Chapter 22

Elementary Mensuration -1

Triangle

Rule 1 Si find the area of a triangle if its base and height are Jpen.

1 J *ra of a triangle = x Base x Height

trative Example The base of a triangular field is 880 metres and its height 550 metres. Find the area of the field. Also calculate the charges for supplying water to the field at the rate of Rs 24.25 per sq hectometre.

hectare. Find the base and the height of the lawn.

Area of the field = Base x Height

880x550 440x550 = sq metres = . . . , . _ sq hectometres.

100x100 = 24.20 sq hectometres. C03J of supplying water to 1 sq hectometre = Rs 24.25

cost of supplying water to the whole field = Rs 24.20 x 24.25 =Rs 586.85

erase Find the area of a triangle in which base is 1.5 m and height is 75 cm. a) 5625 sq cm b) 5265 sq cm c)5635sqcm d)5525sqcm Find the area of a triangle whose one angle is 90, the hypotenuse is 9 metres and the base is 6.5 metres. a)20sqm b)20.5sqm c)20.15sqm d)21 sqm The base of a triangular field is three times its altitude. I f the cost of cultivating the field at Rs 24.60 per hectare is Rs 332.10, find its base and height. a) 250 m, 650 m b) 300 m, 900 m c) 350,850 m d) None of these A lawn is in the form of a triangle having its base and

height in the ratio 2 : 3 . The area of the lawn is

6.

7.

9.

a) 55 m, 34 m

10.

4 1 b)50m, 33 j m

c) 50 m, 35 m d) Data inadequate The base of a triangular field is three times its height. I f the cost of cultivating the field at Rs 36.72 per hectare is Rs 495.72, find its base and height. a) 900 m, 300 m b) 600 m, 300 m c) 900 m, 600 m d) Can't be determined Find the area of a triangle in which base is 36.8 cm and height is 7.5 cm. a) 128 sq cm b)148sqcm c)130sqcm d)138sqcm I f the area of a triangle with base x is equal to the area of a square with side x, then the altitude of the triangle is:

a) b)x c)2x d)3x

[I tax & Central Excise 1988] I f the area of a triangle is 150 sq m and base: height is 3 : 4, find its height and base. a)20m, 15m b)30m, 10m c) 60 m, 5 m d) Data inadequate

[GIC Exam 1983] The base of a triangular field is three times its height. I f the cost of cultivating the field at Rs 1505.52 per hectare is Rs 20324.52 find its base and height. a) 900 m, 300 m b) 300 m, 100 m c) 600 m, 200 m d) Data inadequate The base of a triangular field is 880 metres, and its height 550 metres. Find the area of the field. Also calculate the charges for supplying water to the field at the rate of Rs 242.50 per sq hectometre. a) Rs 5688.50 b)Rs 5868.50 c) Rs 6858.50 d) None of these

Answers l .a

J_ 2. c; Hint: Height of the triangle = ^92-(6.5)2 = V38.75/W 12

5 1 2 PRACTICE BOOK ON QUICKER MATHS

=6.2m

.-. Area of the triangle = -xBase* Height

(a + b + c) then, Area of the triangle

x6.5x6.2 12

sqm = 20.15 sqm

332.10 3. b; Hint: Area of the field - . . =13.5 hectares

24.01) = (13.5 x 10000)= 135000sqm Let, altitude be x metres. Then, base = 3x metres.

Area =

3xl

x base x altitude | = ]_

xxx3x 3xz

sqm

= 135000 or, 135000x2

= 300 2 - 3

Hence, altitude = 300 metres and base = 900 metres (Also see Rule - 68)

4. b; Hint: Let the base be 2x metres and height 3x metres.

Then | ( 2 x x 3 * ) = ^ U l 0 0 0 0 [ v 1 hectare =10000 sqm]

or, x = 10000x2._ 100 50

6 ~ 3

Base =

6x12

2x50 1 -33 metres,

3 3

3x50 Height = = 50 metrees. [See Rule - 68]

495 72 5.a;Hint: Area= 3 6 7 2 x 1 0 0 0 0 ^ 135000sqm

| * 3 x x x =135000

.-. x = 300 ie height = 300 m and base = 300 x 3 = 900 m [See Rule 68]

1 , 2 7. c;Hint: -xxxh = x .-. h = 2x

8. a;Hint: ^ x 3 x x 4 x = 150 or, x = 5 2

.-. base = 3 * 5= 15 m and height = 4x = 4 x 5 = 20 m

Rule 2

If a, b, c are the lengths of the sides of a triangle and S = ^

= y]s(s-a)(s-b)(s-c)

Illustrative Example Ex.: Find the area of a triangle whose sides are 50 metres,

78 metres, 112 metres respectively and also find the perpendicular from the opposite angle on the side 112 metres.

Soln: Here a = 50 metres, b = 78 metres, c = 112 metres.

.-. s= j (50 + 78+ 112)

= - - x 240 metres = 120 metres. 2

.-. s-a = (120-50)metres = 70metres s - b = (120 - 78) metres = 42 metres s - c = (120 -112) metres = 8 metres

area = Vl20x70x42x8 = 1680 sq metres

Perpendicular = 2Area 1680x2

metres Base 112 = 30metres. [SeeRule-1]

Exercise 1. Find the area of a triangle with two sides equal, each

being 5.1 metres and the third side 4.6 metres. a) 10sqm b) 10.5sqm c) 10.46sqm d) 11.46sqm

2. Find the area of a triangle in which a = 25cm,b=17cm and 0=12 cm. a)90sqcm b)80sqcm c)85sqcm d)75sqcm

3. I f the sides of a triangle are doubled, its area a) remains same b) becomes doubles c) becomes 3 times d) becomes 4 times

[Railway Recruitment Board Exam, 1991) 4. The sides of a triangular field are 949,1095,1022 metres.

It is let at Rs 10000 per hectare. Find the rent of the field, a) Rs 447636 b)Rs 446736 c) Rs 447663 d) Data inadequate

5. The sides of a triangular field are 165 metres, 143 metres and 154 metres, find its area. a) 10164 sqm b) 10146 sqm c) 10614 sq m d) None of these

6. Two sides of a triangular field are 85 metres and 154 metres respectively and its perimeter is 324 metres. Find (i) the area of the field

a) 2882 sqm b) 2782 sqm c) 2772 sqm d) 2672 sqm

(ii) the perpendicular from the opposite angle on the side 154 metres a) 36 metres b) 18 metres c) 45 metres d) 27 metres

(iii) the cost of levelling the field at the rate of Rs 5 per sq

Elementary Mensuration - I

m a) Rs 12860 b)Rs 13760 c)Rs 13860 d)Rs 13960

7. The sides of a triangle are 51,52,53 cm, find the perpen-dicular from the opposite angle on the side of 52 cm. Also find the areas of the two triangles into which the original triangle is divided. a) 45 cm, 560 sq cm, 640 sq cm b) 45 cm, 540 sq cm, 630 sq cm c) 48 cm, 540 sq cm, 630 sq cm d) 48 cm, 530 sq cm, 640 sq cm

Answers l.c 2.a

3. d; Hint: Let the original sides be a, b, c then

1 s= - (a + b + c)

Area of this triangle = J s(s - a)(s - b)(s - c)

For new triangle, the sides are 2a, 2b, 2c & S = 2S.

.-. Area of new triangle = y]S(S - 2a)(S - 2b)(S - 2c)

= J2s(2s - 2a)(2s - 2b%2s - 2c)

- ^ / l 6s(s - a)(s - b)(s - c)

= s(s - a)(s - b)(s - c) =4 x (area of original triangle).

1 4. a; Hint: S = - (949 + 1095 +1022) = 1533 m

Area= ^1533x584x438x511

= V511x3xl46x2x3xl46x511 = 511 x3 x 146x2sqm

10000x511x3x146x2 required rent =

10000 = Rs 447636

3. a Hint: The third side of the triangle = 324 - (154 + 85) = 85 metres

Now find the area by applying the given rule. (i) c; Area = 2772 sq m

2x2772 (ii) a; perpendicular distance = = 36 metres

(iii) c; the required cost = 2772 * 5 = Rs 13860

51 + 52 + 53 7b; Hint:S = = 78 cm

Area= ^78(78 - 51)(78 - 52)(78 - 53)

51 cm/ \ 5 3 cm \ 5 3 cm

45 cm \

B D 52 cm C

BD= ^ 5 1 2 ^ 4 5 2 =24cm

A A B D = - x 2 4 x 4 5 =540sqcm

DC = 52-24=28cm

A ADC= - x 2 8 x 4 5 =630sq cm

Rule 3

V 3 2

Area of an equilateral triangle = x (side) and perim-

eter of an equilateral triangle = 3 x side.

Illustrative Example Ex.: Find the area of an equilateral triangle each of whose

sides measures 8 cm. Also find perimeter of the equi-lateral triangle.

Soln: Applying the above formula, V3

Area of an equilateral triangle = x (8)2

x8x8 = 16V3 sq cm

V78x 27x26x25 =1170sqcm

Perimeter of an equilateral triangle = 3 x side = 3 x 8 = 24 cm

Exercise 1. Find the area of an equilateral triangle each of whose

sides measures 12 cm.

a) 36VJ sqcm b) 18^3 sqcm

c) 24V3 sq cm d) 30^3 sq cm

2. Find the area of a triangle in which each side measures 8 cm.

a) 2V3 s q c m b) 8>/3 sq cm

c) 16V3 sq cm d) 1 2 V J sq cm 3. Each side of an equilateral triangle is increased by 1.5%.

The percentage increase in its area is:

514 PRACTICE BOOK ON QUICKER MATHS

5.

a) 1.5% b)3% c)4.5% d)5.7% (Railway Recruitment Board Exam, 1991)

I f the perimeter of an equilateral triangle is 12 metres, find its area.

a)4V3/M 2 b ) i 6 V 3 m 2 c ) *Sm2 d) 6m 2 (LIC Exam 1986)

The side of an equilateral triangle is 7 metres. Calculate its area correct to three places of decimals. a)21.218sqm b)21.281sqm c) 21,128 sqm d) None of these

Answers La 2.c 3. a; Let original length of each side = a

& i Then, area = a = A

4 New area

4 101.5 N

a 100 4 I , 20 J I , 20 J

Increase in area= | Ax xlOO |% = l.5/0

4. a; Hint: 3 xside= 12 m 12

side = = 4 m

Area = x 4 x 4 = 4-\/3 sqm

5.a

Rule 4

(I) Area of an isosceles triangle = ^ 4a2 -b2

b/2 b/2

(ii) Height(h) = j ^ - [ f ] " " \ ^ a 2 - b 2 (Hi) Perimeter = (2a + b)

Illustrative Example Ex.: The base and the other side of an isosceles triangle is

10 cm and 13 cm respectively. Find its area and perim-eter.

Soln: Applying the above formula,

Area = ^ M l 3 ) 2 - ( l 0 ) 2

= V676-109-= x 2 4 = 60 cm2 4 4

Perimeter=2x 13 + 10=36cm.

