
Sol: Let the value of each prize = Rs x
Each prize is Rs 20 less than its preceding prize
The next prizes are x – 20, x – 40 ……
The value of 7 prizes from an AP
x, x – 20, x – 40, x – 60, ………
Here a = x, d = a2 – a1 = x – 20 – x = -20, S7 = 700, n = 7
Sum of n terms of an AP is
Sn = [2a + (n – 1) d]
700 = [2x + (7 – 1) (-20)]
700 = [2x -120]
700 = 2 (x – 60)
700 = 7 (x -60)
100 = (x – 60)
100 + 60 = x
x = 160
The seven prizes in rupees are 160, 140, 120, 100, 80, 60, 40
5. The sum of the 4th
and 8th term of an AP is 24 and the sum of the 6
th and 10
th
terms is 44. Find the first three terms of the AP?
Sol: nth term of the AP is an = a + (n – 1)d
4th
term of the AP is a4 = a + 3d
8th
term of the AP is a8 = a + 7d
According to the problem
a4 + a8 = 24
a + 3d + a + 7d = 24
2a + 10d = 24
2(a + 5d) = 24
a + 5d = 12 (1)
6th
term of the AP is a6 = a + 5d
10th
term of the AP is a10 = a + 9d
According to the problem
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a6 + a10 = 44
a + 5d + a + 9d = 44
2a + 14d = 44
2(a + 7d) = 44
a + 7d = 22 (2)
Solving (1) and (2) we get
a + 7d = 22
a + 5d = 12
2d = 10
d =
d = 5
(1) a + 5 (5) = 12
a + 25 = 12
a = 12 - 25
a = -13
1st term of the AP is a1 = a = -13
2nd
term of the AP is a2 = a1 + d = -13 + 5 = -8
3rd
term of the AP is a3 = a2 + d = -8 + 5 = -3
First three terms of the AP are -13, -8, -3
6. Two APs have the same common difference the difference between their
100th
terms is 100. What is the difference between their 1000th
terms?
Sol: Let the common difference of two APs = d
Let x and y be the first terms of the two APs respectively
nth term of the AP is an = a + (n – 1)d
100th
term of the 1st AP is a100 = x + 99d
100
th term of the 2
nd AP is a100 = y + 99d
According to the problem
(x + 99d) – (y – 99d) = 100
x + 99d – y – 99d = 100
x- y = 100 (1)
1000th
term of the 1st AP is a1000 = x + 999d
1000
th term of the 2
nd AP is a1000 = y + 999d
The difference between the 1000th
terms of the two APs
= (x + 999d) – (y + 99d)
= x + 999d – y – 999d
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= x- y
From equation (1) and (2) we get
x – y = 100
Difference between the 1000th
terms of two APs = 100
Choose the Correct answers
1. Common difference for the AP = , , ……… is ________[ ]
a) 4 b) c) d) -
2. Which of the following is an AP [ ]
a) 2, , 3, b) 0.2, 0.22, 0.222,… c) a, a2, a
3, a
4 d) , ,
,
3. 10th
term of the AP 5, 1, -3, -7,….. is [ ]
a) -25 b) -12 c) 0 d) -31
4. In an AP of a1 = 2, a3 = 26, then a2 = _________ [ ]
a) 16 b) 14 c) 12 d) 18
5. Which of the following is not a GP [ ]
a) 1, 4, 9, 16 b) 1, -1, 1, -1,…
c)-4, -20, -100, -500 … d) , , ,……..
6. In a GP if a1 = 7, r = then a7 = ________ [ ]
a) b) c) d)
7. The arithmetic mean of the number 2 and 8 is _________ [ ]
a) 10 b) 16 c) 5 d) 4
8. The 6th
term of the AP x + b, x+3b, x +5b,_____ is _______ [ ]
a) x +13b b) x+8b c) x+11b d) x+7b
9. If x, y, z are in AP then 2y= ______ [ ]
a) x – 2 b) x + z c) d)
10. If the nth term in an AP is 2n + 5 then its first term is ______ [ ]
a) 9 b) 11 c) 5 d) 7
11. If the nth term of a GP is 2(0.5)n-1
then the common ratio is _____[ ]
a) 2 b) 1 c) n-1 d) 0.5
12. If a, b, c are in GP then ________ [ ]
a) b-a = c –b b) b = a+c c) b2 = ac d) b =
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II Fill in the blanks
1.The A.M of the three quantities a+2, a, a-2 is _________
2.If tn = then t9 =________
3.The next term of = ____________
4.The common difference of the series 2a – b, 4a – 3b, 6a – 5b, …. Is_______
5.The number of odd numbers between 0 and 100 is _______
6.The sum of first „n‟ even natural numbers is ________
7.The third term of the sequence whose nth term is given by an = ( )n-1
is _______
8.The 9th
term of the series -3 + 6 – 12 + 24 – 48 + …. Is ________
9.The nth term of the series (a – 1) + (a – 2) + (a – 3) + ….. is ______
10.13 + 2
3 + 3
3 + 4
3 + 5
3 + ……… +50
3 = ___________
III Match the following
A B
1. nth
term of an ApP [ ] a)
2. nth
term of an GP [ ] b)
3. Sum of n terms of an AP [ ] c) a + (n – 1)d
4. Sum of n terms of an GP [ ] d) [2a+(n-1)d]
5. Sum of first n natural numbers [ ] e)
[ ] f)
[ ] g)
Answers
I 1.c 2.a 3.d 4.b 5.a 6.d
7.d 8.c 9.b 10d 11.d 12.c
II 1.a 2. 3. 4. 2a – 2b 5. 50
6. n (n + 1) 7. 8. – 768 9. a – n 10. 1625625
III 1.c 2.b 3.d 4.a 5.f
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CHAPTER 7
COORDINATE GEOMETRY 1. The distance between the points (x1, 0) and (x2, 0) =
2. The distance between the points (0, y1) and (0, y2) =
3. The distance between the points (x1, y1) and (x2, y2) =
4. Section formula: The coordinates of the point that divides the line joining
the points (x1, y1) and (x2, y2) internally in a ratio m1 : m2 =
If the ratio is k : 1 then coordinates =
5. Centroid of the triangle whose vertices are (x1,y1), (x2,y2) and (x3, y3) =
6. Area of the triangle whose vertices are (x1,y1), (x2,y2) and (x3, y3)
=
7. Heron‟s formula
Area of triangle A = where S =
8. Slope of line joining (x1, y1) and (x2, y2) is m =
1 Mark questions
1. What is the distance between the points (-4, 0) and (6,0)?
Sol: Distance between (-4, 0) and (6,0) =
=
=10 Units
2. Find the distance between A (2,0), B(0,4)?
Sol: According to Pythagoras theorem,
AB2 = 0A
2 + 0B
2
= 22 + 4
2
= 4 + 16
=20
AB =
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=
=2 Units
3. Find the radius of the circle whose centre is (3, 2) and passes through (-5, 6)
Sol: Radius of the circle
= Distance between (3, 2) (-5, 6)
=
=
=
=
=
=
= 4
4. Find the centroid of a triangle whose vertices are (-4, 6), (2, -2) and (2,5)
Sol: The centroid of the triangle whose vertices are (-4, 6), (2, -2) and (2,5)
=
=
= = (0,3)
5. Determine x so that 2 is the slope of the line through P(2,5) and Q (x, 3)
Sol: Given points = P(2,5) and Q (x, 3)
Slope of PQ = = 2
= 2 = 2
= x – 2
-1 = x – 2
x = -1 + 2 = 1
2 Marks Problems
1. If the distance between two points (x,7) and (1,15) is 10. Find the value of
x?
Sol: Distance between the points (x,7) and (1,15) =
= = 10
= = 10
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= = 10
Squaring on both sides we get
= 102 = 100 – 64 = 36
1 – x = = 6
(a) If 1 – x = 6 then –x = 6-1 = 5 x = -5
(b) If 1 – x = -6 then –x = -6-1 = -7 x = 7
2. Find the point which divides the line segment joining the points (3, 5) and
(8, 10) internally in the ratio 2 : 3?
Sol: Given points (3, 5), (8, 10) Ratio : 2:3
Required points =
=
=
=
= (5, 7)
3. Lahari said that the point (-4, 6) divides, the line segment joining the points
A(-6, 10) and B (3, -8) in the ratio 2: 7 but Rupa said that the ratio is 7 : 2
whom do you support? Why?
Sol: Let the ratio m1: m2
The point that divides the line segment joining (-6, 10), (3, -8) in m1: m2
ratio
= = (-4, 6)
= = (-4, 6)
Equalize the x coordinates we get
= = -4
3m1- 6m2 = - 4m1- 4m2
3m1+4m1 = -4m2 + 6m2
7m1= 2m2
=
The ratio = 2 : 7
So I support Lahari
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4. The x and y coordinates of a point are equal the point is equidistant from (1,
0) and (0, 3) find the point?
Sol: Let the point (x, x)
Distance between (x, x) and (1, 0) =
=
=
Distance between(x, x) and (0, 3) =
=
But (x, x) is equidistant from (1, 0) and (0, 3).
= =
Squaring on both sides
= =
= = - = 0
= (1 – x + 3 – x) (1 – x – 3 + x) = 0 [a2 – b
2 = (a+b)(a-b)]
= (4 – 2x) (-2) = 0
= 4 – 2x = 0 -2x = -4 x = = = 2
The required point = (2,2)
5. Sreeja found an old map in her house that has the information about an
Indian treasure. According to the given scale in the map the treasure point
and three statues which are at the points (3, 5) (-5, -4) and (7, 10) form a
parallelogram. But in the map at the treasure point was damaged. Help her to
find the treasure point?
Sol: Let the treasure is at the point A(x, y)
Given points = B (3,5), C (-5, -4), D(7, 10)
ABCD is a parallelogram
So midpoint of the diagonals are equal
Midpoint of AC =
=
=
Midpoint of BD =
=
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=
According to the data = =
= = 5 and =
= x – 5 = 2 5 = 10 and y – 4 = 2 = 15
= x = 5 + 10 = 15 and y = 15 + 4 = 19
The treasure point = (15, 19)
6. Find the tri sectional points of the line joining (-3, -5) and (-6, -8)?
Sol: Let A = (-3, -5) and B = (-6, -8)
The trisection point divides AB in 1 : 2 and 2 : 1 ratio
Given points = (-3, -5), (-6, -8) Ratio = 1 : 2
Required point =
=
=
= = (-4, -6)
(b) (x1, y1) = (-3, -5), (x2, y2) = (-6, -8), m1: m2 = 2 : 1
Required point =
=
= = (-5, -7).
The trisection points = (-4, -6) and (-5, -7).
7. Keerthana said that (1, -1) (4, 1) (-2, -3) are collinear. Samyuktha said that
these points form a triangle who is correct?
Sol: Given points = (1, -1) (4, 1) (-2, -3)
Area of the triangle formed by these three points
=
=
=
=
=
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= = 0
The points are collinear so keerthana is right
8. Find the value of K if the points (k, k), (2,3), (4,-1) are collinear (or lie on
same straight line)
Sol: Let A = (k, k) B= (2,3), C = (4,-1)
Area of ABC =
=
=
=
=
If A,B,C are collinear then area of ABC = 0
=
= 6K – 14 = 0 6K = 14
K = =
Note: The above problem may also be asked as given below.
“A point is on the line joining the points (2,3) and (4,-1) x and y coordinates
of the point are same. Find the point”
9. Find the ratio in which y axis divides the line segment joining the points (5, -
6) and (-1, -4) also find the point of intersection?
Sol: Let the ratio = K : 1 Points (5, -6), (-1, -4)
The points that divides the line joining (5, -6) and (-1, -4) is a ratio K : 1
=
=
=
This points lies on y axis so x coordinate = 0
= 0 k = 5 K : 1 = 5 : 1
Required ratio = 5 : 1
By substituting K = 5 we get the point of intersection =
=
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=
=
10. In the adjoining figure P(2,3) is mid point of line segment AB. Write down
the coordinates of A and B?