Exercise 1. The perimeter of an isosceles triangle is equal to 14 cm:

the lateral side is to the base in the ratio 5 to 4. The area. in cm2 , of the triangle is:

10

a ) ~ i / 2 L 2

b) V 2 T d) 2 V 2 T

(CDS Exam 1989)

2. A plot of land is in the shape of a right angled isosceles

triangle. The length of hypotenuse is 5 0 ^ 2 m - The cost of fencing is Rs 3 per metre. The cost of fencing the plot will be: a) less than Rs 300 b) less than Rs 400 c) more than Rs 500 d) more than Rs 600

(CDS Exam 1991) 3. In an isosceles right-angled triangle, the length of one

leg is 10 metres. Find its area and its perimeter. a)50sqm,34.15m b) 50 sqm, 44.14 m c) 50 sq m, 34.41 m d) Data inadequate

Answers 1. d; Hint: Let lateral side = 5x & base = 4x

Then,5x + 5x + 4;c = 14 => x = l .-. The sides are 5 cm, 5 cm, 4 cm

Now, 2 = 5 2 -22 = V 2 T

Area = - x 4 x V 2 1 2

cm2 = ijllcm2

2. c; Hint: Let each of the equal sides be a metres long.

Then, a 2 + a 2 = (50^2) 2 = 5000

=>a2 =2500=>a = 50

.-. Perimeter of the triangle = (50 + 50 + 50 J2 )

= 100 + 50x1.4146= 170.73 m .-. Cost offencing = Rs( 170.73 x 3) = Rs 512.19

3. a; Hint: Area= ^ x l O x l O =50 sqm

10 m

C 10m Perimeter = AC + CB + AB

B

Elementary Mensuration - I 515

[AB= V l O 2 +10 2 = V 2 0 0 = 1 0 V 2 ]

= 1 0 + 1 0 + 1 0 7 2 =20+14.14 = 34.14m

Rule 5 Theorem: The perimeter of an isosceles triangle is given as P cm. Now consider the following cases. Case I: If the base of the isosceles triangle is given by 'b'

(P-b) cm, then the length of the equal sides is cm.

\ / Case II: If the length of equal sides is given by 'a' cm, then

the length of the base is(P- 2a) cm. Illustrative Examples Ex. 1: The perimeter of an isosceles triangle is 120 cm. I f the

base is 60 cm, find the length of equal sides. Soln: Applying the above formula, (case -1)

120-60 o n Length of equal sides = = 30 cm

Ex. 2: The perimeter of an isosceles triangle is 100 cm. I f the length of the equal sides is given by 32 cm, find the length of the base.

Soln: Applying the above formula (case - I I ) , Length of the base = 100 - 2 x 32 = 36 cm

Exercise 1. The perimeter of an isosceles triangle is 60 cm. I f the

base is 30 cm, find the length of equal sides. a) 30 cm b)15cm c)12cm d)20cm

2 The perimeter of an isosceles triangle is 45 cm. I f the base is 25 cm, find the length of equal sides. a)20cm b)10cm c)8cm d)15cm

3. The perimeter of an isosceles triangle is 32 cm. I f the base is 18 cm, find the length of equal sides. a) 7 cm b)9cm c)14cm d)8cm

4. The perimeter of an isosceles triangle is 1 2 0 cm. I f the length of the equal sides is given by 50 cm, find the length of the base. a) 25 cm b)20cm c)15cm d)30cm

5. The perimeter of an isosceles triangle is 50 cm. I f the length of the equal sides is given by 12 cm, find the length of the base.

a)26cm b)24cm c)36cm d)16cm

Answers L b 2.b 3.a 4.b 5.a

Rule 6 Theorem: To find the height of the equilateral triangle when the length of its side is given.

s Height of the equilateral triangle = -Tk side.

Illustrative Example Ex^ Length of the side of an equilateral triangle is 4 ^3

cm. Find its height. Soln: Applying the above rule, we have height

= x 4A/3 = 6 cm. 2

Exercise 1. Height of an equilateral triangle is 6 cm. Find its side,

a) 4 cm b) 3^/3 cm c) cm d) 5^/3 cm

_2_

2. Length of the side of an equilateral triangle is cm.

Find its height.

a) 1 m b)1.5m c ) ^ m d)0.5m

3. Length of the side of an equilateral triangle is 3 ^ 3 cm.

Find its height.

a)4.5m b)4m c)5m d)5.5m

Answers

1. c; Hint: 6 cm = x side 2

6x2 . ir Side= t=- = 4V3 C m

V3"

2. a 3.a

Rule 7 Theorem: To find the area of an equilateral triangle If its height is given.

{Height)2

Area of the equilateral triangle = ^=

Illustrative Example Ex.: Height of an equilateral triangle is 6 cm. Find its area. Soln: Detail Method: Let the base of an equilateral triangle

be x and the height be h. V3 2 1 , Now, x = - x x x / i

or, x =

4 2

2x/ j 2x6 12

- x base* height J

s s s

V3 3 V3 12x12 j r Area = x x = x - = 12V3 sqcm.

4 4 3 Quicker Method: Applying the above theorem, we have


6x6 6x2xV3xV3 , IT Area = - 7 = - = 7= = 1 2 V 3 sq cm.

Exercise 1. Height of an equilateral triangle is 12 cm. Find its area,

a) 48VJ sq cm b) 36-^3 sq cm

c ) 12^3 sq cm d) Data inadequate

2. Height of an equilateral triangle is 9 cm. Find its area,

a) 27 V J sq cm b) 36-^3 sq cm

c ) 54^3 sq cm d) Data inadequate

3. Area of an equilateral triangle is 75-^3 sq cm. Find its

height. a) 12 cm b) 25 cm c) 15 cm d) Data inadequate

Answers l .a 2. a 3.c

Rule 8 Theorem: The perimeter of a square is equal to the perim-eter of an equilateral triangle. If the diagonal of the square is'd' units, then

d (i) the side of the square = J T units,

Ad (ii) the side of the equilateraltriangle = ^ c - units,

(Hi) the area of the square = square units, and

Z A* (Iv) the area of the equilateral triangle = 3 " ^ * ^

sq units.

Illustrative Example Ex.: Perimeter of a square and an equilateral triangle is

equal. I f the diagonal of the square is 12 V2 cm, then

find the area of the equilateral triangle.

Soln: Detail Method: Diagonal of the square = 12V2

cm. or, side x J % = I 2 V 2 .-. Side of the square = 12 cm Perimeter of an equilateral triangle = Perimeter of the square or, 3 x side of the triangle = 4 x 1 2 .. side of the triangle = 16

.-. area= ^ - x l 6 x l 6 = 16-73 sqcm.

Quicker Method: Applying the above theorem, we have the

2 area of the equilateral triangle : r x l 2 x l 2 x 2

3V3 '

= 64-\/3 sqcm.

Exercise 1. Perimeter ofa square and an equilateral triangle isequal.

I f the diagonal of the square is \^/2 c m > then find the (i) side of the square.

a) 12 cm b) 15 cm c) 12V2 c m


the rectangle are given. Breadgh = Area

Length

Illustrative Example Ex.: Area of a rectangular field of length 12 m is 120 sq m.

Find the breadth of the field. Soln: Applying the above formula, we have Breadth

120 12

= 10 cm.

(iii) To find the length of a rectangle, if area and breadth of

Area the rectangle given. Length : Breadth

Illustrative Example Ex.: Area of a rectangular field of breadth 10 cm is 120 sq

m. Find the length of the field. Soln: Applying the above formula, we have Length

120 10

12 cm

Exercise 1. Find the area and perimeter of a rectangular plot whose

length is 24.5 metres and breadth is 16.8 metres. a)411.6sqm,82.6m b)412sqm,83m c) 416.1 sq m, 86.2 m d) None of these

2 Calculate the area of a rectangular field whose length is 13.5m and breadth is 8 m. a)180sqm b)108sqm c) 140 sq m d) None of these

3. The length and breadth of a rectangular field are in the ratio 5 : 3. I f the cost of cultivating the field at 25 paise per square metre is Rs 6000, find the dimensions of the field. a)250mby 100m b)50mby30m c) 200 m by 120 m d) Can't be determined

4. A room 15 m long requires 7500 tiles,'each 15 cm by 12 cm, to cover the entire floor. Find the breadth of the room. a) 10m b)12m c)6m d)9m

5. A lawn in the form of a rectangle is half as long again as . 2

it is broad. The area of the lawn is ~ hectares. The length of the lawn is:

a) 100 m b) 33^ m z*2 c) ooy m '100^

m

6. The width of a rectangular hall is of its length. I f the

area of the hall is 300 sq m, then the difference between its length and width is:

a) 3 m b)4m c) 5 m d) 15 m IBank PO Exam-1990|

7. A room 8 m * 6 m is to be carpeted by a carpel 2 m wide. The length of carpet required is a) 12m b)36m c)24m d)48m

[Railway Recruitment, 1990| 8. The length of a plot of land is 4 times its breadth. A

playground measuring 1200 sq m occupies one-third of the total area of the plot. What is the length of the plot, in metres? a) 90 b)80 c)60 d) None of these

(Bank PO Exam-1990) 9. Length of a room is 6 m longer than its breadth. I f the

area of the room is 72 sq m, its breadth will be: a) 12m b)6m c)8m d)10m

10. I f only the length of a rectangular plot is reduced to

2

rd of its original length, the ratio of original area to

reduced area is: a)2:3 b)3:2 c ) l : 2 d) None of these

|Railway Recruitment 19911 11. The cost of carpeting a room 15 m long with a carpet 75

cm wide at 30 paise per metre is Rs 36. The breadth of the room is: a)8m b)12m c)9m d)6m

12. Calculate the area of a rectangle 23 metres 7 decimetres long and 14 metres 4 decimetres 8 centimetres wide. a)343sqm b)363sqm c)334sqm d)365sqm

13. The sides of a rectangular field of 726 sq metres are in the ratio 3 :2. Find the sides. a)33m,22m b)30m,20m c) 45 m, 30 m d) Can't be determined

14. The length of a room is 3 times its breadth and its breadth is 5 m 5dm. Find the area of its floor. a)90.75sqm b)81.12sqm c) 80.75 sqm d) 90.25 sqm

Answers 1. a; Hint: Area = (24.5) x (16.8) sq m = 411.6 sq m

Perimeter=2 x (24.5 + 16.8) m = 82.6 m (See Rule -11) 2. b

3. c; Hint: Area = 6000x100

25 = 24000 sqm

Let the length be 5x and breadth be 3x

or, 5x x 3x=24000 .-. x = ^ / l 600 ='40

.-. Length = 5 x 40 = 200 m and breadth = 3 x 40 = 120 m

( 1 5 12^ 4. d; Hint: Area of 1 tile = I Yoo"X Too" J s t ' m

5 1 8 - PRACTICE BOOK ON QUICKER MATHS

Area of the floor = 7500x 15 12

x J =135sqm

Breadth of the room = Area

Length '135^ V 15 y

m = 9 m

5. a; Hint: Let breadth = x metres. Then, length = x metres.

3 2 xx~x=-x 10000

2 3

01- . v 2 . [ 1x10000j S o , x = ^

L e n g t h ^ * 3 J 3 200

= 100 metres.

6. c; Hint: Let length = x metres. Then breadth = x metres

3 2 300x4 A n n . x x - x = 300=>x = = 400 o r x = 20

4 3

: ( 3 } . (length - breadth) = 20 x 20 = 5 metres.

7. c; Hint: Length of the carpet 8x6

m = 24 m

(Also see Rule - 53)

8. d; Hint: xxAx = 3600 => x2 = 900 => x = 30 .-. Length = (4x30)m=120m

9. b;Hint: x(x-6) = 72=>x2-6x-12 = 0

=> (x-12)(x + 6) = 0 => x = 12(neglectingx=-6) .-. Breadth = (12-6)m = 6m

10. b; Hint: Let length = x & breadth = y

New length = x & breadth = y

Original area _ xy _ 3 2 Reduced area 2

-xy

3600 11. d; Hint: Length of carpet = ^ = 120 m

75 1^ 120x V 100 Area of the carpet

.-. Area of the room = 90m 2

m2 = 90m 2

f90^ So, breadth of the room = i j 5 m = 6 m

12. a; Hint: Length = 23.70 metres [since 10 decimetres = 1 m] Breadth = 14.48 metres .-. Area=23.70x 14.48=343.176 343sqm

13. a;Hint:3xx2x = 726 ' or, x = 11 .-. sides = 3 x H=33mand2x = 2 x l l = 2 2 m

14. a

Rule 10 To find the length of diagonal of a rectangle, if length and breadth are given.

(Diagonal)2 = (Length)2 + (Breadth)2

Or, Diagonal = -^(Length)2 + (Breadth)2

Illustrative Example Ex.: Find the length of diagonal of a rectangle of length 8

cm and breadth 6 cm. Soln: Applying the above theorem, we have

Diagonal = ^(length)2 + (breadth)2

= A / (8) 2 +(6) 2 = V64 + 36 = Vl00 = 10 cm.

Exercise 1. Two roads XY and YZ of 15 metres and 20 metres length

respectively are perpendicular to each other. What is the distance between X and Z by the shortest route? a) 35 metres b) 30 metres c) 24 metres d) 25 metres

(SBI Associates PO -1999) 2. The length of the longest rod which can be laid across

of floor of a rectangular room 12 m in length and 5 m in breadth will be: a) 17m b)7m c)2.4m d)13m

3. The legs of a right triangle are in the ratio of 1 :2 and its area is 36. The hypotenuse of the triangle is:

a) 3 b ) V J c )V3 d ) 6 A / 5 \y Recruitment 1991)

4. Find the diagonal of a rectangle whose sides are 12 metres and 5 metres a) 13 metres b) 14 metres c) 16 metres d) Can't be determined

5. A ladder is placed so as to reach a window 63 m high. The ladder is then turned over to the opposite side of the street and is found to reach a point 56 m high. I f the ladder is 65 m long, find the width of the street. a)49m b)45m c)40m d)59m


Answers I d : Hint:

15 m

X Z = V i 5 2 +20 2 =V625 =25 m 2.d

3. d: Hint: *xx.2x = 36 => x = 6 2

Hypotenuse = ^ 6 2 +(12) 2 =Vl80 =6>/5

4. a 5. a; Hint:

B o C

OB = V 6 5 2 - 6 3 2 =16m OC= ^tf^X? =33m

.-. width of the street = OB + OC = 16 + 33 = 49 m

Rule 11 7o /inrf f/re perimeter of a rectangle if length and breadth are given.

/ Perimeter = 2(length + breadth)

Illustrative Example Ex.: Find the perimeter of a rectangle of length 8 cm and

x breadth 6 cm. Soln: Applying the above theorem, we have

Perimeter=2(8 + 6)=28 cm Exercise 1. The length of a rectangular plot is 20 metres more than

its breadth. I f the cost of fencing the plot at the rate of Rs 26.50 per metre is Rs 5300, what is the length of the

plot (in metres)? a) 40 b)120 c)50 d) None of these

(Bank of Baroda PO 1999) 2. The length and breadth of a playground are 36 m and 21

m respectively. Flagstaffs are required to be fixed on all along the boundary at a distance 3 m apart. The number of flagstaffs will be: a) 37 b)38 c)39 d)40

(I Tax & Central Excise 1989)

3. A rectangle is having 15 cm as its length and 150 sq cm

. 1 ' as its area. Its area is increased to 1 times the original

area by increasing only its length. Its new perimeter is: a)50cm b)60cm c)70cm d)80cm

(Bank PO Exam 1989) 4. The length and breadth of a rectangular piece of land are

in the ratio of 5 : 3. The owner spent Rs 3000 for sur-rounding it from all the sides at the rate of Rs 7.50 per metre. The difference between length breadth is: a)50m b)100m c)200m d)150m

(BSRB BankPO Exam 1991) 5. The sides of a rectangular park are in the ratio 3 : 2 and

its area is 3750 m 2 . The cost of fencing it at~50 paise per metre is: a)Rs312.50 b)Rs375 c)Rs 187.50 d)Rsl25

6. The length of the rectangular floor is twice its width. I f

the length of a diagonal is 9J5 m, then perimeter of the

rectangle is: a)27m b)54m c)81m d)162m

7. The area of a rectangular field is 27000 sq m and the ratio between its length and breadth is 6 : 5. Find the cost of the wire required to go four time round the field at Rs 740 per km of length of the wire. a) Rs 1953.60 b)Rs 448.40 c) Rs 1963.50 d) Data inadequate

8. The perimeter of a rectangle is 640 metres and the length is to the breadth as 5 : 3. Find its area. a) 2400 sqm b) 24000 sqm c) 24 hectare d) Can't be determined

9. The length of a rectangular field is twice its breadth. I f the rent of the field at Rs 3500 a hectare is Rs 28000, find the cost of surrounding it with a fencing at Rs 5 per metre.

a)Rs6000 b)Rs7000 c)Rs6500 d)Rs8000

Answers /, -v, 5300

1. d;Hint: V + = =200m

.-. 1 + b = 100 m (i) and 1 - b = 20 m (given) (ii) From eqn (i) and (ii), we have

, 120 / = = 6 0 m

2 2. b; Hint: Petimeter=2 (36 + 21) = 114 m

required no of flagstaffs = 114

= 38

3. b; Hint: Original length = 15 cm & breadth = 150 15

10 cm


New area = I50x-\m2 = 200/n2 3

New area _ 200

N e w = Original breadth ~ 7o~ = 2 0 c m

New perimeter = 2 (20 + 10) = 60 cm

3000 4. a; Hint: Perimeter of the field

7.50 = 400m

.-. 2(5x + 3x) = 400 => x = 25 So, length = 125 m & breadth = 75 m Difference between length & breadth

= (125-75)m = 50m 5. d;Hint:3xx2x = 3750 r^>x2=625 => x=25

.-. Length = 75 m & breadth = 50 m Perimeter=[2 x (75 + 50)] m=250 m

.-. Cost of fencing = Rs [250x^j = Rs 125 6. b; Hint: Let breadth = x metres and length = 2x metres

Then, x 2 + ( 2 x ) 2 = (9V5) 2

==>5x2 =405-=>x = V 8 l = 9 [SeeRule-10] .-. Perimeter = 2 (18 + 9) = 54 m

7. a; Hint: 6x x 5x=27000 or, x = 30 .-. length = 180 m, breadth = 150 m Length of wire required to go round the field four times = [4x 2 (180 + 150)] = 2.64 km .-. required cost = Rs (2.64 * 740) = Rs 1953.60

8. b;Hint:2(5x + 3x) = 640 or,x = 40 .-. length = 200 m and breadth = 120 m .-. area = 200 x 120 = 24000 sq m = 2.4 hectare

28000 9. a; Hint: Area of the rectangular field =

= 8 hectare = 80000 sqm 2 x x x = 80000 [See Rule-9]

.'. x = V40000 = 200 m and length = 400 m Perimeter=2 (400 + 200) = 1200 m .". Cost of fencing the rectangular field = 1200 x 5

= Rs6000

Rule 12 To find the area and perimeter of a rectangle, if its one side and one diagonal are given.

(i) Area of rectangle = ^1*4 d2 -I2 j sq units

(ii) Perimeter of rectangle = 2(1+ ^d2-!2) units.

Illustrative Example Ex.: One side and the diagonal of a rectangle are 40 m and

50 m respectively. Find its area and perimeter. Soln: Applying the above formula, we have

Area of the rectangle = 4fjx-\/502 - 4 0 2 = 1200 m 2

2| 40 + -/50 40 Perimeter of the rectangle;

= 140 metres.

Exercise 1. Find the area of a rectangle whose one side is 3 metres

and the diagonal is 5 metres. a) 12 sqm b)8sqm c) 16 sqm d) 14 sqm

2. Calculate the area of a rectangular field whose one side is 12 m and the diagonal is 13 m. a) 70 sq m b) 60 sq m c) 45 sq m d) 75 sq m

3. One side of a rectangular field is 4 metres and its diago-, nal is 5 metres. The area of the field is:

a) 12 m2 b)20 m2 c) 15 m2 d) 4 /^5 m2

4. A man walked 20 m to cross a rectangular field diago-nally. I f the length of the field is 16 m, the breadth of the field is: a)4m b)16m c) 12 m d) Can't be determined

(Railway Recruitment 1991)

Answers L a 2.b 3.a

4. c; Hint: Area = 16XV20 2 - 1 6 2 =16x12

breadth = 16x12

16 12m

Rule 13 To find the area of a rectangle when Its perimeter and di-agonal are given.

Area of a rectangle = (Perimeter)2 (Diagonal)2

8 2 sq

units.

Illustrative Example Ex^ I f the perimeter and diagonal of a rectangle are 14 cm

and 5 cm respectively. Find its area. Soln: Detail Method: Let the length and width of the rect-

angle be x and y cm respectively. 2(x+y) = Perimeter = 14cm .-. x + y = 7cm (i)

ijx2 +y2 = diagonal = 5 cm

.-. x +y =25 cm....(ii)


Now, squaring equ (i)

{x + yf = 49

=>x 2 +y2 +2xy=49 => 25 + 2xy = 49 .-. xy = area of the rectangle

49-25 24 .-. xy = = = 12 sq cm.

2 2 Quicker Method: Applying the above theorem, we have

14x14 5x5 49-25 required area5

8 12 sq cm.