Sol: Let A = (x, 0) and B = (0,y)
Mid point of AB =
=
=
Given that midpoint = (2, 3)
= = (2, 3)
= 2 and = 3
x = 4 and y = 6
A = (4, 0) and B = (0, 6)
11. Find the distance of the point (1, 2) from the midpoint of the line segment
joining the points (6, 8) and (2, 4)?
Sol: Midpoint of the line segment joining (6, 8) and (2, 4)
=
=
= = (4, 6)
Distance between (1, 2) and (4,6) =
=
=
=
=
= 5 units
12. Among the points P(8,4), Q(6,7), R(9,0) which is closer to the origin?Why?
Sol: Distance between origin and P(8,4) =
=
=
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=
Distance between origin and Q(6,7) =
=
=
= units
Distance between origin and R(9,0) =
=
=
= units
= 9 units
< <
So P (8, 4) will be closer to the origin.
4 Marks Problems
1. A point is on y – axis, Mythili and Leela calculated the distance of the point
from (6,5) and (-4, 3) respectively and found both got same value. By using
distance formula find the point.
Sol: let the point on y – axis = p (0, y)
A = (6, 5) B = (-4, 3)
PA =
=
=
=
PB = =
=
=
PA = PB =
Squaring on both sides we get
= =
= - = 16-36
= (5 - y + 3 – y) (5 – y – 3 + y) = -20 [
= (8 – 2y) (2) = -20
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= 8 – 2y = = -10
= - 2y = -10 – 8 = -18
y = = 9
Required point = (0, 9)
2. Show that A (a, 0) B (-a, 0), C(0, a )
Distance between A (a, 0) B (-a, 0) =
=
=
= 2a
AB = 2a units
Distance between B (-a, 0) and C(0, a )
=
=
=
=
=
= 2a
BC = 2a units
Distance between A (a, 0) and C(0, a )
=
=
=
=
=
= 2a
AC = 2a units
AB = BC = AC = 2a units
Given points form an equilateral triangle.
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3. Prove that the point (-7, -3), (-5, 10), (15, 8) and (3, -5) taken in order are the
corners of a parallelogram and find its area?
Sol: Let A= (-7, -3), B = (-5, 10), C = (15, 8), D = (3, -5)
AB =
=
=
=
=
= Units
BC =
=
=
= Units
CD =
=
=
= Units
DA =
=
=
=
= Units
AB = CD and BC = DA
ABCD is a parallelogram
[By proving midpoint of AC = midpoint of BD also you can prove
that ABCD is a parallelogram]
To find the area: =
=
=
=
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=
= 154
= 77 sq. Units
Area of parallelogram ABCD = 2 Area of ABC
= 2 77 sq. units
= 154 Sq. units
4. Find the coordinates of the points which divides the line segment joining the
point A(-2, 2) and B(2, 8) into 4 equal parts?
Sol: Given points are A(-2, 2) and B(2, 8)
Required points = The points that divides AB in the ratios 1 :3, 2 : 2, and 3 :
1
(a) A(-2, 2), B(2, 8) Ratio : 1 :3
Point =
=
=
=
=
(b) (-2, 2) (2, 8) Ratio : 2 : 2 = 1 : 1
Point = Midpoint of AB
=
=
=
=
(c) (-2, 2) (2, 8) Ratio : 3 : 1
Point =
=
=
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=
Required points = (0, 5) and
5. Prasad has a site in the shape of an equilateral triangle. He observed that
from a fixed point the current can be represented by the points (0, -1), (2, 1),
(0, 3). He wants to construct a swimming pool by joining the midpoint of the
edges of the site. How much area will the pool occupies? What is the ratio of
the area of the pool and the total site?
Sol: Let A=(0, -1), B=(2, 1), C = (0, 3).
Area of ABC =
=
=
=
= 4 Sq units
Midpoint of BC =
=
=
=
D = (1, 2)
Midpoint of AB =
=
E =
Midpoint of AB =
=
F =
Area of swimming pool = Area of DEF
=
=
=
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75
=
=
= 2
= 1 Sq units
Ratio of the swimming pool area of total area = 1 : 4
6. If (1, 2) (4, y) (x, 6) and (3, 5) are the vertices of a parallelogram taken in
order find x, y?
Sol: Let A = (1, 2) B=(4, y) C=(x, 6) and D = (3, 5)
ABCD is a parallelogram
Midpoints of the diagonals are equal
Midpoint of AC =
=
=
=
Midpoint of AC =
=
Midpoint of AC = Midpoint of BD =
= and 4 =
1 + x = 7 and 8 = y + 5
x = 7 – 1 and 8 – 5 = y
x = 6 and y = 3
Practice these problems also:
1. If the distance between (2, -3) and (10, y) is 10 units find y?
2. Find a relation between x and y such that (x, y) is equidistance from (7, 1)
and (3, 5)?
3. Find the point on x – axis which is equidistance from (2, -5) (-2, 9)
4. If A (-5, 7),B(-4, -5) C (-1, -6) D (4, 5) are the vertices of a quadrilateral
then find its area
[Hint: required area = are ( ABC) ar (ACD)]
5. Show that (-4, -7), (1, 2), (8, 5), (5, -4) are vertices of a Rhombus.
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6. The point (2, 3), (x, y), (3, -2) are vertices of a triangle. If the centroid of this
triangle is origin find (x, y)
7. Find the coordinates of a point A, where AB is the diameter of a circle
whose centre is (2, -3) and B is (1, 4)
8. Find the value of „b‟ for which (1, 2), (-1, b), (-3, -4) are collinear.
9. Find the area of the triangle formed by the points (8, -5), (-2, -7) and (5, 1)
by using Herones formula.
[Hint: Let A (-8, -5), B=(-2, -7), C = (-5, 1)
Find a = BC, b = CA, c = AB and s =
Now area = – – –
10. Find the perimeter of the quadrilateral whose vertices are J (-2, -4) K (-2, 1),
L (6,7) and M(6, -4)
Bits
1. Coordinate geometry introduced by _________
2. The point of intersection of medians of a triangle whose vertices are
(6,2)(0,0)and (4, -7) is ________
3. Slope of the line joining (a, 0) and (0,b) is ________
4. The centre of a circle is (0, 0) diameter has one end point at (3,2) then other
end point is at ______
5. If (7, -2) (5, 1) and (3,k) are collinear then k = __________
6. Two vertices of a triangle are (-3, 1) and (0, -2) if the centroid of the
triangle is origin, third vertex = __________
7. A point on y – axis _________
A) (2,3) B) (2,0) C) Above two D) None of these
8. A point if at a distance of 5 units from (0, 1) its y coordinate is -3 then x
coordinates is ________
9. End points of a diameter of a circle are (3, 2) and (7, 2) radius of the circle is
______ units.
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8.SIMILAR TRIANGLES
You have studied in your earlier classes basic concepts of geometry which
are essential to study the further course in geometry. Hence, you must acquire the
requisite preliminary knowledge with regard to the concepts such as “the
congruent”, corresponding angles, alternate angles etc.,
(a)A two parallel lines are intersected by a transversal then the following angles are
equal.
(i) The corresponding angles are equal.
(ii) The alternate angles are equal.
i.e., l//m and „p‟ in a transversal
(Corresponding angles) and (alternate angles)
(b)(i) The height (or) attitude of an acute angled triangle lies inside of the triangle.
(ii) The height (or) attitude of an abtuse angled triangle lies inside of the triangle.
(iii) The height (or) attitude of a right angled triangle is one of the two sides.
(iv) In any case, the area of the triangle = ½ base x (height drawn to the base)
(c) In an isosceles triangles, the sides opposite to the equal angles are equal or
the angles opposite to the equal sides are equal.
(i) In ABC, if AB = AC
(d) A quadrilateral in which two opposite sides are parallel is called a trapezium.
i.e., AB // DC then ABCD is a Trapizium AC, BD are its diagonals.
(Alternate angles)
(e) In the triangles having the same base and are lying between the same parallel
lines, then their areas are equal triangles ABC, ABD, ABE are on the same
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base „AB‟ and are between the same parallel lines l,m. Then ar ( ) = ar
( = ar (
(f) If the variables a,b,c are proportion to x,y,z then = =
Similar of geometrical figures
Two Polygons with same number of sides are similar if
(i) all the corresponding angles are equal and
(ii) All the corresponding sides are in proportion.
Ex: Place a polygonal cardboard between a bulb and a table as shown in the
figure mark outline shadow quadrilateral A1B
1C
1D
1. It is the enlargement of
magnified quadrilateral ABCD. By measuring the angles and the sides, we
can verify
(i)
(ii) = = = = (Scale factor)
So, A1B
1C
1D
1
Note: 1. All squares are similar
2. All equilateral triangles are similar.
3. All circles are similar
Note: 1. If the scale factor k < 1 then the figure will diminished
2. If the scale factor k > 1 then the figure will enlarges
3. If the scale factor k = 1 then we get the congruent figure.
Similarity of triangles:-
If three angles in one triangle are equal to the corresponding three angles of
another triangle then the triangles are similar.
(or)
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If the corresponding sides of two triangles are proportion then the two
triangles are similar.
i.e. If ABC PQR
(or) If = = ABC PQR
Note: 1. In two (or) more triangles, if either the three angles are equal or the
sides are in proportion, then the triangles will have the same shape and hence the
triangles are similar.
Note: „Congruence‟ is the property pertaining to both shape and size.
1. AAA criterion for (similarity) of two triangles:
In two triangles, if the three angles of one triangle are equal to the
corresponding three angles of another triangle, then they are similar and
hence the corresponding sides are proportion.
I,e., = =
2. AA criterion for (similarity) of two triangles:
In two triangles, if two angles of one triangle are equal to the
corresponding two angles of one triangle are equal to the corresponding two
angles, another triangle, then the triangles are similar.
Explanation: In ABC, let ,
1800, 180
0
,
,
and thus ABC
Note: AAA criterion is not alone sufficient for the congruence of triangle.
Note: Here after we use AA criterion of similarity instead of AAA criterion
of similarity.
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80
3. SSS Criterion of similarity: If the corresponding sides of two triangles are
proportion, then they are similar and hence the corresponding, angles are
equal.
If = =
4. SAS Criterion for similarity of two triangles:
If one angle of a triangle is equal to one angle of the other triangle and
the sides including these angles are in proportion, then the two triangles are
similar.
= and ABC PQR
Example: A man sees the top of a lower in a mirror which is at a distance of
87.6cm from the tower. The mirror is on the ground facing upwards
the man is 0.4m away from the mirror and his height is 4.5m how tall
is the tower?
Solution: = = 900.
=
(Angle of incidence = angle of reflection)
By AA criterion of similarity, ABC EDC
= =
i.e., =
i.e., 40h = 15 876
h = = 328.5 m.
Example: A person 1.65m tall casts 1.8m shadow. At the same instance, a lamp
posts casts a shadow of 5.4m find the height of the lamppost.
Solution: AB = Height of the man = 1.65m
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BC = Shadow of the man = 1.8m
PQ = height of the lamp post = hm say
QR = Shadow of the lamp post = 5.4m
In ABC and
900.
(sun rays are parallel at any instance).
Note: Congruent geometrical figures are always similar.
Example: (i) Are the triangles similar? If so, name the criterion of similarity.
Solution: GF // KI
(Alternate angles)
=
(Vertex opposite angles)
By AAA criterion of similarity
(ii) Solution: = 800
= 400
= = 600
By AAA criterion of similarity
Ex:2. Explain why the triangles are similar and then find the value of „x‟?
Sol: Since AB // ZY
(Corresponding angles)
So, by AA criterion of similarity,
= =
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= 30x = 150 + 20x
(By AA countering of similarity)
= =
PQ = = 4.95m.
The height of the lamp post = 4.95m.
Ex:- The perimeter of two similar triangles are 30cm and 20cm respectively? If
one side of the first triangle is 12cm, determine the corresponding side of the
second triangle?
Sol: Let BC = 12cm
We should calculate EF (Corresponding side of BC)
Given
AB + BC + CA = 30 (1)
DE + EF + DF = 20 (2)
Since = = =
= = =
AB = 12. , CA = 12
Substituting in (1) and using (2),
12. + 12 + 12. = 30
12 (DE + EF + DF) = 30EF (Multiplying both sides by EF)
12 20 = 30EF
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EF = = 8cm.
Example: A girl of height 90c, is walking away from the base of a lamp post at a
speed of 1.2 m/sec if the lamp post is 3.6m above the ground, find the
length of her shadow after 4 seconds.
Solution: Initially the girl is a B.
AB = Weight of the lamppost = 3.6m = 360cm
DB = Distance travelled by the girl in 4sec at a speed of 1.2 m/sec
= Time Speed
= 4 1.2 100cm
= 480cm
ED = Shadow of the girl when she is at „D‟ = xcm (say)
CD = Height of the girl = 90cm.
= 900.
= (Common angle)
=(by AA similarity criterion)
= =
= 4x = x + 480
3x = 480
x = 160cm (or) 1.6m
Example: A flag pole 4m tall casts a 6m, shadow at the same time, a nearby
building casts a shadow of 24m. How tall is the building?
Solution: AB = Height of the flag pole = 4m
BC = Length of the shadow of the flag pole = 6m
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84
DE = Height of the building = hm say
EF = Length of the shadow of the building = 24m.
Since , =
= h = = 16m
The height of the building = 16m.
Example: In the given figure, = then show that (i)
(ii) If AD = 3.8cm, AE = 3.6cm
BE = 2.1cm, BC = 42cm
Find DE
Solution: (i) = (Given)
= (common angle)
By AA similarity criterion
(ii) = ( )
=
5. Diagonals AC and BD of a trapezium ABCD with AB // DC intersects each
other at the point „O‟. Using the criterion of similarity for two triangles,
show that =
Sol: // (Given)
(alternate angles)
=
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(by the AA criterion of similarity)
= (Corresponding sides are in proportion)
Example: AB, CD, PA are perpendicular to the AB = x, CD = y and PQ = z
prove that =
Sol:
i.e., =
= (1)
Similarity
i.e., =
= (2)
(1) + (2) + = = = 1
z ( + ) = 1 + =
Example: AX and DY are attitudes of two similar triangles
prove that AX : DY = AB : DE
Solution: (Given)
= (
= = 900.
( AX BC & DY EF)
(By AA criterion of similarity)
=
AX : DY = AB : DE
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86
Theorem: The ratio of the areas of the similar triangles is equal to the ratio of the
squares of their corresponding sides.
Given
RTP : .
= ( )2 = ( )
2 = ( )
2
Construction: Draw AM & PN such that AM BC and PN QR
Proof: = . = . (1)
In , ,
(
= 900. (Construction)
(By AA criterion of similarity)
= (2)
From (1), using the property = then (1) becomes,
= =
= ( using (2)
= )2 (3)
But = =
Using this in (3)
= )2
= )2
= )2
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Example: Prove that if the areas of two similar triangles are equal, then they are
congruent.
Solution: and =
= )2
= )2
= )2
But = =1
)2
= )2
= )2 = 1
AB2 = PQ
2, BC
2=QR
2, AC
2 = PR
2
AB = PQ, BC=QR, AC = PR.
(SSS axiom of congruency)
Example: The areas of two similar triangles are 81cm2, and 49cm
2, respectively.
If the attitude of the bigger triangle is 4.5cm find the corresponding
attitude of the similar triangle?
Solution: Let = 81cm2
= 49cm2
Let AX, DY are the height of and respectively, i.e.,
farmer is a bigger and the later is the smaller.
( 900.
= (and by AA similarity centurion)
= (1)
But = )2
(The ratio of the areas of two similar triangles in the ratio
of the squares of their corresponding sides)
= )2
= =
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88
From (1) = DY = = 3.5cm.
Example: BC=3cm, EF = 4cm, and area of = 54cm2.
Determine the area of .
Hint: Use = )2
Example: D,E,F are the midpoint of sides BC, CA, AB of Find the ratio
of areas of and .
Solution: DE // FA and DE = ½ AB = FA (1)
( In a triangle, a line segment connecting the midpoints of two sides
in parallel to the third side and is equal to half of the third side)
AEDF is a parallelogram.
( In a quadrilateral, a pair of opposite sides are parallel and equal
then the quadrilateral is a parallelogram)
A = (opposite angles of the //gm AEDF)
Similarly BDEF, DCEF are parallelograms.
B = and =
= )2
= = (using (1)
Example: Prove that the ratio of area of two similar triangles is equal to the
squares of the ratio of their corresponding medians.
Solution:
= and =
= (Since AX, DY are the medians of the sides BC, EF
respectively)
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89
= and =
(By SAS criterion of similarity)
= (1)
But = )2 = )
2 (using (1))
And the concludes the problem.
Example: Prove that the area of the equilateral triangle described on the side of a
square in half the area of the equilateral triangle described on its
diagonal.
Solution: In the square ABCD, BDE is an equilateral triangle described on the
diagonal BD and is another equilateral triangle described on the
side „BC‟.
= BD2.
But, in square length of the diagonal = length of a side
= ( BC)2 = 2( BC
2) = 2
=
Theorem: If a perpendicular is drawn from the vertex of the right angled triangle
to the hypotenuse, then the triangles on both sides of the perpendicular
are similar to the whole triangle and to each other.
Given: In ABC, = 900 and BD AC
RTP (i)
(ii)
(iii)
Proof: (i) = (Common to
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90
= 900 (Given)
(By AA criterion of similarity) (1)
(Ii) = (Common to )
= 900 (Given)
(2)
From (1) and (2) , ,
(I.e., Triangles similar to the same triangle are similar)
PYTHAGORAS THEOREM (BAUDHAYAN THEOREM)
Theorem: In a right triangle, the square of hypotenuse is equal to the sum of the
squares of the other two sides.
Given: AC2 = AB
2+BC
2
Construction: BD AC
Proof: (i) = (Common to )
= 900 (Given)
(By AA criterion of similarity)
= (sides of similar triangles are in proportion)
AD.AC = AB2 (1)
Similarly
=
DC.AC = BC2 (2)
(1) + (2) (AD + DC).AC = AB2 + BC
2.
AC.AC = AB2 + BC
2.
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AC2 = AB
2 + BC
2
Applications of Pythagoras theorem:
Example: ABC is a right angled triangle and right angle. Let BC = a, CA = b,
AB = c and let „p‟ be the length of perpendicular from „c‟ on „AB‟
prove that
(i) PC = ab
(ii) = +
Solution: = base height
=
= ab
= AB CD
=
= cp
Since =
ab = cp
cp = ab
(ii) By Pythagoras theorem:
c2 = a
2 + b
2 ( is a right angled triangle)
Substituting (1) in c2 = a
2 + b
2
= a
2 + b
2
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Inverse of Pythagoras theorem:
Theorem: In a triangle, if square of one side is equal to the sum of squares of the
other two sides, then the angle opposite to the first side is a right angle
and the triangle is a right angled triangle.
Given: In , AC2 = AB
2+BC
2.
RTP: = 900.
Construction: Construct such that PQ = AB, QR= BC and = 900
Proof: PR2 = PQ
2 + QR
2 ( is right angle triangle right angle at Q)
= AB2 + BC
2 ( Construction)
= AC2 (Given)
PR = AC
In and
AB = PQ (by construction)
BC = QR (by construction)
AC = PR (Proved)
(by SSS axiom of congruence)
but = 900.
= 900.
Example: In , = 900 and CD AB, prove that
=
Proof: We have proved that
and
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93
= =
Taking the first two fractions
= CD2 = AD.BD (1)
Taking the next two fractions
= squaring, it becomes =
= (using 0)
=
=
Example: In the given figure, if AD BC, prove that AB2+CD
2 = BD
2+AC
2
Solution: In = 900 (AD BC)
AD2 = AB
2 – BD
2 (1)
In = 900 (AD BC)
AD2 = AC
2 – CD
2 (2)
(1) and (2) due to Pythagoras theorem
AB2 – BD
2 = AC
2 – CD
2
AB2 + CD
2 = AC
2 – BD
2
Example: Prove that the sum of the squares of the sides of a rhombus is equal to
the sum of the squares of its diagonals.
Solution: Since ABCD is a rhombus
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AB = BC= CD = DA and
OB = OD = BD,
OA = OC = AC
is a right angled triangle
AB2 = OA
2 + OB
2
4AB2 = 4OA
2 + 4OB
2
AB2 + AB
2 + AB
2 + AB
2 = 4 ( AC)
2 +4( BD)
2
AB2 + BC
2 + CD
2 + DA
2 = AC
2 + BD
2.
Example: „O‟ is any point inside a rectangle ABCD prove that OB2 + OD
2 =
OA2 + OC
2.
Solution: Through „O‟ draw PQ // BC so that p lies on AB and Q lies on DC.
PQ // BC PQ AB and
PQ CD ( = 900)
is a right angled triangle
B2 = OP
2 + PB
2 (1)
Similarly OD2 = OQ
2 + QD
2 (2)
OA2 = OP
2 + PA
2 (3)
OC2 = OQ
2 + QC
2 (4)
1 + 2 OB2 + OD
2 = OP
2 + PB
2 + OQ
2 + QD
2
= OP
2 + QC
2 + OQ
2 + PA
2
= (OP2
+ PA2) + (OQ
2 + QC
2)
= OA2 + OC
2 (using 3 & 4)
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95
Example: is an isosceles triangle right angled at C. Prove that AB2 = 2AC
2
Solution: = 900 and therefore
AC = BC
By Pythagoras theorem:
AB2 = AC
2 + BC
2
= AC2 + AC
2
AB2 = 2AC
2
Example: In an equilateral triangle ABC, D is point on side BC such that BD =
BC. Prove that 9AD2 = 7AB
2.
Solution: Draw AE BC
AE bisects BC at E
BE = CE = BC (1)
In , = 900 (construction)
AE2 = AD
2 – DE
2 (By Pythagoras theorem)
= AD2 – (BE – BD)
2
= AD2 – ( BC - BC)
2 (using 1 and given)
= AD2 - BC
2
AE2 = AD
2 - AB
2 (2) ( is an equilateral triangle)
In = 900 (construction)
AE2 = AB
2 – BE
2 (By Pythagoras theorem)
= AB2 – ( BC)
2
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96
= AB2 – BC
2
= AB2 – AB
2 ( is an equilateral triangle)
AE2 = AB
2 (3)
= AD2 – AB
2= AB
2 (using 2 and 3)
= 36AD2 - AB
2 = 27AB2
36AD2 = 28AB2
9AD2
= 7AB2
Example: ABC is a right triangle right angled at „B‟ let „D‟ and „E‟ be any
points on „AB‟ and „BC‟ respectively. Prove that AE2 + CD
2 = AC
2 +
DE2
Solution: In , 900 (Given)
AE2 = AB
2 + BE
2 (by Pythagoras theorem)
In , 900
CD2 = BC
2 + BD
2 (By Pythagoras theorem)
1+ 2 AE2 + CD
2 = (AB
2 + BC
2) + (BE
2 + BD
2)
= AC2 + DE
2 ( are right angled
triangle)
Example: ABD is a triangle right angled at A and AC BD show that
(i) AB2 = BC. BD
(ii) AC2 = BC. DC
(iii) AD2 = BD. CD
Solution: = (common angle)
= 900 (AC BD & = 90
0)
(By AA similarity criterion)
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97
= (ratios of corresponding sides)
(i) AB2 = BC. BD
= (common angle)
= 900 (AC BD & = 90
0)
(By AA similarity criterion)
= (ratios of corresponding sides)
(iii) AD2 = BD. CD
By part (1) & (2)
= AC2 = BC. DC
Example: In the given figure, ABC is a triangle right angled at „B‟. D and E are
points on BC trisect it. Prove that 8AE2 = 3AC
2 + 5AD
2
Solution: BD = DE = EC = BC
(DE are trisecting BC)
L.H.S. = 8AE2
= 8(AB2 + BE
2) ( right angled triangle)
= 8 AB2 + 8 (2BD)
2 and by Pythagoras theorem)
= 8AB2
+ 32BD2
= 8AB2 + 32 ( BC)
2
= 8AB2 + BC
2
R.H.S = 3AC2 + 5AD
2
= 3(AB2 + BC
2) + 5(AB
2 + BD
2)
(By Pythagoras theorem on rt angled triangle ABD,
ABC)
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98
= 8AB2 + 3BC
2 + 5 ( BC)
2
= 8AB2 + 3BC
2 + BC
2
= 8AB2 + BC
2
L.H.S. = R.H.S
Example: Two poles of heights 6m and 11m stand on a plane ground. If the
distance between the feet of the poles is 12m find the distance
between their tops.