Exercise 1. A rectangular carpet has an area of 120 m and a perim-

eter of 46 m. The length of its diagonal is: a)15m b)16m c)17m d)20m

(Railway Recruitment 1991) 2. I f the perimeter and diagonal of a rectangle are 16 cm and

4 cm respectively. Find its area. a) 32 sq cm b) 26 sq cm c) 24 sq cm d) Data inad-equate

3. I f the perimeter and diagonal of a rectangle are 24 cm and 6 cm respectively. Find its area. a) 72 sq cm b) 54 sq cm c) 45 sq cm d) Data inadequate

4. A rectangular carpet has an area of 96 sq m and a diago-nal of 8 m. Find the perimeter of the carpet. a)32m b)16m c)24m d)28m

Answers

l .c;Hint: 120 = 46x46 (Diagonal)

2.c

8 2

Or, 46x 46-4(Diagonal)2 =120x8

Or, 4 x (Diagonal)2 =1156

.-. diagonal = ^289 =17m

3.b

(Perimeter)2 8x8 4. a; Hint: 96 =

Or, (Perimeter)2 =(96 + 32)8=1024

.-. Perimeter = ^1024 =32m

Rule 14 To find length and breadth of a rectangle If its area and perimeter are given. (i) Length of the rectangle

(Perimeter)2 . Perimeter 1Area+

16 4 units.

(ii) Breadth of the rectangle

Perimeter {(Perimeter) 16

- Area units.

Illustrative Example Ex.: I f the area and perimeter of a rectangle are 240 cm2

and 68 cm respectively, find its length and breadth. Soln: Detail Method: Let the length and breadth of the rect-

angle be x and y cm. Area of the rectangle = xy = 240 cm2 Perimeter of the rectangle = (x + y) 2 = 68 cm .-. x + y = 34 cm (i)

(x-yf =(x + y)2-4xy

= (34) 2 -4x240 =1156-960 = 196

:.x-y = Vl96 =14 ....(ii)-By adding equ (i) and equ (ii), we have 2x = 48 .-. x = 24 cm

2y = 20 :.y = 10 cm Quicker Method: Applying the above theorem, we have the

68x68 . . . 68 length of the rectangle = ^ 240 +

^289-240 + 17 = 7 + 17 = 24 cm

68 68x68 Breadth of the rectangle = ^ ^ 240

= 17-V49 = 17-7 = 10cm.

Exercise 1. When the length of a rectangular plot is increased by

four times its perimeter becomes 480 metres and area 12800 sq m. What was its original length (in metre)? a) 160 b)40 c)20 d) Can't be determined

(BSRB Bhopal PO - 2000) 2. Calculate the area of a rectangular field whose length is

66 m and perimeter is 242 m. a) 3630 sqm b) 3360 sqm c) 3560 sqm d) None of these

3. The cost of fencing a rectangular field at Rs 3.50 per metre is Rs 595. I f the length of the field be 60 metres,


find the cost of levelling it at 50 paise per square metre. a)Rs700 b)Rs860 c) Rs 750 d) Data inadequate

4. The perimeter of a rectangle is 82 m and its area is 400 sq m. The breadth of the rectangle is: a) 14m b)16m c)18m d)12m

Answers 1. b; Hint: 4x original length of the rectangular plot

- - , 2 8 0 0 + l * - = 4 0 + 1 2 0 = 1 6 0 16 4

460 .-. Original length = = 40 metres

2a ; Hint: 66 = (242T

16 Area +

242

or, (242)2

16 Area =66-60.5 = 5.5

A r e a = 242x242 _ ( 5 5 ) ( 5 5 ) 16

= 366025-30.25 =3630 sqm

3. c; Hint: Perimeter of the rectangular field =

Now, applying the given rule,

170

595 3.50

= 170m

60 = (170)' 16

Area + -

/I70x or, ft

170" 1 ^ 170 , n , Area =60 = 17.3

16 4 170x170 n c

Area= - (17.5x17.5) 16

= 1806.25 - 306.25 = 1500 sq m

1500x50 Cost of levelling = = Rs 750

4. b; Hint: Required breadth

100

82 82x82

i 4 V 16 =20.5-4.5 = 16 m

400

Rule 15 To find the difference in length and width of a rectangle when perimeter and area are given. Difference in length and width of a rectangle

units.

Illustrative Example Ex.: Perimeter and area of a rectangle are 82 cm and 400 sq

cm. Find the difference in length and width. Soln: Detail Method: Let the length be 1 cm and width be b

an

As per the question, (/ + b)l = 82 cm.

or, l + b = A\m ... .(i) and / x = 400 sqcm...(ii)

(l + bf ={l + bf-Alb = ( 4 l ) 2 - 4x400

l~b= V1681-1600 = V81 =9'cm Quicker Method: Applying the above rule, we have the re-

quired answer

' 8 2 ^

2 J 4x400 =- / (4 l ) 2 -1600 = V 8 l = 9 cm.

Exercise 1. I f the width of a rectangle is 2 m less than its length, and

its perimeter is 32 m, the area of the rectangle is: ,

a)224 m2 b) 108 m2 c)99 m1 d)63 m2 2. A man drives 4 km distance to go around a rectangular

park. I f the area of the rectangle is 0.75 sq km, the differ-ence between the length and the breadth of the rect-

'WgieisC ':, a)1j^25km b) 0.5 km c) 1 km d) 2.75 km

3. Th'e breadth of a rectangular tennis court is 7 metres less than its length and its perimeter 138 metres. Find its area. a)1178sqm b) 1187sqm c)1168sqm d) 1278sqm

Answers

l . d ; Hint: 2 = 32

- 4 x Area

4 = 256-4 x Area area = 256-4 252

= 63 sqm 4 4 2. c; Hint: Perimeter=4 km, Area=0.75 sq km

By applying the given rule find the difference between length and breadth.

3. a; Hint: 7 =

or, (69)2 -4x / f rea = 49

4 x Area


.(ii)

4761-49 or, Area = = 1178 sq m

Rule 16 Theorem: If length of a rectangle is increased by 'x' units and due to this increase, area of the rectangle also increases

by 'y' sq units, then width is given by units.

Illustrative Example Ex.: I f increasing the length of a rectangular field by 5

metres, area also increases by 30 sq metres, then find the value of its width.

Soln: Detail Method: Let the length and breadth of the rect-angular field be / m and b m respectively. In first case area = lb sq m In second case area = (1 + 5) b = lb + 30 or, lb + 5b=lb + 30 .-. b = 6 metres. Quicker Method: Applying the above theorem, we have

30 ,

the width of the rectangular field = = & m.

Exercise The area of a rectangular courtyard is 100 sq metres. Had the length of the courtyard been longer by 2 metres, the area would have been increased by 10 sq metres. Find the length and breadth of the courtyard. a)20m,5m b)25m,4m c)30m, 3 0 j m d) Data inadequate

1 If increasing the length of a rectangular field by 4 metres, area also increases by 16 sq metres, then find the value of its width. a) 4 m b)8m c)6m d) Data inadequate If increasing the length of a rectangular field by 8 metres, area also increases by 32 sq metres, then find the value of its width. a)4m b)6m c)9m d)12m

- I f increasing the length of a rectangular field by 9 metres, area also increases by 54 sq metres, then find the value of its width.

a)6m b)8m c)9m d)15m

j Answers :

La;Hint: Width = 10

= 5m

100 ' .-. length = r- = 20 m

3.a

5 4. a

[See Rule-9]

Rule 17 Theorem: There is a rectangle of area 'A' sq unit. If the sum of its diagonal and length Is n times of its breadth, then the

length and breadth of the rectangle are An' -1

2n and

2An - j units respectively, n - 1

Illustrative Example Ex.: There is a rectangular field of area 60 sq cm. Sum of its

diagonal and length is 5 times of its breadth. Find the breadth of the rectangular field.

Soln: Detail Method: Let the length and breadth ofthe rect-angular field be x cm and y cm respectively. As per the question,

xy = 60 sq cm (i) and yjx2 +y2 + x = 5y .... (ii)

or, x2 +y2 =(5y-x)2

or, x2 +y2 = 2Sy2 +x2-\0xy

or, 24^ 2 =10x60 or, y2 = 5 x 5

y = 5 cmandx : 60

12 cm

.-. Length and breadth of the rectangular field are 12 cm and 5 cm respectively. Quicker Method: Applying the above theorem, we have the length of the rectangular field

J60x(52 - l ) 160x24 = J * 1 = , = 12 cm and

V 2x5 V 10 the breadth of the rectangular field

2 x 6 0 x 5 = 5 cm.

Exercise 1. There is a rectangular field of area 48 sq cm. Sum of its

diagonal and length is 3 times of its breadth. Find the length and the breadth of the rectangle. a) 8 cm, 6 cm b) 12 cm, 4 cm c) 16 cm, 3 cm d) Data inadequate

2. There is a rectangular field of area 120 sq cm. Sum of its diagonal and length is 4 times of its breadth. Find the perimeter of the rectangle. a)46m \ b)15m j c)8m d) Data inadequate

3. There is a rectangular field of area 420 sq cm. Sum of its


diagonal and length is 6 times of its breadth. Find the diagonal of the rectangle. a) 35 cm b)37cm c)33cm d)32cm

Answers 1. a 2. a; Hint: First find the length and breadh.

Length = 15 m and breadth = 8 m .-. perimeter = (15 + 8)2 = 46m

3. b; Hint: First find the length and breadth of the rectangle. Lengths 35 cm, breadth = 12 cm

.-. diagonal = J352 +\22 = Vl369 =37m

Rule 18 Theorem: There is a rectangle. Its length is 'x'units more than its breadth. If its length is increased by 'y' units and its breadth is decreased by 'z' units, the area of the rectangle is unchanged. Length and breadth of the rectangle are

y-z J - I y-z (x + z)y

and units respectively.

Illustrative Example Ex.: Length of a rectangular blackboard is 8 cm more than

that of its breadth. I f its length is increased by 7 cm and its breadth is decreased by 4 cm, its area remains unchanged. Find the length and breadth of the rect-angular blackboard.

Soln: Detail Method: Let the breadth of the blackboard be x cm, then length = (x + 8) cm As per the question,

(x + $ + 7)(x-4)=(x + S)x

or, (x + l5)(x-4)=(x + S)x or, 3x = 60 :. x = 20 :. Breadth = 20 cm and Length = 20 + 8 = 28 cm Quicker Method: Applying the above theorem, we have

the length of the blackboard : (8 + 4)7 : cm and

/ '

the breadth of the blackboard = (8 + 7)4

7 - 4 = 20 cm.

:4x5

Exercise 1. Length of a rectangular blackboard is 16 cm more than

that of its breadth. I f its length is increased by 14 cm and its breadth is decreased by 8 cm, its area remains un-changed. Find the length and breadth of the rectangular blackboard. a) 28 cm, 20 cm b) 56 cm, 40 cm

c) 26 cm, 10 cm d) Data inadequate 2. Length of a rectangular blackboard is 12 cm more thar

that of its breadth. I f its length is increased by 13 cm and its breadth is decreased by 8 cm, its area remains ur-changed. Find the length and breadth of the rectangular blackboard. a) 52 cm, 40 cm b) 48 cm, 42 cm c) 26 cm, 20 cm d) Data inadequate

3. Length of a rectangular blackboard is 15 cm more than that of its breadth. I f its length is increased by 9 cm and its breadth is decreased by 6 cm, its area remains un-changed. Find the length and breadth of the rectangular blackboard. a) 60 cm, 40 cm b) 63 cm, 48 cm c) 64 cm, 48 cm d) Data inadequate

4. Length of a rectangular blackboard is 20 cm more thar. that of its breadth. I f its length is increased by 15 cm anc its breadth is decreased by 10 cm, its area remains un-changed. Find the perimeter of the black board. a) 150 cm b) 280 cm d) 270 cm d) 160 cm

5. Length of a rectangular blackboard is 10 cm more than that of its breadth. I f its length is increased by 8 cm anc its breadth is decreased by 5 cm, its area remains un-changed. Find the area of the black board. a) 1200 sqcm b) 1250 sqcm c) 1320 sq cm d) Data inadequate

Answers l . b 2.a 3.b 4.b 5.a

Rule 19 Theorem: Length of a rectangle is increased by 'a' units and breadth is decreased by 'b' units, area of the rectangle remains unchanged. If length be decreased by 'c' units and breadth by increased by'd' units, in this case also area 0/ the rectangle remains unchanged. Length and breadth of

the rectangle are given by ac d + b

ad-be and bd\

f a + c N \ad-bc

units respectively.

Illustrative Example Ex.: Length of a rectangular field is increased by 1 metres

and breadth is decreased by 3 metres, area of the field remains unchanged. I f length be decreased by 7 metres and breadth be increased by 5 metres, again area re-mains unchanged. Find the length and breadth of the

. rectangular field. Soln: Detail Method: Let the length and breadth of the field

be x m and y m respectively. As per the question, In the first case,

(x + 7) (y - 3) = xy or, xy + ly - 3x - 21 = xy


or, 3 x - 7 v = -21 ....(i) In the second case,

(x - l)(y + 5) = xy or,xy-7y + 5x-35=xy

or, 5x-7>> = 35 ....(ii)

From equ (i) and equ (ii), we get x = 28 m and

y = 15 m /, Length and breadth of the rectangular field are 28 metres and 15 metres respectively. Quicker Method: Applying the above theorem,'we have

( 7 x 5 + 7x3"! 56 Length = | ~c ^ , | x 7 _ T T x 7 ~ 2 8 metres 35-21 14

Breadth ; 21 + 21 35-21

x5 = 3x5 = 15 metres.

Exercise 1. Length of a rectangular field is increased by 14 metres

and breadth is decreased by 6 metres, area of the field remains unchanged. I f length be decreased by 14 metres and breadth be increased by 10 metres, again area re-mains unchanged. Find the perimeter of the rectangle, a) 172m b)192m c)162m \) Data inadequate

2 Length of a rectangular field is increased by 8 metres and breadth is decreased by 4 metres, area of the field remains unchanged. I f length be decreased by 6 metres and breadth be increased by 5 metres, again area re-mains unchanged. Find the area of the rectangle, a) 283.5 sqm b) 284 sqm c) 285 sq m d) Data inadequate

3. Length of a rectangular field is increased by 21 metres and breadth is decreased by 9 metres, area of the field remains unchanged. I f length be decreased by 21 metres and breadth be increased by 15 metres, again area re-mains unchanged. Find the length of diagonal of the rectangle. a)90m b)64m c)95.3m(approx) d) 64.8 m (approx)

Answers l.a . 2.a 3. c; Hint: Firstfind the length and breadth of the rectangle.

Length = 84 m and breadth = 45 m.

diagonal = V84 2 +45 2 = V9081 * 95.3 m

Square

Rule 20 To find the area of a square if length of one of the sides is gixen. ;

Area of a square = (sidef

Illustrative Example Ex.: Find the area of a square whose length of the side is

5 cm. \ Soln: Applying the above formula, we have \

Area of a square = (s) 2 = 25 sq cm

Exercise 1. The length of a rectangular plot is 144 m and its area is

same as that of a square plot with one of its sides being 84 m. The width of the plot is: a) 7 m b) 49 m c) 14 m d) Data inadequate

2. The length of a rectangular hall is 16 metres. I f it can be partitioned into two equal square rooms, what is the length of the partition? a) 16 m b) 8 m c) 4 m d) Data inadequate

|UTI Exam 1990| 3. Find the area of a square whose side is 75 metres.

a)5625sqm b)5265sqm c)5635sqm d)5675sqm 4. The side of a square field is 89 metres. By how many

square metres does its area fall short of a hectare? a)2179sqm b)2099sqm c)2079sqm d)7921 sqm

5. Two carpets are made at the same price per sq m. One of them is 25.6 m long and 8.1 m broad, and costs Rs 14400, the other which is square costs Rs 28900. What is the length of each side of the square carpet? a)20.4m b)21m c) 21.5 m d) Data inadequate

6. Find, to the nearest cm, the length of the side of a square ' ' " * ' * 1 '

piece of ground whose area is of a hectare.

a) 31 m 62 cm c) 30 m 62 cm

b) 31 m 52 cm d) 32 m 62 cm

7. Find the side of a square whose area is 68 sq m.

1 3 1 3 a) 7m b) 8m c) 8 m d) 7 m

8. Find the sides of two squares, which contain together 12.25 hectares, the sides of the squares being in the ratio of 3:4. a)210m,280m b)90m, 120m c)150m,200m d)180m,240m

Answers

1. b; Hint: Required answer = (84)1 144

= 7x7 =49m

2. b; Hint: Let the length of partition be.x m

or, 1 6 X * = J C 2 + x 2 . . x = 8 m


3. a 4. c; Hint: 1 hectare = 10000 sq m

14400 5. a; Hint: Rate of the carpet per sq m = = Rs 69.4

0.1 X Z 3 . D

28900 Area of the square = ^ ^ = 416.42 sq m

.-. length of the square = ^416.42 = 20.4 m

6. a; Hint: 1 hectare = 10000 sq m

1 hectare = 1000 sq m

.-. length of the side = 7l000 = 31.62 m = 31 m 62 cm

7. c 8. a;Hint: ( 3x ) 2 +(4x ) 2 =12.25x10000 = 122500 or,x=70

.-. sidesare3x = 3 x70 = 210mand4x = 4x70 = 280m.

Rule 21 To find the area of a square if length of the diagonal is given.

Area of the square = {diagonal)2

Illustrative Example Ex.: Find the area of a square whose length of diagonal is

6 cm. Soln: Applying the above formula, we have

Area of a square = ] - x (6) 2 = 18 sq cm.

Exercise 1. Area of a square field is 18 sq m find its length of diago-

nal. a) 5 m b)6m c)4m d) Data inadequate

2. What would be the length of the diagonal of a square plot whose area is equal to the area of a rectangular plot of 45 m length and 40 m width a) 42.5 m b)60m c) 4800 m d) Data inadequate

(Bank of Baroda PO -1999) 3. Find the area of a square whose diagonal is 2.9 metres

long. a)4.5sqm b)5sqm c) 4.205 sqm d) Can't be determined

4. Find the area of a square field, the length of whose di-agonal is 36 metres. a)648sqm b)678sqm c)684sqm d)668sqm

5. Find the length of the diagonal of a square of;area 200 square centimetres. f a) 30 cm b)25cm c)20cm , d)24cm

6. The area of a square field is 8 hectares. How long woutr a man take to cross it diagonally by walking at the rate of 4 km per hour. a)5min b)6min c)4min d)8min

7. The diagonal of a square is 3.2 m. Its area is:

a)10.24m 2 b )2 .56 / 2 c)3.41 m2 d)5.12 m2

8. Area of a square field is hectare. The diagonal of the

square is:

a)50m b)100m c)250m d) 50^/2 m

9. A square field of 2 sq km is to be divided into two equal parts by fence which coincides with a diagonal. Find the length of the fence. a)2km b)3km c ) l k m d) 1.5 km

10. What is the area of a square whose diagonal is 15 metres" a) 225 sqm b) 112.5 sqm c) 115 sqm d) 125 sqm

11. The area of a square 11370.32 sq metres. Find the lengti of its diagonal.

a) 158.8 m b) 148.8 m c) 150.6 m d) 150.8 m

Answers 1 2

1. b; Hint: - x (diagonal) =18 .-. diagonal = 6 m

2. b; Hint: Area of the square plot = 45 x 40 = 1800 sq m [See Rule - 9]

^(diagonal)2 =1800

.-. diagonal = Vl 800x2 =60m

3. c 4. a

5. c;Hint: ^.(diagonal)2 = 200

.-. diagonal = ^400 =20 cm

6. b; Hint: Area = 8 hectares = 8 x 10000 = 80000 sq m

l x (diagonal)2 =80000

.-. diagonal = Vl 60000 =400m

60x400 .-. required time = =6min.

7. d

1 , 1 1 8. b; Hint: - x (diagonal)2 = - x 10000

=> diagonal = VL0000 = 100 m.

9. a; Hint: Area=2X100X100X100 = 2000000 sq m

\ Diagonal = V2x 2000000 = 2000 m = 2 km.

10. b 11.d


Rule 22 (i) To find perimeter of a square if its length ofside is given.

Perimeter of a square = 4* side

Illustrative Example Ex.: Find the perimeter of a square field of length of one of

its sides 4 m. Soln: Applying the above formula, we have

perimeter of the square field = 4 x 4 = 16 m. (ii) To find the length of the side of a square if perimeter of the square is given.

Perimeter Length of the side of a square =

Illustrative Example Ex.: Perimeter of a square field is 16 m. Find the length of

its sides. Soln: Applying the above formula, we have

, . , 1 6 length of sides = = 4 m.

Exercise 1. Find the perimeter of a square field of length of one of its

sides 5 m. a) 25 m b)10m c)20m d) Data inadequate

2 Find the perimeter of a square field of length of one of its sides 6 m. a) 24 m b)12m c)30m d) None of these

3. Perimeter of a square field is 28 m. Find the length of its sides. a) 7 m b)6m c)8m d)5m

4 Perimeter of a square field is 24 m. Find the length of its sides.

a)6m b)5m c )9m d) None of these

Answers 1. c 2. a 3. a 4. a

Rule 23 To find the diagonal of a square whose sides are given.

Length of the diagonal of a square = x side

Dlustrative Example I x : Find the diagonal of a square field whose side is of 10

m length. Soln: Applying the above formula, we have

Length of the diagonal = ^2 x 10= io72 m -

Exercise L Find the diagonal of a square field whose side is of 5 m

length.

a ) 5V2 m D ) I0V2 m c) 10 m d)None ofthese 2. Find the diagonal of a square field whose side is of 4 m

527

length. ' /

a) 4^ /2 m b) 16-\/2 m c ) 16 m d) Data inadequate 3. Find the length of sides of a square field whose diagonal

is 12V2 m.

a) 12 m b) 10 m c) 24 m d) Data inadequate

Answers L a 2.a

3. a; Hint: 12V2 = 72 x side .-. side = 12 m.

Rule 24 To find the diagonal and the perimeter of a square if its area is given.