Solution: AB = Height of the smaller pole = 6m
CD = Height of the taller pole = 11m
BD = The distance between the poles = 12m
CE = CD – DE
= CE – AB ( DE = AB)
= 11 – 6 = 5m
AC2 = AE
2 + CE
2 ( AEC is a rt angled triangle)
= BD2 + CE
2 ( ABCD is a rectangle)
= 122 + 52 = 144 + 25
AC2 = 169 AC = = 13
The distance between their tops = 13m.
Example: The hypotenuse of a right triangle is 6m more than thrice of the
shortest side. If the third side is 2m, less than the hypotenuse, find the
sides of the triangle.
AB = Length of the smaller pole = xm
BC = Length of the third side
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99
AC = Length of the hypotenuse.
By the problem:
AC = (2AB + 6)m = (2x+6)m
BC = (AC – 2)m = (2x + 6-8)m
= (2x + 4)m.
Since is a right angled triangle By Pythagoras theorem
AB2 + BC
2 = AC
2
x2 + (2x+4)
2 = (2x + 6)
2
5x2 + 16x+16 = 4x
2+24x +36
x2 – 8x – 20 = 0
x2 – 10x + 2x – 20 = 0
x (x – 10) + 2(x-10) = 0
x = 10, x = -2
AB = 10m, BC = 2 10 + 4 = 24m, AC = 2 10 + 6 = 26m.
Example: A ladder 15m long reaches a window which is 9m above the ground
on one side of a street. Keeping its foot at the same point, the ladder is
turned to other side of the street to reach a window 12m high. Find the
width of the street.
Solution: CE = AE = Length of the ladder
= 15m
A, C are the windows
AB = 9m, CD = 12m.
ABE is right triangle right angle at B.
BE2 = AE
2 – AB
2 (By Pythagoras theorem)
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100
= 152 – 9
2
= 225 – 81
= 144
BE = = 12m.
CDE is right triangle right angle at D.
DE2 = CE
2 – CD
2 (By Pythagoras theorem)
= 152 – 12
2
= 225 – 144
= 81
DE = = 9m.
Width of the street = BE + DE = 12 + 9 = 21m.
Construction of a triangle similar to a given triangle with given scale factor:-
Construction: Construct a triangle similar to a given triangle ABC with its
sides equal to of corresponding sides of
Construction steps:-
(i) Draw a triangle ABC with the given measurements
(ii) Draw a ray .
(iii) Locate the points B1, B2, B3 and B4 on such that BB1 = B1B2
= B2B3 = B3B4 and join B4C.
(iv) Draw a line through B3, parallel to B4C and the point
intersection by this line on BC be „D‟.
(v) Again draw a line through D and parallel to CA. Let the point
of intersection by this line on BA be „E‟.
is the required [similar triangle]
BD : BC = 3 : 4 (Construction)
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DE // CA (Construction)
=
= (Corresponding angles)
ABC (By AA criterion of similarity)
(corresponding sides are in proportion)
But =
EB = AB, BD = BC, ED = AC.
Note: < 1, so when the scale factor < 1, then the similar triangle so formed
is smaller in size. When the scale factor > 1, then the similar triangle
thus formed be bigger in size.
Construction: II Construct a triangle shadow similar to the given , with
its sides equal to of the corresponding sides of the triangle ABC.
Construction steps:-
(i) Draw a triangle „ABC‟ with the given measurements
(ii) Draw a ray .
(iii) Locate the points B1, B2, and B3 on such that BB1 = B1B2 =
B2B3 and join B3C.
(iv) Draw a line through B2, parallel to B3C and the point
intersection by this line on BC be „D‟.
(v) Locate the points C1, C2, on produced BC such that CC1 = C1C2
= CD.
(vi) Draw a line through C2 and parallel to „CA‟ let the point of
intersection by this line on the produced BA be „A‟.
A1BC2 is the required similar triangle with the given scale factor.
Since > 1, the similar triangle so has been constructed is bigger than
that of the given triangle.
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102
ABC A1BC2
= = =
Construction: Construct an Isosceles triangle whose base is 7cm and attitude
is 3.5 cm. Then, draw another triangle whose sides are times the
corresponding sides of the isosceles triangle.
Solution: For the problem scale factor = =
Construction steps:
(i) Construct Isosceles with the measurements BC = 7cm,
AD = 3.5cm.
(ii) Bisect BC at D, BD = CD
(iii) Locate E on produced BC such that CD = CE
(iv) Draw a line through E, and parallel to CA let the point of
intersection by this line on produced BA be „F‟.
(v) BEF is the required shadow similar Isosceles triangle to the
given
Proof: BE : BC = 3 : 2
Since CA // EF, there fore
= = = AC.
FB = AB, BE = BC, FE = AC.
Exercise:
1. Construct a triangle of sides 4cm and 6cm. Then, construct a triangle
similar to it, whose sides are of the corresponding sides of the first
triangle
2. Construct an Isosceles triangle whose base is 8cm and attitude is 4cm.
Then draw another triangle whose sides are 1 times the
corresponding sides of the isosceles triangle.
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103
3. Construct a triangle with AB = 5cm = 600. AC = 4.5cm then draw
another triangle with scale factor .
Example: Construct a triangle with AB = 5.3cm, = 400, = 50
0, then draw
another triangle with its sides are of the corresponding sides of the
given triangle.
Steps:
1. Construct with AB = 5.3cm, = 400, = 50
0
2. Draw a ray and mark on it A1, A2, ………., A7 such that AA1 =
A1A2 = ……. = A6A7 and join A7,B.
3. Draw a line through A3 and parallel to 7 let the point of intersection
on AB be „B1‟
4. Draw again a line through B1 and parallel to let the intersection
point on AC be „C1‟.
AB1C1 and therefore AB1C1 is the required similar
triangle.
Basic Proportionality theorem: (Thales Theorem)
Theorem: If a line is drawn parallel to one side of a triangle to intersect the other
two sides in distinct points, then the other two sides are divided in the
same ratio.
Given: In ,
RTP: =
Construction: DM AC and EN
Proof: BDE, are the triangles with the same base DE and are lying
between two parallel lines
=
Dividing by on both sides
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104
=
=
=
By inverted, =
Note: Basic proportionality theorem can also be slated as follows:
(i) =
=
=
(or) =
Note: Basic proportionality theorem can also be proved using the concept of
similarity. In the figure DE // BC and
ADE (By AA criterion of similarity)
= (1)
Also 1 - =
= (2)
Dividing (1) by (2) =
=
Which is Thales Theorem
Converse of Basic Proportionality Theorem:
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105
In a line divides two sides of a triangle in the same ratio, then the line
is parallel to the third side in ABC,
DE divides AB and AC such that =
Then DE // BC
Ex:- E and F are points on the sides PQ and PR respectively of state
EF // QR or not?
(i) PE = 3.9cm, EQ = 3cm, PF = 3.6cm, and FR = 2.4cm
(ii) PE = 4cm, QE = 4.5cm, PF = 8cm, and RF = 9cm
(iii) PQ = 1.28cm, PR = 2.56cm, PE = 1.8cm and PF = 3.6cm.
Solution:
EF // QR EF // QR EF // QR
( using the converse of Thales Theorem)
Example: In the given figure LM // AB, AL = (x – 3) AC = 2x, BM = (x-2) and
BC = (2x+3) find the value of „x‟
Solution: LM // AB (Given)
= (by Thales Theorem)
=
2x2 2x
2 – 4x
-3x – 9 = -4x
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106
x = 9
Example: What values of „x‟ will make DE // AB in the given figure?
AD = (8x + 9), CD = (x + 3), BE = 3x+4, CE = x
Solution: DE // AB (Assume)
= (by Thales Theorem)
=
8x2
8x2 3x
2+4x+9x+12
5x2 - 4x – 12 = 0
5x2-10x+6x-12 = 0
-3x – 9 = -4x
5x(x-2)+6(x-2) = 0
(x-2) (5x + 6) = 0
x = 2, or x =
Exercise: In DE // BC, AD = x, DB = (x – 2), AE = (x + 2) and EC = (x –
1) find „x‟?
Solution: Use Thales Theorem Ans: x = 4
Ex: In the given figure, LM // CB and LN // CD prove that =
Solution: In ABC, LM // CB, (Given)
= (1) (Thales Theorem)
In LN // CD
= (2) (Thales Theorem)
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From 1 & 2 =
Ex:- In the given figure AB // CD // EF given AB = 7.5cm, DC = y cm, EF
= 4.5cm, BC = x cm, calculate the values of „x‟ and „y‟?
Solution: AB // FE (Given)
BAC = CEF (Alternate angles)
ABC = CEF
ACB = ECF (vertex opposite angle)
ABC
= =
= 3 = 5
In BEF, CD // FE (Given)
= ( BDC )
=
=
y = = 2.8125
Ex: Prove that a line joining the midpoints of any two sides of a triangle is
parallel to the third side.
Solution: Let D,E be two mid points of the sides AB and AC
AD = BD
And AE = CE
So,
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108
DE // BC (by the converse of Thales Theorem)
Ex:- ABCD is a trapezium in which AB // DC and its diagonals intersects
each other at point „O‟ show that =
Proof: Draw EO parallel to AB
In ABD, EO //AB (Construction)
By BPT: = (or) = (1)
In ADC, EO //DC
By BPT: = (2)
From 1 & 2 = (or) =
Ex:- Prove that a line drawn through the midpoint of one side of a triangle
parallel to another side bisects the third side.
Solution: D is the midpoint of AB (Given)
DE // BC (Also given)
By Thales Theorem
=
= 1 ( AD = BD)
AE = CE E is the midpoint of AC.
Construction: Draw a line segment of length 7.2cm and divide it in the ratio
5:3 measure the two parts.
Construction Steps:
(i) Draw a line segment AB = 7.2cm
(ii) Draw a ray .
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109
(iii) Locate A1, A2,……..A8 on such that AA1 = A1A2 =……
A7A8
(iv) Join B, A3.
(v) Draw a line through A5, and parallel to A3B Let the intersection
on AB be B1.
AB1 :BB1 = 5 : 3
Proof: In ABA8 A5B1 // A3 B (by construction)
= = (Thales Theorem)
After measurement AB1 = 4.5cm, BB1 = 2.7cm.
Circle Tangent Secant:
A circle and a line in a plane can be depicted in three ways.
1. The circle and the line have no common point.
2. The circle and the line have only one common point.
3. The circle and the line have exactly two common point.
In the second case the line is called the largest and the common point
is called the point of contact
In the third case the line is called the secant of the circle and the
common points are called points of intersection.
Theorem: The tangent at any point of a circle is perpendicular to the radius
through the point of contact.
Given: XY is the tangent to the circle with centre „o‟ at „p‟
RTP: XY OP
Proof: Assume XY is not perpendicular to „OP‟ if we take point on XY i.e. Q
on XY then it must be outside of the circle, otherwise Q lies inside of
the circle, and consequently, XY becomes secant
OP < OQ
For every point on XY other than P,
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110
OP < OQ
Note: Using the theorem we can construct a tangent to the given circle.