(i) Length of diagonal of a square = 72 x area

(ii) Perimeter of a square = 7l6x area

Illustrative Example Ex.: Find the length of the diagonal and the perimeter of a

square plot i f its area is 400 square metres. Soln: Applying the above formulae, we have

(i) length of diagonal of the square

= 72 x 7400 = 2072 metres. (ii) perimeter of the square

= 716x400=4x20 = 80 metres.

Exercise 1. In order to fence a square Manish fixed 48 poles. I f the

distance between two poles is 5 metres then what will be the area of the square so formed?

a)2600 cm2 b)2500 cm2

c)3025 cm2 d) None ofthese [BSRB Bangalore PO 2000]

2. The cost of cultivating a square field at the rate of Rs 160 per hectare is Rs 1440. Find the cost of putting a fence around it at the rate of 75 paise per metre. a)Rs900 b)Rs850 c)Rs950 d)Rs940

3. I f the ratio of areas of two squares is 9 : 1, the ratio of their perimeters is: a ) 9 : l b )3 :4 c ) 3 : l d) 1:3

(Asstt. Grade 1990) 4. How long wi 11 a man take to walk round the boundary of

a sq field containing 9 hectares at the rate of 6 km an hour? a)12min b)10min c) 24 min d) Can't be determined

5. The perimeter of a square field is 400 m. What is its area? a) 1 hectare b) 0.845 hectare c) 1.2 hectare d) Can't be determined


6. The area of a square field is 6.25 hectares. How long will it take a man to walk round the outside of it at the rate of

2-j km/hr? (1 hectare = 10000 sq metres)

a)32min b)24min c)28min d)20min 7. Find in km the length of the wire required to go 10 times

round a square field of 6- j hectares.

a)8km b )5km c)10km d)16km 8. Calculate the cost of surrounding with a fence a square

field of 16 hectares at Rs 20 per metre. a) Rs 30000 b)Rs 24000 c)Rs 32000 d)Rs 36000

9. The cost of levelling and turfing a square cricket field at Rs 16000 per hectare is Rs 262440. Find the cost of sur-rounding it with a railing costing Rs 25 per metre. a) Rs 45000 b)Rs 40500 c)Rs 42500 d)Rs 48500

10. How long will it take to run round a square field contain-ing 1681 sq m at the rate of 4 km an hour? a)3min b)2.45min c)3.46min d)2.46min

Answers 1. d; Hint: Perimeter = 48 * 5 = 240 metres

'Perimeter^2 ' 240^

{ 4 J - I 4 J Area = = 60x60 = 3600 sqm [See Rule-24(H)]

( Total cost \ 1440 2. a; Hint: Area= [ R a t e / h e c t a r e j ~ ^

= 9 hectares = 90000 sq m

Perimeter = Vl6x90000 = 1200 m

. 1200x75 . .-. Cost of fencing = = Rs 900

9 3. c; Hint: Required ratio = A | = 3 : 1

[ v Perimeter a 4Area , See Rule - 24 (ii)]

4. a; Hint: Area=9x 10,000 sq m

.-. perimeter= Vl6x9x l0000 =4 x 3 x 100 = 1200m

60x1200 .-. required time = _ , = 12 min

6x1000

5.a;Hint: J\6xArea =400

400x400 Area =

16 10000 sqm = 1 hectare

.-. required time =

25

60 2500

xlOOO =24 min

7. c; Hint: Area = x 10000 = 62500 sq m

Perimeter= Vl6x625000 = 1000 m = 1 km

.-. required length of the wire = 10 x 1 = 10 km

8. c; Hint: Perimeter = Vl6x 16x10000 = 1600 m

.-. required cost = 1600 x 20 = Rs 32000

9.b; Hint: Area = 262440 16000

= 16.4025 hectare = 164025 sq m

Perimeter = ^16x164025 = 1620 m

.-. required cost = 1620 * 25 = Rs 40500

10. d; Hint: Perimeter = -^16x1681 = 164 m

60x164 .-. time required = - , . . . =2.46 min

4x1000

Rule 25 To find the perimeter of a square if its diagonal is given.

Perimeter of the square = (2 V2 x Diagonal)

Illustrative Example Ex.: The diagonal of a square is 10 cm. Find its perimeter

and area.

Soln: Applying the above formulae, we have

Perimeter = 2 ^2 * 10 = 20 V2 cm \ (10)2

Area = 50 sqcm. (SeeRule-21)

6. b; Hint: Area = 6.25 hectare = 6.25 x 10000 = 62500 sq m .'. Diagonal =

Exercise 1. The diagonal of a square is 5 cm. Find its perimeter,

a) 5V2 cm b)6V2"cm c) 10^2 cm d) 1572 cm 2. The diagonal of a square is 6 cm. Find its perimeter.

a ) I2V2 c m D ) 6>/2 cm c)24cm d) Data inadequate

3. Perimeter of a square is 24V2 cm. Find its diagonal,

a) 12 cm b) 12V2 c m c ) 6V2 c m d)8cm

Answers l .c 2.a

3. a; Hint: 24-J2 = 2^2 x Diagonal

24^2

2^2 12 cm.

Perimeter = Vl 6x62500 =4 x 250 = 1000 m

Elementary Mensuration - I 5 2 9

Rule 26 Theorem: If the diagonal of a square becomes x times, then

the area of the square becomes x2 times.

Illustrative Example Ex.: The diagonal of a square increases to its thrice. How

many times will area of the new square become? Soln: Detail Method: Let the diagonal of the original square

bexm.

Original square, Diagonal = x. Area

New square, Diagonal = 3x, Area =

2

N 2

Area of the new square = = 9 times of the x T

original square. Quicker Method: Applying the above theorem, we have

the required answer = (3 ) 2 ~ 9 times.

Exercise 1. The ratio of areas of two squares, one having double its

diagonal then the other is a ) 2 : l b ) 3 : l c)3:2 d) 4 :1

2 The diagonal of a square increases to its twice. How many times will area of the new square become? a) 2 times b) 6 times c) 4 times d) Data inadequate

3. The diagonal of a square increases to its 4 times. How many times will area of the new square become? a) 16 times b) 2 times c) 8 times d) None ofthese

Answers

1. d; Hint: Diagonal a (Area)2 2.c 3.a

Rule 27 Theorem: If the ratio of the areas ofsquare A and square B is a: b, then

(i) the ratio of their sides = 4a : 4b >

(ii) the ratio of their perimeters = 4a :4b and

(iii) the ratio of their diagonals = 4a :4b

Illustrative Example Ex.: Ratio of the areas of the two squares is 16 : 9. Find

(i) the ratio of their sides, (ii) the ratio of their perimeters and

(iii) the ratio of their diagonals. Soln: Detail Method:

Let the sides of the first square be x and the second square be y.

i 16

Ratio of sides = -> y 16

= 4:3

Ratio of perimeters = 4x:4y = x : y = 4:3

Ratio of diagonals = x-fl: yyfl = x : y = 4 :3

Quicker Method: Applying the above theorem, we have

the required ratios = 4 :3 .

Exercise 1. Ratio of the areas of the two squares is 25 : 9. Find the

ratio of their perimeters. a)5:9 b)9:5 c)5:3 d)3:5

2. Ratio of the areas of the two squares is 9 : 4. Find the ratio of their diagonals. a)3:2

Answers l .c

b)3: 1 c)9:2 d)9:

2. a

Rule 28 Theorem: If the perimeter of a square is equal to the perim-

eter of a circle, then the side of the square is 7 I X -2)

and

radius of the circle is 2x

Where, x is the side of the

square and r is the radius of the circle.

Illustrative Examples Ex: 1. There is a square of side 22 cm. Find the radius of the

circle whose perimeter equals the perimeter of the square.

Soln: Applying the above theorem, we have the

radius of the circle = 2 ^ 2 2 = 14 cm. 22 7

Ex: 2. There is a circle of radius 7 cm. Find the side of the square whose perimeter equals the perimeter of the circle.

Soln: Applying the above theorem,,we have

22 7 the side of the square = x = 11 cm.

M 7 2

Exercise 1. There is a square of side 44 cm. Find the radius of the

circle whose perimeter equals the perimeter of the square.


a) 7 cm b) 14 cm c) 28 cm d) Data inadequate 2. There is a square of side 11 cm. Find the radius of the

circle whose perimeter equals the perimeter of the square, a) 7 cm b)21cm c)12cm d)9cm

3. There is a circle is radius 21 cm. Find the side of the square whose perimeter equals the perimeter of the circle, a) 11 cm b) 22 cm c) 33 cm d) Data inadequate

Answers l .c 2.a 3.c

Rule 29 Theorem: If the side of a square is increased by 'x' units and its area becomes 'y' square units, the side of the square

is given by V

units, its area is given by sq units

and its perimeter is given by 41 ^ x units.

Note: I f the side of a square is increased by 'x ' units and its area increases by 'y ' units then the side of the square is

given by 1

units.

Soln:

Illustrative Example Ex.: Length of a square is increased by 8 cm. Its area be-

comes 208 sq cm. Find its perimeter. Detail Method: Let the side of the square be x cm.

As per the question, (x + 8)x = x2 + 208

or, x 2 + 8 x = x 2 + 2 0 8 .-. x = 26 cm .-. Perimeter = 4x = 4 * 26 = 104 cm Quicker Method: Applying the above theorem, we have

the required answer = 4 x 208 8

104 cm

Exercise 1. I f the side of a square be increased by 4 cm, the area

increases by 60 sq cms. The side of the square is: a) 12 cm b)13cm c) 14 cm d) None of these

2. Length of a square is increased by 4 cm. Its area be-comes 44 sq cm. Find the area of the square. a) 11 sqcm b) 44 sqcm c) 121 sqcm d) Data inadequate

3. Length of a square is increased by 9 cm. Its area be-comes 135 sqcm. Find its perimeter. a) 60 cm b)30cm c) 45 cm d) None of these

Answers

1. d; Hint: Required answer = [ t * \~Z = 5.5 cm [See Nc 60

2.c 3.a

Rule 30 Theorem: A square room is surrounded by a verandah the outside of the square room) of width 'd' metres. If area of the verandah is 'A' sq metres, then the area of

sq metres and obviously side of room is A-4dl

4d

square room is given by A-4d2)

4d metres.

Illustrative Example Ex.: A square room is surrounded by a verandah of \

2 metres. Area of the verandah is 64 sq metres, the area of the room.

Soln: Detail Method: Let the side of the room ABCD metres.

Area of the room ABCD = x * x = x2 sq m, Width of the path = 2 metres (given) Sides of the figure A B C D ' =x + 2 + 2

= (x + 4) metres.

Area of the figure A ' B ' C ' D ' = (x + 4)2 sqm.

A' A B

2 m

D ( r

B'

D' C

As per the question, Area of the path = 64 sq metres

or, (x + 4)2-x2 =64

or, x2+\6 + 8x-x2 =64 or, 8x = 48 .. x = 6 metres.