Construct of a Tangent:
Construction steps:
1. Draw a circle with radius „OP‟ let „O‟ be the its centre.
2. Locate the points A,B on such that AP = BP
„P‟ is the midpoint of
3. Draw a perpendicular bisector to and ensure that it passes
through „P‟.
OP XY XY is a tangent.
Length of the Tangent:
OP = Distance of the centre point „P‟ = d
OA = Radius = r
AP = Length of the tangent from an external point „P‟
AP2 = OP2 – OA2 (By Pythagoras Theorem)
AP =
Ex:- Find the length of the tangent to a circle with centre „O‟ and radius 6cm
from a point „P‟ such that OP = 10cm
OP = 10cm
Here OP = d = 10cm
OA = r = 6cm
Length of the tangent AP =
=
=
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111
= = 8cm.
Construction of Tangents to a circle from an external point:
Construction Steps:
1. Draw a circle with given radius and let its centre be „O‟ let „P‟
be an external point so that OP > r
2. The perpendicular bisector to „OP‟ intersects OP at „Q‟
3. Draw a circle with radius ½ OP = PQ = OQ and centre at „Q‟
let the points of intersection of two circles be A, B
4. Join P, A and P, B PA, PB are two tangents.
Proof: OAP = 900 (Angle in the semi circle)
OA PA (At the end of a radius of the circle becomes the tangent)
Similarly OBP = 900
OB PB
PA, PB are two tangents from „P‟ to the circle.
Since PA2 = OP2 – OA2
= OP2 – OB2 (
= PB2
PA = PB
i.e., The two tangents drawn from an external point to the given circle are
equal.
Ex:- If a circle touches all the four sides of a quadrilateral ABCD at P,Q,R,S
Show that AB + CD = BC + DA
Proof: AP = AS
BP = BQ
CR = CA
Page 112
112
DR = DS Tangents to the circles drawn from the external points
A,B,C,D.
AP + BP + CR + DR = AS + BQ + CQ + DS
(AP + BP) + (CR+DR) = (AS+DS) + (BQ + CQ)
AB + CD = AD + BC
Example: Draw a pair of tangents to a circle of radius 5cm which are inclined to
each other at an angle of 600.
Solution:
To draw the tangents, we must know the location the outside point „p‟
i.e., we must know the distance between „O‟ and „P‟.
OA = OB (Radii)
OP = OP (Common side)
PA = PB (Tangents)
OAP (SSS axiom of congruence)
APO = BPO = APB
= 600.
= 300.
In OAP OA PA
Sin 300 = =
= (Sin 300 = )
OP = 10cm.
Construction Steps:
Page 113
113
1. Draw a circle with radius OA = OB = 5cm and let „O‟ be the centre
of the circle. Locate the point „P‟ which is 10cm away from „O‟.
2. The perpendicular bisector to „OP‟ intersects OP at „M‟.
3. Draw a circle with radius = ½ OP = 5cm and centre at „M‟ let the
points of intersection of two circles be A,B
4. Join P,A and P, B. PA, PB are the required Tangents.
CHAPTER 11
TRIGONOMETRY 4 Marks Questions
1. If A and x are acute angles such that cos A = cos x then show that A =
x?
Sol: In ABC and XYZ
B = = 900
cos A = cos x (Given)
=
=
Let = be K
= = k AB = K.XY and AC = K.XZ
By using Pythagoras theorem
AB2 + BC
2 = AC
2
BC2 = AC
2 – AB
2
= (K.XZ)2 – (K.XY)
2
= K2 (XZ)
2 – K
2(XY)
2
= K2 (XZ
2-XY
2)
Page 114
114
= K2YZ
2
BC = = K.Y.Z
= K
Hence = = then ABC XYZ
A = x
2. Page 276 Example 2
3. In a right angled triangle ABC right angle is at B, if tan A = then find the
value of (i) sin A cos C + cos A sin C (ii) cos A cos C – sin A sin C
Sol: In ABC = 900
Tan A =
=
=
BC : AB = : 1
Let BC be and AB be 1K
In ABC AC2 = AB
2 + BC
2 (By Pythagoras theorem)
= K2 +
2
= K2 + 3
2
= 4K2
AC = = 2K
sin A = = =
cos A = = =
Page 115
115
sin C = = =
cos C = = =
(i) sin A cos C + cos A sin C = . + . = + = = = 1
(ii) cos A cos C – sin A sin C = . - - = - = 0
4. If sin (A – B) = , cos (A+B) = , 00
A+B = 900, A > B find A and B?
Sol: sin (A – B) =
But sin 300 = (from the table)
A-B = 300 …… (1)
cos (A+B) =
But cos 600 = (from the table)
A + B = 600 …… (2)
Let‟s solve (1) & (2)
A - B = 300
A + B = 60
0
2A = 90
0
A = = 450.
Substituting A = 450
in (2) we get
450
+ B = 600
B= 600 – 45
0 = 15
0.
A = 450, B = 15
0.
Page 116
116
5. Prove that = cosec + cot
Sol: L.H.S. =
Multiplying Nr and Dr with 1+ cos we get
=
= [(a-b)(a+b) = a2 – b
2]
=
= [sin2
+ cos2
= sin2
= 1- cos2
= +
= csc + cot
= R.H.S
L.H.S = R.H.S
6. Page 292, Q.no:3 Exercise 11.4
7. If csc + cot = K then prove that cos =
Sol: csc + cot = K
R.H.S. = =
= [(a+b)2 = a
2+2ab+b
2]
Page 117
117
=
=
=
=
= = = = cos = L.H.S
2 Marks Questions
1. The sides of a right triangle PQR are PQ = 7cm, PR = 25cm and Q = 900
respectively then find tan P – tan R?
Sol: In PQR Q = 900, PQ = 7cm, PR = 25cm
PQ2 + QR
2 = PR
2 (Pythagoras theorem)
72 + QR
2 = 25
2
49 + QR2 = 625
QR2 = 625 – 49 = 576
QR = = 24
tan P = = =
tan R = = =
tan P – tan R = = =
2. If 3 tan A = 4 then find sin A and cos A?
Sol: 3 tan A = 4
tan A =
Page 118
118
tan A = =
Opp side : adj side = 4 : 3
Let opp side to be 4K and adj side to be 3K
In ABC = 900.
AC2 = AB
2 + BC
2 (Pythagoras theorem)
= (3K)2 + (4K)
2
= 9K2+16K
2 = 25K
2.
AC = = 5K
sin A = = =
cos A = = =
3. Page 277 Q.No:4, Exercise 11.1
4. Page 282 example 4
5. A chord of a circle of radius 6 cm is making an angle 600 at the centre. Find
the length of the chord?
Sol: Radius OA = OB = 6cm
= 600.
OC is height from „O‟ upon AB and it is an angle bisector
= 300
In COB, sin 300 =
= =
= 2BC = 6 Length of the chord AB = 2BC = 6cm.
6. Page 284, example 6
Page 119
119
7. A evaluate sin 600 cos 30
0 + sin 30
0 cos 60
0, what is the value of sin (60
0 +
300) what can you conclude?
Sol: From the table sin 600 = , cos 60
0 = , sin 30
0 = , cos 30
0 = , sin 90
0 = 1
Now sin 600 cos 30
0 + sin 30
0 cos 60
0 = + = + = = = 1
sin (600 +30
0) = sin 90
0 = 1
sin (600 +30
0) = sin 60
0 cos 30
0 + sin 30
0 cos 60
0
Taking 600 = A, 30
0 = B we get
sin (A + B) = sin A cos B + sin B cos A
8. Show that cos 360 cos 54
0 – sin 36
0 sin 54
0 = 0
Sol: cos 360
= cos (90
0 - 54
0) = sin 54
0 ( cos (90
0 – ) = sin )
sin 360
= sin (90
0 - 54
0) = cos 54
0 ( sin (90
0 – ) = cos )
Now cos 360 cos 54
0 – sin 36
0 sin 54
0 = sin 54
0 cos 54
0 - cos 54
0 sin 54
0 = 0
9. If A, B and C are interior angles of a triangle ABC, then show that tan ( )
= cot
Sol: The sum of the interior angles is ABC = A + B + C = 1800.
Dividing on both sides by 2, we get
=
+ = 900.
= 90
0 -
tan ( ) = tan (900 )
tan ( ) = cot ( tan (900 – ) = cot )
Page 120
120
10. Page 288, example 12
11. Show that tan2
+ tan4
= sec4
-sec2
?
Sol: L.H.S = tan2
+ tan4
= tan2
(1+ tan2
) (sec2
– tan2
= 1
= tan2
. sec2
sec2
= 1+ tan2
= (sec2
– 1) + sec2
sec2
– 1 = tan2
= sec4
- sec2
= R.H.S
= L.H.S = R.H.S
12. Simplify sec A (1-sin A) (sec A + tan A)
Sol: sec A (1-sin A) (sec A + tan A) = (sec A – sec A sin A) (sec A + tan A)
= (sec A - sin A) (sec A+ tan A)
= (sec A – ) (sec A + tan A)
= (sec A – tan A) (sec A + tan A)
= sec2A – tan
2A
= 1
sec A (1-sin A) (sec A + tan A) = 1
13. Prove that (sin A + cosec A)2 + (cos A + sec A)
2 = 7+tan
2A + cot
2A?
Sol: L.H.S. (sin A + cosec A)2 + (cos A + sec A)
2
= sin2A+cosec
2A+2sinA cosecA+cos
2A+sec
2A+2cosA secA
= (sin2A + cos
2A)+cosec
2A+sec
2A+2sinA. +2cosA.
= 1+ cosec2A+sec
2A+2+2 [ cosec
2A-cot
2A = 1
= 5+ cosec2A+sec
2A cosec
2A = 1+cot
2A
= 5+1+cot2A+1+tan
2A sec
2A-tan
2A=1
Page 121
121
= 7+tan2A + cot
2A sec
2A=1+tan
2A]
14. Page 292, Q.No:8, exercise 11.4
1 Mark Questions
1. Evaluate sin 450+cos45
0?
Sol: sin 450
= , cos450 =
sin 450+cos45
0 = +
= =
= =
2. What can you say about cot00 = it is defined? Why?
Sol: cot00 = = it is not defined because division with „0‟ is not defined.
3. Evaluate 2tan245
0 + cos
230
0 – sin
260
0.
Sol: We know that tan450 = 1, cos30
0 = , sin60
0 =
2tan245
0 + cos
230
0 – sin
260
0 = 2(1)
2 + ( )
2 - ( )
2
= 2 + -
= 2
4. Is it right to say that sin (A + B) = sin A + sin B? Justify your answer?
Sol: Let‟s take A = 600, B = 30
0
Then sin (A + B) = sin (600 + 30
0) = sin 90
0 = 1
sin A + sin B = sin 600 + sin 30
0 + =
sin (A+B) sin A + sin B
Page 122
122
It is not right to say that sin (A+B) = sin A + sin B
5. Evaluate cos 120 – sin 78
0?
Sol: cos 120 – sin 78
0 = cos (90
0-78
0) – sin 78
0
= sin 780
– sin 780 [ cos(90
0- )=sin ]
= 0
6. If tan2A = cot (A-180) where 2A is an acute angle, find the value of A?
Sol: tan 2A = cot (A-180)
cot (900-2A) = cot (A-18
0) [ cot (90
0-A)=tanA]
900 – 2A = A-18
0
900+18
0 = A+2A
108=3A
A=1080
A = = 360
A = 360.
7. If tan A = cot B, where A & B are acute angles, prove that A + B = 900?
Sol: tan A = cot B
= tan (900-B) [ tan (90
0- )= cot ]
A = 900 – B
A+B = 900
8. Express tan 750 + cos65
0 in terms of trigonometric ratios of angles between
00 and 45
0?
Sol: sin 750 = sin (90
0-15
0) = cos 15
0 [ sin (90
0- )= cos ]
Cos 650 = (90
0 – 25
0) = sin 25
0 [ cos (90
0- )= sin ]
Page 123
123
sin 750+cos65
0 = cos15
0+sin25
0
9. If sin C = then find cos C?
Sol: sin C =
We know that sin2C+cos
2C=1
( )2 + cos
2C = 1
+ cos2C = 1
cos2C = 1- = =
cos C = =
cos C =
10. Evaluate (sin + cos )2 + (sin - cos )
2
Sol: (sin + cos )2+(sin - cos )
2 = sin
2 + cos
2+2 sin .cos + sin
2 + cos
2-2
sin .cos
= 1+1 = 2
(sin + cos )2 + (sin - cos )
2 = 2
11. Evaluate (sec2
-1) (cosec2
-1)?
Sol: (sec2
-1) (cosec2
-1) = (tan2
)(cot2
) [ sec2
-tan2
= 1 sec2
-1=tan2
= tan2
. cosec2
-cot2
=1 cosec2
-
1=cot2
]
= 1
(sec2
-1) (cosec2
-1) = 1
12. Example 13, page 290
Page 124
124
13. Page 292, Q.No:5, Exercise 11.4
14. Page 292, Q.No:9, Exercise 11.4
Fill in the blanks
1. In the adjacent triangle cos A = ________
2. cos A = , sin A = __________
3. = ________
4. = __________
5. = ________
6. = ________
7. = ________
8. = ________
9. cosec 310 – sec 59
0 = ___________
10. sin 150 sec 75
0 = ________
11. tan 260. tan 64
0 = ________
12. if tan x = , sec x = ________
13. cosec = then cot = _________
14. sin 50 cos 85
0 + cos5
0 sin 85
0 = ________
15. If sec + tan = P then sec - tan = ________
16. cosec + cot = P then cosec – cot = __________
17. If sin = then tan = ________
Page 125
125
18. sin2500 + cos
2500 = _________
19. sec21000 – tan
21000 = ________
20. cosec2750 – cot
2750 = _______
21. If sin = cos then = ________
22. sin = then = _________
23. cos (A+B) = then A + B = ________
24. sin100 sec80
0 = _______
25. = _____________
Answers
1. 2. 3. Sec 4. 1 5.
6.0 7. 8.1 9.0 10.1
11.1 12. 3. 14. 1 15.
16. 17. 18.1 19.1 20.1
21.450 22.60
0 23.30
0 24.1 25.1
Page 126
126
CHAPTER - 10
MENSURATION 4 Marks Questions
1. The radius of a conical tent is 7 meters and its height is 10 meters calculate
the length of canvas used in making the tent if width of canvas is 2m?
Sol: Radius of the conical tent (r) = 7 meters
Height (h) = 10 meters
Slant height (l) = ?
l2 = r
2 + h
2
= 72 + 10
2
= 49+100
= 149
l =
= 12.2 mts
Surface area of the tent =
= 7 12.2 m2
= 268.4 m2
Area of canvas used = 268.4 m2
Width of canvas = 2m (given)
Length of the canvas used = = = 134.2 meters.
2. An oil drum is in the shape of cylinder having the following dimensions
diameter is 2m and height is 7 meters. The painter charges 3 per m2 to paint
the drum. Find the total charges to be paid to the painter for 10 drums?
Page 127
127
Sol: Diameter of the oil drum = 2m
Radius = = 1m
Height = 7m
Total surface area of the drum = 2 (r + h)
= 2 1 (1+7)
= 2 .
=
= 50.28m2.
Painting charges per 1m2 = Rs 3
Total charges to be paid to the painter for 10 drums = 50.28 3 10
= Rs. 1508.40
3. A sphere, a cylinder and a cone are of the same radius and same height. Find
the ratio of their curved surface areas?
Sol: Let the radius of a sphere, a cone and a cylinder be „x‟
Height of the sphere = Its diameter
= 2x
As per the problem
Height of the cone = 2x and
Height of the cylinder = 2x
Slant height of the cone l =
=
=
Page 128
128
=
= x
Curved surface area of the sphere = 42 = 4
2
Curved surface area of the cylinder = 4 = 2
=
4
2
Curved surface area of the cone =
= x = 2
Ratio of curved surface areas = 42
: 42 :
2
= 4 : 4 :
4. Page 250, Q.No:5, exercise 10.1
5. A heap of rice is in the form of a cone of diameter 12m and height 8m. Find
its volume. How much canvas cloth is required to cover the heap (use =
3.14)?
Sol: Radius of the cone (heap of rice) (r) = 6m
Height (h) = 8m
Slant height (l) =
=
=
=
= 10m
Volume of the heap of rice =
= 3.14 6 6 8
= 301.44 m2.
Page 129
129
Area of canvas cloth required to cover the heap =
= 3.14 6 10
= 188.4 m2.
6. A toy is in the form of a cone mounted on a hemisphere. The diameter of the
base and the height of the cone are 6cm and 4cm respectively. Determine the
surface area of the toy. (use = 3.14)?
Sol: Diameter of the base of the cone (d) = 6cm
Radius (r) = = = 3cm
Height (h) = 4cm
Slant height l =
=
=
=
= 5 m
Surface area of the toy =
Curved surface area of the cone + curved surface area of the
hemisphere
= + 22
= (1 + 2r)
= 3.14 3(5+2 )
= 9.42 (5 + 6)
= 9.42 (11)
= 103.62
Page 130
130
Surface area of the toy = 103.62
7. Page 255, Q.No:2, Exercise 10.2
8. A medicine capsule is in the shape of a cylinder with two hemispheres stuck
to each of its ends. The length of the capsule is 14mm, and the width is
5mm. Find its surface area?
Sol: The width of the capsule =
Diameter of the hemi-sphere = 5mm
Radius (r) = = 2.5mm
Length of the capsule = 14mm
Length of the cylindrical part = 14 – 2
= 14 – 5 = 9mm
Curved surface area of the cylindrical part = 2
= 2
= 141.3 mm2
Curved surface area of the 2 hemi sphere = 2 22
= 2
= 78.5 mm2
Total surface area of the capsule =
Curved surface area of the cylindrical part + curved surface area of the
hemisphere
= 141.3 + 78.5 = 219.8 mm2
9. Page 255, Q.No:5, Exercise 10.2
10. Page 255, Q.No:8, Exercise 10.2
Page 131
131
11. An iron pillar consists of a cylindrical portion of 2.8m height and 20cm in
diameter and a cone of 42cm height surrounding it. Find the weight of the
pillar if 1 cm3 of iron weight 7.5g?
Sol: Radius of the cylindrical portion (r) = 10cm
Height (h) = 2.8m = 2.8 100cm = 280cm
Volume of the cylinder = 2h
= 10
= 88000 cm3
Radius of the conical part (r) = 10cm
Height (h) = 42cm
Volume of the cone =
= 10 42
= 4400 cm3.
Total volume of the iron pillar =
Volume of the cylindrical portion + Volume of the conical portion
= 88000 + 4400
= 92400 cm3.
The weight of 1cm3 of iron = 7.5 g
Weight of the pillar = 92400 7.5
=
=
Weight of the pillar = 6930 kg.
Page 132
132
12. Page 260, Q.No:2 exercise 10.3
13. A cylindrical tub of radius 5cm and length 9.8cm is full of water. A solid in
the form of right circular cone mounted on a hemisphere is immersed into
the tub. The radius of the hemisphere is 3.5cm and height of cone outside the
hemisphere is 5cm. Find the volume of water left in the tub. (Take = )
Sol: Radius of the cylinder (r) = 5cm
Height (length) of the cylinder = 2h
=
= 770 cm3.
Radius of the hemisphere (r) = 3.5cm
Volume of the hemisphere = 3
=
Radius of the base of the cone ® = 3.5cm
Height of the cone (h) = 5cm
Volume of the cone = 2h
= cm3.
Total volume of the solid =
Volume of the hemisphere + Volume of the cone
=( + (
= ( )
= ( )
Page 133
133
= ( )
=
= 154 cm2.
The volume of water left in the tub =
Volume of cylindrical tub – volume of the solid immersed
= 770 – 154
= 616 cm2.
14. Page 261, Q.No:6, Exercise 10.3
15. Page 261, Q.No:7, Exercise 10.3
16. A metallic sphere of radius 4.2 cm is melted and recast into the shape of a
cylinder of radius 6cm. Find the height of the cylinder?
Sol: Radius of the sphere (r) = 4.2cm
Volume of the sphere =
= cm3
Radius of the cylinder (r) = 6cm
Height of the cylinder (h) = ?
Volume of the cylinder =
= cm3
According to problem
Volume of the cylinder = Volume of the sphere
=
Page 134
134
h =
= 2.74 cm
Height of the cylinder = 2.74cm
17. Page 267, Q.No:3, Exercise 10.4
18. Page 267, Q.No:6, Exercise 10.4
19. Page 265, Example – 17
20. Page 266, Example – 19
2 Marks Questions
1. A company wanted to manufacture 1000 hemispherical basins from a thin
steel sheet. If the radius of hemispherical basin is 21cm, find the required
area of steel sheet to manufacture the above hemispherical basins?
Sol: Radius of the hemispherical basin (r) = 21 cm
Surface area = 2
= 2
= 2772 cm2.
The steel sheet required for one basin = 2772 cm2.
The steel sheet required for 1000 basin = 2772 cm2.
= / cm2.
= 277.2 / cm2.
2. Page 249, Example 6.
3. Page 250, Example 7
4. Page 250, Q.No:1, Exercise 10.1
5. Page 250, Q.No:2, Exercise 10.1
Page 135
135
6. A cylinder and cone have bases of equal radii and are of equal height show
that their volumes are in the ratio of 3:1?
Sol: Radii of cylinder and cone are equal
Height of cylinder and cone are equal
The ratio of their volumes = :
= 1 :
= 3 : 1
7. Page 250, Q.No:7, Exercise 10.1
8. Page 251, Q.No:9, Exercise 10.1
9. Two cubes each of volume 64 cm3 are joined end to end together. Find the
surface area of the resulting cuboids?
Sol: Volume of the cube = 64cm3
a3 = 64
a3 = 4
3
a = 4 cm.
The cubes are joined together
Length of the resulting cuboid = 4 + 4 = 8cm
Height of the cuboid = 4 cm
Breadth of the cuboid = 4cm
The surface area of the resulting cuboid = 2 (lb + bh + lh)
= 2 (8 4 + 4 4 + 8 4)
= 2(32+16+32)
= 2 80
Page 136
136
= 160 cm2.
10. Page 260, Q.no:3, Exercise 10.3
11. Page 262, Example 14.
1 Mark Questions
1. Find the volume of right circular cone with radius 6cm and height 7cm?
Sol: Radius (r) = 6cm
Height (h) = 7cm
Volume =
= 6
= 264 cm3.
2. Page 250, Q.No:4, Exercise 10.1
Fill in the blanks
1. The volume of the solid formed by joining two basic solids is the ______ of
the constituents.
2. Curved surface area of a cylinder = _________
3. Curved surface area of a cone = _________
4. Curved surface area of a sphere = _________
5. Curved surface area of a hemisphere = _________
6. Total surface area of a cylinder = _________
7. Total surface area of a cone = _________
8. Total surface area of a hemisphere = _________
9. Volume of a cylinder = ________
10. Volume of a cone = ________
Page 137
137
11. Volume of a sphere = ________
12. Volume of a hemisphere = ________
13. The volume of a cone with radius of the base 7cm and height 24cm is
_______
14. The total surface area of a hemisphere is 115.5cm2, then its radius is
_______
15. If the slant height and height of a cone are 25 cm and 20cm, then its radius is
_____
Answers:
1. Sum of the volumes 2. 2 3. 4. 42
5. 22
6. 2 7. 8. 32
9. 2h 10.
11. 12. 13. 1232 cm3 14. 3.5 cm 15. 15 cm
CHAPTER 12
APPLICATIONS OF TRIGONOMETRY
1. Two men on either side of a temple of 30 meter height observe its top at the
angles of elevation 300 and 60
0 respectively. Find the distance between the
two men?