Area = x2 = 6 x 6 = 36 sq metres. Quicker Method: Applying the above theorerr. i have

the area of the room = / 6 4 - 4 x 2 2 " '

4x2


64-16 (6 ) 2 =36 sq metres.

531

Area of the figure A 'B C D ' = (x-4)2 sq m.

Exercise 1.

2.

5.

A square room is surrounded by a verandah of width 3 metres. Area of the verandah is 96 sq metres. Find the area of the room. a) 36 sqm b) 25 sqm c) 49 sq m d) Data inadequate A square room is surrounded by a verandah of width 4 metres. Area of the verandah is 160 sq metres. Find the area of the room. a) 42 sqm b) 49 sqm c) 36 sqm d) None ofthese A square room is surrounded by a verandah of width 2 metres. Area of the verandah is 72 sq metres. Find the area of the room. a) 49 sqm b) 64 sqm c) 81 sqm d) 36 sqm A square room is surrounded by a verandah of width 1 metre. Area of the verandah is 24 sq metres. Find the area of the room. a) 25 sqm b) 16 sqm c) 36 sqm d) 30.25 sqm A path 2m wide running all round a square garden has an area of 9680 sq m. Find the area of the part of the garden enclosed by the path.

a) (1208)2 sq m

c) (2208)2 sq m

b) (1028)2 sqm

d) (1308)2 sqm

Answers l .b 2.c 3. a 4. a 5. a

Rule 31 Theorem: If a square room has a verandah of area 'A' sq metres and width'd' metres all round it on its inside, then

the area of the room is rA + 4d2^

4d sq metres and obvi-

ously side of the square room is given as A + 4d2\

Ad metres.

Illustrative Example Ex.: A square room has a verandah of area 64 sq metres

and width 2 metres all round it on its inside. Find the area of the room.

Soln: Detail Method: Let the side of the room ABCD be x metres.

Area of the room ABCD = x2 sq metres Width of the path = 2 metres (given) Sides ofthe figure A ' B ' C ' D '

= x - (2 + 2) = (x - 4) metres

D C As per the question, Area of the path = 64 sq metres

or, x2 -{x-4)2 =64

or, x2-x2-16 + 8x = 64 or, 8x = 80

.. x = 10 metres. Area of the room = 10 x 10 = 100 sq metres. Quicker Method: Applying the above theorem, Area of the room

64 + 4 x 2 2 l

4x2 80

10x10 = 100 sq m.

Exercise 1. A square field contains 2.89 hectares. It has to be fenced

all-round and a path 10 m wide has to be laid out all-round close to the fence inside. I f the cost of fencing is Rs 50 per m and the cost of preparing the path is Rs 10 per sq metre. Find the total expenses. a) Rs 64000 b)Rs 34000 c)Rs 94000 d)Rs 98000

2. A square room has a verandah of area 96 sq metres and width 3 metres all round it on its inside. Find the area of the room. a) 121 sqm b)132sqm c) 25 sq m d) Data inadequate

3. A square room has a verandah of area 160 sq metres and width 4 metres all round it on its inside. Find the area of the room. a)196sqm b)169sqm c)256sqm d)36sqm

4. A square room has a verandah of area 24 sq metres and width 1 metre all round it on its inside. Find the area of the room. a)49sqm b)25 sqm c) 64 sq m d) Data inadequate

Answers 1. d; Hint: Area of the square = 2.89 hectares = 28900 sq m

Perimeter = ^16x28900 = 680 m [See Rule - 24] .-. Cost of fencing the square field = 680 x 50 = Rs 34000

Now applying the given

^ + 4 x l Q 2

4x10

rule we have

V28900 =170


Or, A = 6400 sq m = Area of the path .-. Cost in preparing the path = 6400 * 10 = Rs 64000 .-. total expenses = Rs 34000 + Rs 64000 = Rs 98000

2. a 3. a 4. a

Area of the four walls of a room

Rule 32 (i) To find the area of thefour walls of a room, if its length, breadth and height are

given. Area of the four walls of a room = 2* (Length + Breadth) x Height Illustrative Example Ex.: A room is 8 metres long, 6 metres broad and 3 metres

high. Find the area of the four walls of the room. Soln: Applying the aboyp formula, we have

Area of the four walls of a room = 2 x (8 + 6) * 3 = 84 sq m.

(it) To find the height of a room, if area offour walls of the room and Its length and breadth are given.

Height = Area of four walls of the room

2(Length + Breadth) metres.

Illustrative Example E X J Area of a hall, whose length is 16 metres and breadth

is half of its length, is 576 sq metres. Find the height of the room.

Soln: Applying the above formula, we have

the height of the room = x = 12 metres. 5 2(16+8)

Exercise 1. A room is 13 metres long, 9 metres broad and 10 metres

high. Find the cost of carpeting the room with a carpet 75 cm boardat the rate of Rs 2.40 per metre. What will be the cost of painting the four walls of the room at Rs 4.65 per sq metre, it being given that the doors and windows occupy 40 sq metres? a) Rs 375.50, Rs 1850 b) Rs 374.40, Rs 1860 c) Rs 376, Rs 1875 d) Rs 374.04, Rs 1806

2. The cost of papering the walls of a room 12 metres long at the rate of 45 paise per square metre is Rs 113.40 and the cost of matting the floor at the rate of 35 paise per square metre is/Rs 37.80. Find the height of the room. a)9m b)8m c)6m d)12m

3. The length and breadth of a room are in the ratio 4:3 and its height is 5.5 metres. The cost of decorating its walls at Rs 6.60 per square metre is Rs 5082. Find the length and breadth of the room. a)40m,30m b)50m,40m c)30m,25m d)40m,20m

4. Area of four walls of a room is 77 m 2 . The length and

breadth of the room are 7.5 m and 3.5 m respectively. The height of the room is: a) 7.7 m b)3.5m c) 6.77 m d)5.4m

5. Area of four walls of a room is 168 m 2 . The breadth and height of the room are 8 m and 6 m respectively. The length of the room is: a) 14m b)12m c)3.5m d)6m

6. The cost of papering four walls of a room is Rs 48. Each one of length, breadth and height of another room is double that of the room. The cost of papering the walls of this new room is: a)Rs96 b)Rsl92 c)Rs384 d)Rs288

7. A hall, whose length is 16 metres and breadth twice its height, takes 168 metres of paper 2 metres wide for its four walls. Find the area of the floor. a) 192sqm b) 196sqm c) 129sqm d) 190sqm

8. Find the cost of painting the walls of a room of 5 metres long, 4 metres broad and 4 metres high at Rs 8.50 per sq metre. a)Rs610 b)Rs216 c)Rs512 d)Rs612

9. The cost of painting the walls of a room 7 metres 6

long, 4 metres wide at Rs 16.20 per sq metre is Rs 1940.40. How high is the room?

a) 4 m 4

10.

11.

12.

13.

14.

b) 4 ! m 4

c) 4 m 3

d) 4- m 3

How many metres of wall paper 2 metres wide will be required for a room 8.3 metres long, 4.2 metres wide and 4 metres high? a)40m b)50m c)45m d)75m The area of the four walls of a room is 5940 sq dm and the length is twice the breadth, the height being 33 dm. Find the area of the ceiling. a) 18 sqm b) 1.8 sqm c) 16 sqm d) 1.6 sqm A rectangular room is 6 m wide and 3 m high. I f the area of its walls is 81 sq m, find the length, a) 6.5 m b)5m c)6m d)7.5m A room is 10.5 metres long and 6.25 metres broad. The cost of papering the walls with paper 1.5 m wide at Rs 24 per metre is Rs 2680. Find the height of the room, a) 5 metres b) 6 metres c) 8 metres d) 10 metres

The length of room is 1^- times its breadth. The cost of

carpeting it at Rs 150 per sq metres is Rs 14400 and the cost of white washing the four walls at Rs 5 per sq metre is Rs 625. Find the length, breadth and height of the room.

, 1 a) 12 m, 8 m, 3 m

8 b) 12^ m , 8^ m ; 3 - m

Elementary Mensuration - I 5 3 3

3 2 2 c) 12 m, 8 m, 3 m d) Data inadequte

15. The length of a room is 6.5 metres. The cost of painting the walls at Rs 56 per sq metre is Rs 4928 and the cost of carpeting the room at Rs 224 per sq metre is Rs 6552. Find the height and width of the room. a)4m,5m b)4.5m,5.5m c)4 m, 4.5 m d) Data inadequate

16. The length of a room is double the breadth. The cost of colouring the ceiling at Rs 25 per sq m is Rs 5000 and the cost of painting the four walls at Rs 240 per sq m is Rs 64800. Find the height of the room. a)4m b)4.5m c)3.5m d)5m

17. Two square rooms, one a metre longer each way than the other, are of equal height, and cost respectively Rs 33600 and Rs 35280 to paper the walls at Rs 70 per sq m. Find the height.

a) 6 m b)8m c)5m d)4m

Answers

1. b; Hint: Cost of carpet f , n 13x9x

= 2.40x

I 75 100

= Rs 374.40

[See Rule-54] Area of four walls = [2 (13 + 9) x 10] = 440 sq m Area to be painted = Rs (440 - 40) = 400 sq m Cost of painting = Rs (400 x 4.65) = Rs 1860

2. c; Hint: Area of floor

Total cost Rate per sqm

( 3780 35

= 108 sqm

.-. Breadth of the room =

= 9 metres Now, area of four walls

Area of floor _H08 12 Length of the room

Total cost of Papering Rate per sq metre

11340 35 = 252 sqm

Let the height of room be h metres. Then,2x(12 + 9 ) x h = 252

252 .-. Height, h 2x21

3. a; Hint: Area of the four walls =

6 metres.

( 5082^1 = 770 sq m 6.6 )

Now,2(4x+3x)x5.5=770 or,x=10 .-. length = 4x = 4x 10 = 40m and breadth

= 3x=3x 10 = 30m. 4.b 5.d

6. b; Hint: Cost of papering [2(1 + b) x h\n2 = Rs 48

.-. cost of papering [2(2/ + 2b) x 2h\n2

ie 4[2(/ + 6 ) x A ] = 4 x 4 8 =RS192

7. a; Hint: 2(16 + 2h)h = 168x2 or, (8 + h)h = 84

or, h2 +8/7-84 = 0 By solving the above equation we get h= 6 and -14 .-. height = 6 m and breadth = 12 m (neglecting negative value of h) .-. area of the floor = 16 x 12 = 192sqm

8.d

9. c; Hint: Area of the four walls = 1940.40

16.20 sqm

47 c ) 19404 1078

2 " T + 5 = " ^ 2 - = ^ S q m

h = 1078

9x2| y + 5 = 4 m. 3 3

2x4(8.3 + 4.2) 10. b; Hint: Required answer = - = 50 m.