Sol: From the adjacent fig
Height of the temple BD = 30 mts
Angle of elevation of one person BAD = 300
Angle of elevation of another person BCD =
600
Let the distance between the first person and
the temple AD = x
Distance between the second person and the temple CD = y
From ABD 300 =
=
Page 138
138
x = 30 (1)
From BCD 600 =
=
CD =
y = DC = (2)
From equation (1) and (2)
The distance between the two persons AC = AD + DC = x + y
= +
=
= .
2. A straight highway leads to the foot of a tower; Ramaiah standing at the top
of the tower observes a car at an angle of depression 300. The car is
approaching the foot of the tower with a uniform speed. Six seconds later,
the angle of depression of the car is found to be 600. Find the time taken by
the car to reach the foot of the tower from this power?
Sol: From the adjacent fig
Let the distance travelled by the car in 6 seconds AB = x meters
Height of the tower CD = h meters
The remaining distance to be travelled by the car BC = d meters
AC = AB + BC = (x + d) meters
PDA = DAB = 300
PDC = DBC = 600
From BCD
600 =
=
h = d (1)
From ACD
300 =
=
Page 139
139
h =x + d
h = (2)
From equation 1 and 2
= d
x + d = 3d
x = 2d
d =
Time taken to travel x meters = 6 seconds
Time taken to travel the distance of „d‟ meters = = seconds
= 3 seconds
3. A TV tower stands vertically on the side of a road from a point on the other
side directly opposite to the tower, the angle of elevation of the top of tower
is 600. From another point 10m away from this point, on the line joining this
point to the foot of the tower, the angle of elevation of the top of the tower is
300. Find the height of the tower and the width of the road?
Sol: In the adjacent figure
AB denotes the height of the tower
BC denotes the width of road
CD = 10mts
ACB =600 and ADC = 30
0
In ABC, B = 900
600 =
=
h = d (1)
In ABD, B = 900
300 =
=
h = d +10
h = (2)
From equation (1) and (2)
Page 140
140
d =
3d = d + 10
3d – d = 10
2d = 10
d = 5 meters
Width of the road = 5mts
Hence the height of the tower h = d
= 5
= 5 mts
4. A status stands on the top of a 2 meters tall pedestal. From a point on the
ground, the angle of elevation of the top of the statue is 600 and from the
same point, the angle of the elevation of the top of the pedestal is 450 find
the height of the status?
Sol: From the adjacent figure
Height of the pedestal BD = 2mts
Let height of the statue CD = h mts
BC = BD + DC
= (2 + h) mts.
Let distance between AB = x mts
BAD =450 and BAC = 60
0
In ABD, B = 900
450 =
1 =
x = 2 mts
AB = x = 2 mts (1)
In ABD, B = 900
BAC = 600
600 =
=
2 + h = 2
h = 2 – 2
= 2 1.732 – 2
= 2 (1.732 – 1)
Page 141
141
= 2 0.732
h = 1.464 mts
Height of the statue CD = h = 1.464 mts
5. The angle of elevation of the top of a building from the foot of the tower is
300 and the angle of elevation of the top of the tower from the foot of the
building is 600. If the tower is 30m high, find the height of the building?
Sol: From the adjacent figure
Height of the tower AB = 30 mts
Let height of the building CD = hmts
Distance between from the foot of the tower to foot of the building = BD
ADB = 600
CDB = 900
60
0 =
=
BD = 30
BD = (1)
In BCD, BDC = 900
300 =
= . (From equation (1) BD =
=
3h = 30
h =
h = 10
height of the building CD = h = 10 mts.
2 Marks Questions
1. A tower stands vertically on the ground from a point which is 15 meter away
from the foot of the tower, the angle of elevation of the top of the tower is
450 what is the height of the tower?
Sol: From the adjacent figure
Let height of the tower BC = h mts
Distance between observation point to the foot of the tower AB = 15mts
Page 142
142
In ABC, B = 900
450 =
1 =
h = 15mts
The height of the tower BC = h = 15mts
2. Length of the shadow of a 15 meter high pole
is 5 meters at 7 “0” clock in the morning
then what is the angle of elevation of the sum rays with the ground at the
time?
Sol: From the adjacent figure
Height of the pole BC = 15mts
Length of the shadow of a 15 meters high pole AB = 5
Let PCA = CAB =
=
=
=
=
But = tan 600
= = tan 60
0
= 600
Hence the angle of elevation of the sum rays with the ground at the time =
600
3. Suppose you are shooting an arrow from the top of a building at an height of
6 meters to a target on the ground at an angle of depression of 600.What is
the distance between and the object?
Sol: From adjacent figure
Height of the building BC = 6mts
Let distance between target on the ground to foot of building AB = xmts
PCA = CAB = 600.
In ABC, ABC = 900
tan 60
0 =
Page 143
143
=
AB =
AB = 2
= 2 1.732
AB = 3.464mts
4. A boat has to cross a river it crosses the river by making an angle of 600 with
the bank of the river due to the stream of the river and travels a distance of
600mts to reach the another side of the river. What is the width of the river?
Sol: From the adjacent figure
Let the width of the river BC = x mts
AB denotes the distance travelled by the boat to reach the another side
= 600
In ABC, = 90
0
Cos 60
0 =
=
x = 300mts
The width of the river BC = x = 300mts
CHAPTER 13
PROBABILITY Probability: Probability means the number of occasions that a particular
statement or event is lifely to occur in a large population of events.
The theoretical (classical) probability of an event E, written as P(E) is
defined as
P (E) =
P(E) + P(E) = 1 where stands for „not E‟
i.e., P ( ) = 1 – P(E)
Page 144
144
The sum of the probability of all the elementary events of an experiment is
„1‟
The probability of an impossible event is „0‟
4 Marks Questions
1. A bag contains a red ball, a blue ball and a yellow ball, all the balls being of
the same size. Manasa takes out a ball from the bag without looking into it.
What is the probability that she takes a (i) yellow ball? (ii) read ball? (iii)
blue ball?
Sol: Manasa takes out a ball from the bag without looking into it So, it equally
that she takes out any one of them.
Let Y be the event „the ball taken out is yellow.
B be the event „the ball taken out is blue.
R be the event „the ball taken out is red.
The number of outcomes favourable to the event y = 1
P (Y) =
=
Similarly
P(B) = , P(R) =
2. Suppose we throw a die once (i) what is the possibility of getting a number
greater than 4? (ii) What is the probability of getting a number less than or
equal to 4?
Sol: In rolling an un based dice.
Sample space S = {1,2,3,4,5,6}
No of outcomes n(s) = 6
Favorable outcomes for number greater than 4 ie n (E) = 2
Page 145
145
Probability P (E) = = =
(ii) Let F be the event getting a number less than or equal to 4
Sample space S = {1,2,3,4,5,6}
No of outcomes n(s) = 6
Favorable outcomes for number greater than 4 ie n(F) = 4
Probability P (F) = =
3. Rahim takes out all the hearts from the cards what is the probability of
(i) Picking out an ace from the remaining pack
(ii) Picking out a diamonds
(iii) Picking out a card that is not a heart
(iv) Picking out the Ace of heart.
All the heads are taken out of 52 cards.
Therefore the remaining cards will be 52 – 14 = 39
(i) The probability of getting on Ace from the remaining pack
P (A) =
= =
(ii) The probability of picking out a diamond
=
= =
(iii) The probability of picking out a card that is not a heart
=
Page 146
146
= =
(iv) The probability of picking out the Ace of hearts
=
= = 0
4. Box contains 5 red marbles, 8 white marbles and 4 green marbles one
marble is taken out of the box at random. What is the probability that the
marble taken out will be (i) red? (ii) white? (iii) not green.
Sol: There are 5 red marbles, 8 white marbles, and 4 green marbles in a bag
The total number of marbles in the bag = 5 + 8 + 4 = 17
Number of all possible outcomes = 17
(i) Let „R‟ be the events that the marble taken out will be red
Total number of red marbles in the bag = 5
Number of outcomes favourable to R ie n(R) = 5
Number of all possible outcomes n(s) = 17
P (R) = =
(ii) Let „W‟ be the events that the marble taken out will be white
Number of outcomes favourable to white ie n(w) = 8
Number of all possible outcomes n(s) = 17
P (W) = =
(iii) Let „G‟ be the events that the marble taken out will be green
Number of outcomes favourable to green ie n(G) = 4
Number of all possible outcomes n(s) = 17
Page 147
147
P (G) = =
Probability that the marble taken out will not be green = P(G)
But P (G) + P ( ) = 1
P ( ) = 1 – P (G) = 1 - =
P ( ) =
5. One card is drawn from a well shuffled deck of 52 cards, find the probability
of getting?
(i) A king of red colour (ii) a face card (iii) a red face card
(iv) The jack of hearts (v) a spade (vi) The queen of
diamonds
Sol: Total number of cards = 52
Number of all possible outcomes n(s) = 52
(i) Let „R‟ be the event of getting a king of red colour
Then the outcomes favourable to „R‟ are kings of diamond and heart
Number of outcomes favourable to king of colour Red
i.e. n (R) = 2
P (R) = = =
(ii) Let „F‟ be the event of getting a face card
Number of outcomes favourable to „F‟ I.e n (F) = 12
Number of all possible outcomes n(s) = 52
P (F) = = =
(iii) Let „R‟ be the event of getting a red face card
Page 148
148
Number of outcomes favourable to red face card
I.e n (R) = 6
n (S) = 52
P (R) = =
(iv) Let „J‟ be the event of getting the jack of hearts
Number of outcomes favourable to jack of hearts
I.e n (J) = 1
n (J) = 52
P(J) =
E be the event of getting a spade
Then the outcomes favourable to a spade ie n (E) = 13
Total outcomes n(s) = 52
P (E) =
=
(iv) Let „d‟ be the event of getting the queen of diamonds
Then the number of outcomes favourable to queen of diamond
i.e., n (d) = 1
n (s) = 52
P (d) =
Page 149
149
2 Marks Questions
1. Define the mutually exclusive events?
Sol: Solution: Two or more events of an experiment, where occurance of an
event prevents occurances of all other events are called mutually exclusive
events.
2. Define the complementary events and probability?
Sol: If an event is denoted by „E‟ in an experiment.
We denote the event „not E‟ by this is called the complement event of
event E.
So P (E) + P ( ) = 1
P ( = 1 – P (E)
In general, it is true that for an event E.
P (E) = 1 – P ( )
3. Sangeeta and Reshma, play a tennis match. It is known that the probability
of sangeeta winning the match is 0.62. What is the probability of Reshma
winning the match?
Sol: Let S and R denote the events that sangeeta wins the match and Reshma
wins the match, respectively.
The probability of sangeeta‟s winning chances P (S) = 0.62 (Given)
The probability of Reshma winning chances P (R) = 1 – P (S)
= 1 – 0.62
= 0.38
4. A lot consists of 144 ball pens of which 20 are defective and the other are
good. The shopkeeper draws one pen at random and gives it to sudha. What
is the probability that (i) she will buy it? (ii) She will not buy it?
Page 150
150
Sol: Total number of Ball pens in a lot = 144
Number of defective pens in that lot = 20
Number of outcomes favourable to draws a good condition pen = 144 – 20
= 124
(i) The probability of sudha will buy a pen =
= =
(ii) The probability that sudha will not buy a pen = 1 –
=
1 Mark Questions
1. If P (E) = 0.05 what is the probability of not „E‟?
Sol: P (E) = 0.05
But we know that P (E) + P( ) = 1
P( ) = 1- P (E)
= 1-0.05
P( ) = 0.95
2. It is given that in a group of 3 students the probability of 2 students not
having the same birthday is 0.992 what is the probability that the 2 students
have the same birthday?
Sol: The probability of any 2 students not having the same birthday P (E) = 0.992
The probability that any two students have the same birthday P( ) = ?
But P (E) + P( ) = 1
P( ) = 1 – P(E)
= 1-0.992
= 0.008
Page 151
151
3. What is the probability of drawing out a red king from a deck of cards?
Sol: Total number of cards in a deck = 52
Number of all possible outcomes = 52
Number of all possible outcomes a red king from a deck of cards
Probability = =
Fill in the blanks
1. The definition of probability was given by ____________
2. A bag contains 6 red, 3 blue and 7 green marbles, a marbles is taken out of
the bag at random, then probability that the marble is blue ________
3. If P (E) = 0.91 the P( ) = _________
4. P (E) + P( ) = _________
5. The probability of a certain events is _________
6. The probability of an impossible event is ________
7. If 0 P (E) m then m = ________
8. P (A1
B1) = _______
9. P (A1
B1) = _______
10. If P(E) = 1/3 then P( ) = _________
11. is a an _________
12. The number of face cards in a deck of playing cards is ___________
13. If two dice are thrown at the same time then the number of all possible
outcomes are __________
14. The probability of a sure events is _______
15. If P(E) = 0.03 then P( ) = _________
Page 152
152
Answers
1. Pirre simon 2. 3. 0.09 4.1 5.1
6. 0 7. 1 8.1-P(A B) 9. 1-P(A B) 10.2/3
11. Impossible event 12.12 13. 36 14.1 15.0.97
CHAPTER 14
STATISTICS Arithmetic Mean:
(a) For ungrouped data, mean =
(b) For grouped data,
(i) Direct method =
(ii) Deviation method = a + [or assumed mean method]
(iii) Minimum deviation (or) step – deviation method
= a + h
2. Mode:
(a) Ungrouped data:
Mode is the observation which occurs more frequently
(b) Grouped data: Mode = l + h
3. Median:
(a) Ungrouped data:
Page 153
153
(i) If number of observations is odd, median = th in
observation
(ii) If n is even, median = Average of , + 1 th
observations
(b) Grouped data:
Median = l + h
4. Relation between Mean, Median and Mode:
Mode = 3 Median – 2 Mean
1 Mark Problems
1. The compare the results of students of different schools in 10th
class
examinations what is the measure that would be best suited? Why?
Sol: Since all the observations are taken into accounts, arithmetic mean is the
best suited measurement.
2. MEK is the most popular TV program being watched by the people of AP
what is the measure used here?
Sol: Mode
3. Find the mean of 16,12,18,8,9,0,5.
Sol: Mean =
=
= = 9.7
4. Find the Median of 35,18,24,30,24,16,40.
Sol: Given observation, in ascending order = 16,18,24,24,30,35,40
Page 154
154
Number of observations = 7
Median = th observation
= = = 4th
observation = 24
5. Median of is 8. Find x?
Sol: Given observation in ascending order =
Number of observations = 7
Median = = = 3rd
observation =
But Median = 8
= 8
x = 8 3 = 24
6. Mean of 50 observations is 38. If two observations 45 and 55 are deleted
what will be the mean of remaining observations?
Sol: Mean of 50 observations = 38
Sum of 50 observations = 38 50 = 1900
If 45 and 55 are deleted then new sum = 1900 – 45 – 55
= 1800
Mean of remaining 48 observations = = 37.5
2 Marks Problems
1. The mean of 9,11,13,P,18,19 is p find the value of p?
Sol: Mean of 9,11,13,P,18,19 =
=
Page 155
155
=
Given that mean = P
P = 6p = 70+p
6p – p = 70
5p = 70
P = = 14
2. In a data frequencies of 1, are 1,2,3,….n find the mean?
Sol: Observations = 1,
Frequencies = 1,2,3,…..,n
Mean = =
=
= = =
3. Median of the observations, which are in ascending order, 4,40,15,x+5,2x-
1,18,23,27 is 17 find x?
Sol: Given observations = 4,40,15,x+5,2x-1,18,23,27
Number of observations = 8
Median = Average of 1th
observations
= Average of 4,5th
observations
= Average of x + 5, 2x – 1
=
Page 156
156
=
Given that Median = 17
= 17
3x + 4 = 34
3x = 34- 4 = 30
x = = 10
4. In a school there are three section A,B,C in a class with 25, 40 and 35
students respectively. In an the average percentage of marks obtained by
A,B,C 70%, 65%, and 50% respectively. Find the average percentage of
class?
Sol: Number of students in sections A,B,C = 25,40,35 respectively
Average percentage of marks of A,B,C = 70%, 65%,50%
Average percentage of class X =
=
=
=
= 61%
5. Write the formula to find the median of a grouped data and explain the terms
involved in it?
6. Write the formula to find the mode of a grouped data and explain the terms
involved in it?
4 Marks Problems
Page 157
157
1. A sample of college students was asked how much they spent monthly on a
cell phone plan (to the nearest rupee) find the mean in all the three methods
and compare?
Monthly
cell phone
plan in
rupees
10-20 20-30 30-40 40-50 50-60 60-70 70-80 80-90 90-100
Number of
students 12 20 26 45 73 54 35 24 11
Sol:
Monthly
cell phone
plan in
rupees
No of
students
(fi)
Class
marks
(xi)
di =xi -
55
ui
= fixi fidi fiui
10-20
20-30
30-40
40-50
50-60
60-70
70-80
80-90
90-100
12
20
26
45
73
54
35
24
11
15
25
35
45
55(a)
5
75
85
92
-40
-30
-20
-10
0
10
20
30
40
-4
-3
-2
-1
0
1
2
3
4
180
500
910
2025
4015
3510
2625
2040
1045
-480
-600
-520
-450
0
54
700
720
440
-48
-60
-52
-45
0
54
70
72
44
=300
16850
Here =300 , a = 55, h = 10, , ,
(i) Mean by direct method =
= = 56.2
(ii) Mean by assumed mean method = a +
Page 158
158
= 55 + = 55+1.2 = 56.2
(iii) Mean by step deviation method = a + h
= 55 + 10 = 55+1.2 = 56.2
Mean in all the three methods = 56.2
Note: Practice some here problems in this model.
2. The following distributions shows the daily pocket allowances of children of
a locality. The mean packet allowance is 18?
Sol:
Daily
packet
allowances
in Rs
11-13 13-15 15-17 17-19 19-21 21-23 23-25
Number of
Childrens 7 6 9 13 f 5 4
Daily
packet
allowances
in Rs
No of
childrens
(fi)
Class
marks
(xi)
di =xi -
18
ui
= fiui
11-13
13-15
15-17
17-19
19-21
21-23
23-25
7
6
9
13
f
5
4
12
14
16
18(a)
20
22
24
-6
-4
-2
0
2
4
6
-3
-2
-1
0
1
2
3
-21
-12
-9
0
f
10
12
= 44 + f
Here = 44 + f , a = 18, h = 2,
Page 159
159
(i) Mean by direct method =
Mean = a + h = 18 (Given)
18+ 2 = 18
= 18 – 18 = 0
(f – 20)2 = 0 (44+f) = 0
f – 20 = 0 f = 20
3. The following table shows the ages of the patients admitted in a hospital
during a year. Find the mode and the mean. Compare and interpret the two
measures of central tendency?
Sol:
Age (in
years) 5-15 15-25 25-35 35-45 45-55 55-65
Number of
Patients 6 11 21 23 14 5
Age in
years
No of
patients
(fi)
Class
marks
(xi)
di =xi -
40
ui
= fiui
5-15
15-25
25-35
35-45
45-55
55-65
6
11
21
23
14
5
10
20
30
40 (a)
50
60
-30
-20
-10
0
10
20
-3
-2
-1
0
1
2
-18
-22
61
-21
0
14 24
10
= 80
Here the model class is 35 – 45
Page 160
160
l = 35, f0=21, f1 = 23, f2 = 14, h = 10
Mode l + h
= 35 +
= 35 +
= 35 + = 35 + 1.8 = 36.8
a = 40, h = 10, = 80
Mean = a + h
= 40 +
= 40 + = 40-4.6 = 35.4
Interpretation:
Mode is 36.8, so maximum numbers of patients have 36.8 years.
Mean is 35.4 so average of the patients is 35.4 years.
4. Find the median of the following data
Marks 0-10 10-20 20-30 30-40 40-50 50-60 60-70 70-80 80-90 90-100
Number of
students 5 3 4 3 3 4 7 9 7 8
Page 161
161
n = 53
= 26.5
Median class is 60 – 70
l = 60, f = 7,cf = 22 h = class size = 10
Median = l + h
= 60 + 10
= 60 + 10
= 60 +
= 60 + 6.4
= 66.4
Median = 66.4 marks
Half of the students got less than 66.4 marks and other half of students got
more than 66.4 marks.
Marks
No of
students
(fi)
Cumulative
frequency
(cf)
= 26.5
In cumulative frequency
26.5 occurs in the class 60 –
70
So 60 – 70 class has to be
taken as median class.
0-10
10-20
20-30
30-40
40-50
50-60
60-70
70-80
80-90
90-100
5
3
4
3
3
4
7 f
9
7
8
5
5+3=8
8+4=12
12+3=15
18
22 cf
29 Median
class
38
45
53
Page 162
162
5. Find the median of 60 observations given below is 28.5, find the values of x
and y?
Sol:
Class
interval
0-10 10-20 20-30 30-40 40-50 50-60
Frequency 5 x 20 15 y 5
Class
interval
Frequency Commulative
frequency
0-10
10-20
20-30
30-40
40-50
50-60
5
x
20 (f)
15
y
5
5
5+x (cf)
25+x
40+x
40+x+y
45+x+y
n = 45+x+y
h = 10, l = 20, f =20, cf = 5+x, n = 45+x+y = 60
x + y = 60 – 45 x + y = 15 ----- (1)
Median = l + h = 28.5
= 20 + 10 = 28.5
= = 28.5-20 = 8.5
= 25 – x = 8.5 2 = 17
= -x = 17 – 25 = -8
= x = 8
Substituting x = 8 in equation (1), we get
8 + y = 15
Page 163
163
y = 15-8 = 7
x = 8, and y = 7
6. A life insurance agent found the following data and ages of 100 policy
holders. Find the median?
Age
(in
years)
Belo
w
20
Belo
w
25
Belo
w
30
Belo
w
35
Belo
w
40
Belo
w
45
Belo
w
50
Belo
w
55
Belo
w
60
Numbe
r of
policy
holders
2 6 24 45 78 89 92 98 100
Age (in years) No of policy
holders
Class Interval Frequency Cumulative
Frequency
Below 20
Below 25
Below 30
Below 35
Below 40
Below 45
Below 50
Below 55
Below 60
2
6
24
45
78
89
92
98
100
Below 20
20 – 25
25 - 30
30 - 35
35 - 40
40 - 45
45 - 50
50 - 55
55 - 60
02
04
18
21
33 f
11
03
06
02
2
6
24
45 cf
78
89
92
98
100
n = 100 n = 100
= = 50
Median class is 35 – 40
h = 5, l = 35, f =33, cf =45
Median = l + h
Page 164
164
= 35 +
=
=
= 35 + 0.76
= 35.76
Interpretation: 50 of them are of the age below 35% years and remaining 50
are above 35.76 years.
Note: In the text book we have 4 graphs. So practice all the graphs
Fill in the Blanks
1. The mean of x1, x
2, x
3,…..,xn observations is . Then the mean of 2x1+7,
2x2+7,…..,2xn+7 observations = _________
2. Range of first „n‟ natural numbers = _______
3. Mean of first „n‟ natural numbers = _________
4. Median of first 7 prime numbers = _________
5. If the mode of the observations 4,5,10,3,5,4,9,3,x,3,4,5 is 5 then x =
_________
6. Class marks are used in finding _________
7. _______ is affected by extreme values of the data
8. When the individual observations are not important then the appropriate
measure of central tendency is ________
9.The x- coordinate of the intersecting point of the two points gives us _______
10. Mode of first 10 natural numbers _________
11. Arithmetic mean of a-2, a, a+2 is _____
Answers
1. 2x+7 2. n – 1 3. 4. 7 5.5 6. Arithmetic mean
7. Mean 8. Median 9. Median 10. Does not exist 11. a
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