11. a; Hint: 2 x 33 (x+20) = 5940 .-. x=30dm = 3mandlength=2x=2 x30 = 60dm = 6m .-. area of the ceiling = 6 x 3 = 18 sqm

12.d;Hint:2x3(x + 6) = 81 15

x = =7.5m 2

13. a; Hint: Length of the paper = 2680

24 m

2680 15 24 10 = sqm

Area of the paper = /

, n t i ' . - ^ 2680 15 Now, as per the question 2 x h(\0.5 + 6.25) = rr-x

167.5 h ~ 1 c ~ - 5 m. 16.75x2

14400 14. a; Hint: Area ofthe room = , e n = 96 sq m

Or. xx x = 96 2

150

x = 8 m = breadth

.-. length = - x 8 = 12m

Area of the four walls

Or,2xh(12 + 8)=125

625 125 sq m

' h " 8 8 m -


15. c; Hint: Area of floor of the room =

6552 Now, 6.5 x x =

224

6552 224

x = 4.5m

4928 Area of the four walls of the room = , = 88 sq m

Or,2xh(6.5+4.5) = 88

56 h = 4m.

16. b 17. a; Hint: Let the side of one square be x m and the other be

(x+ l )m. . . . , 33600 . . .

Now, as per the question, 2n(x + x) - - 480

.-. hx=120 (i)and 35280 r n ,

2(x + l)x2h = = 504 70

.-. xh+h=126 (ii) Putting the value of xh from equ (i) into the equ (ii) _vh = 126-120 = 6m

Parallelogram

Rule 33 Theorem: To find the area of a parallelogram if its Base and Height are given. Area of a parallelogram = Base x Height.

D _C

h (Height)/

Base Illustrative Example Ex.: One side of a parallelogram is 17 cm. The perpendicu-

lar distance between this and the opposite side is 13 cm. Find the area of the parallelogram.

Soln: Here,b= 17cmandh= 13cm Now, applying the above formula, Area of parallelogram = Base * Height = 17 x 13 = 221 cm2.

Exercise 1. Find the area of a parallelogram whose base is 35 metres

and altitude 18 metres. a)630sqm b)650sqm c)730sqm d)660sqm

2. The area of a parallelogram is 338 sq m. I f its altitude is twice the corresponding base, determine the base and the altitude. a)13m,26m b) 14m,28m c) 15m,30m d)12m,24m

3. One side of a parallelogram is 14 cm. Its distance from the opposite side is 16 cm. The area of the parallelogram is:

a) 112c/w2 b)224 cm2 c)56 n cm2 d)210 cm2

4. Find the areas of the following parallelograms (i) Base 26 metres, height 8 metres

a) 208 sqm b) 206 sqm c) 200 sqm d) 205 sqm

(ii) Base 54 metres, height 22 metres a)1288sqm b) 1388sqm c) 1188 sqm d) 1088 sqm

Answrs 1. a 2. a;Hint:338 = xx2x .\x = 26

.-. base =13 metres and the altitude = 26 metres 3. b 4.(i) a (ii)c

Rule 34 Theorem: To find the area of a parallelogram, if the lengths of the two adjacent sides and the length of the diagonal connecting the ends of the two sides are given, (see the figure).

D' b C

A ' B' Where, 'a' and 'b' are the two adjacent sides and 'D' is the diagonal connecting the ends of the two sides.

Area of a parallelogram = 2^s(s - a\s - b\s - D) and S =

a + b + D

Illustrative Example Ex.: The two adjacent sides of a parallelogram are 5 cm

and 4 cm respectively, and i f the respective diagonal is 7 cm then find the area of the parallelogram?

Soln: Required area = 2yjs(s - afc - b\s - D)

Where S = a+b+D 5 + 4 + 7

2^/8(8-5X8-4X8-7)

= 2>/8x3x4 =8^6 =19.6 sqcm.

Exercise 1. Find the area of a parallelogram; i f its two adjacent sides

are 12 cm and 14 cm and i f the diagonal connecting the ends is 18 cm. a) 176.49 sq cm b) 167.49 sq cm c) 167.94 sq cm d) None of these

2. Find the area of a parallelogram wh.ose two adjacent sides are 130 metres and 140 metres and one of the diagonals is 150 metres long. a) 16800 sqm ' b)i7800sqm


c) 18600 sq m d) Can't be determined 3. Find the area of a parallelogram whose two adjacent sides

are 130 metres and 140 metres and one of the diagonals is 150 metres long. Find also the cost of gravelling it at the rate of Rs 10 per square metre, a) 15800sqm,Rs 158000 b) 16800 sqm, Rs 168000 c) 14800 sq m, Rs 148000 d) None of these

Answers he 2. a 3.b

Rule 35 Theorem: In a parallelogram, the sum of the squares of the diagonals = 2* (the sum of the squares of the two adjacent sides)

or, D2+D2 = l(a2+b2)

Where, Dx and D2 are the diagonals and a and b are the

adjacent sides.

Illustrative Example Ex.: In a parallelogram, the lengths of adjacent sides are

12 cm and 14 cm respectively. I f the length of one diagonal is 16 cm, find the length of the other diago-nal.

Soln: In a parallelogram, the sum of the squares of the di-agonals = 2 x (the sum of the squares of the two adjacent sides)

or, D2+D22 = l{a2+b2)

or, 1 6 2 + x 2 =2 ( l2 2 +14 2 )

or, 256 + x2 =2(144+196)

or, x 2 i 680 - 256 = 424 .'. * = V424 = 20.6 c m

Exercise 1. A parallelogram, the lengths of whose sides are 11 cm

and 13 cm has one diagonal 20 cm long. Find the length of another diagonal. a) 15 cm b)18cm c) 20 cm d) Can't be determined

2. A parallelogram, the lengths of whose sides are 11 cm and 8 cm has one diagonal 10 cm long, find the length of the other diagonal. a) 17.78 cm (approx) b) 18.68 cm (approx) c) 17.87 cm (approx) d) Data inadequate

3. In a parallelogram, the lengths of adjacent sides are 24 cm and 28 cm respectively. I f the length of one diagonal is 32 cm, find the length of the other diagonal. a) 41.2 m (approx) b) 31 m (approx) c) 43.2 m (approx) d) None of these

Answers l .c 2. a 3. a

Rule 36 Theorem: To find the sides of a parallelogram if the dis-tance between its opposite sides and the area of the paral-lelogram is given.

h > ^ v { ' o

Here, ABCD is a parallelogram, hx and h2 are the dis-tance between opposite sides, 7' and 'b' are the sides of the parallelogram. 'A' is area of the parallelogram.

A=lhx= bh2

A_ A_ : . l = k andb= ^

Illustrative Example Ex.: A parallelogram has an area of 160 cm2. I f the distance

between its opposite sides are 10 cm and 16 cm. Find the sides of the parallelogram.

Soln: Applying the above formula, we have

1 6 0

Length of the parallelogram = -~r- -1 o cm.

1 6 0 ,n

Breadth of the parallelogram = -rr- H i cm.

Exercise 1. A parallelogram has an area of 150 cm2. I f the distance

between its opposite sides are 15 cm and 25 cm. Find the sides of the parallelogram. a) 10 cm, 6 cm b) 12 cm, 8 cm c) 8 cm, 4 cm d) Data indequate

2. A parallelogram has an area of 144 cm2. I f the distance between its opposite sides are 12 cm and 16 cm. Find the sides of the parallelogram. a) 12 cm, 9 cm b) 10 cm, 6 cm c) 14 cm, 10 cm d) None of these

3. A parallelogram has an area of 196 cm2. I f the distance between its opposite sides are 7 cm and 14 cm. Find the sides of the parallelogram. a) 28 cm, 14 cm b) 14 cm, 7 cm c) 28 cm, 21 cm d) Data inadequate

Answers l .a 2. a 3. a


Rhombus

Rule 37 Theorem: To find perimeter of a rhombus if the length of the two diagonals are given.

Perimeter of the rhombus = ^2^/(rf,2 + 2^)j units.

Where, dx and d2 are the two diagonals.

Illustrative Example Ex.: In a rhombus, the length of the two diagonals are 40

metres and 30 metres respectively. Find its perimeter. Soln: Applying the above formula, we have

perimeter of the rhombus

= 2A/(40) 2 +(30) 2 = 2^2500 = 50x2 = 100m.

Exercise 1. In a rhombus, the length of the two diagonals are 3 metres

and 4 metres respectively. Find its perimeter. a) 14m b)10m c)5m d)7m

2. In a rhombus, the length of the two diagonals are 12 metres and 16 metres respectively. Find its perimeter. a)20m b)40m c)25m d)45m

3. In a rhombus, the length of the two diagonals are 18 metres and 24 metres respectively. Find its perimeter. a)30m b)45m c)60m d)55m

Answers l .b , 2.b 3.c

Rule 38 Theorem: To find area of a rhombus If the side and the height are given.

Area of the rhombus = (side x height) sq units.

Illustrative Example Ex.: The side and the height of a rhombus are 14 cm arid

30 cm respectively. Find its area. Soln: Applying the above formula, we have

area of the rhombus = 14 cm * 30 cm = 420 sq cm. Exercise

I f a square and a rhombus stand on the same base, then the ratio of areas of square and rhombus is: a) greater than 1 b) equal to 1

1 1 c) equal to d) equal to

The side and the height of a rhombus are 15 cm and 25 cm respectively. Find its area, a) 325 sqcm b) 375 sqcm c) 345 sq cm d) None of these The side and the height of a rhombus are 20 cm and 30 cm respectively. Find its area, a) 900 sq cm b) 600 sq cm

1.

2

c) 625 sq cm d) Data inadequate 4. The side and the height of a rhombus are 12 cm and 18

cm respectively. Find its area. a)216sqcm b)261sqcm c) 316 sq cm d) Data inadequate

Answers Area of square _ base x base

1. a, Hint. ^ r e a Q^ r n o m | j U S base x height

2.b

axa a = > 1, since a > h axn n

3.b 4. a

Rule 39 Theorem: To find the side and one of the diagonals of a rhombus if area and one of Us diagonals are given.

2A (i) Diagonal of the rhombus [d2) = ^

if 1 ,2 (ii) Side of the rhombus -

Where, A = area of the rhombus

dx = length of the one diagonal

d2 = length of the other diagonal.

Illustrative Example Ex.: A rhombus of area 24 sq cm has one of its diagonals

of 6 cm. Find the other diagonal and side of the rhom-bus.

Soln: Detail Method: Area = 24 sq cm Length of the diagonal = dx =6 cm

Area = - (Product of its diagonals) = ^-xdxxd2

d2 = 2 Area 24 x 2

= 8 cm

cm Side= -4dx2+d22 = - W 8 2 + 6 2 =5 2 ' 2 Quicker Method: Applying the above formula, we have

2x24 . (0 Diagonal of the rhombus = ~ = cm

1

(ii) Side of the rhombus = 2

10

4x24x24 36.+

6x6

cm


Exercise 1. A rhombus ofarea 6 sqm has one ofits diagonals of 3 m.

Find the other diagonal and side of the rhombus. a) 4 m, 5 m b) 6 m, 8 m c) 4 m, 2.5 m d) Data inadequate

2. A rhombus of area 96 sq m has one of its diagonals of 12 m. Find the other diagonal and side of the rhombus. a) 16m, 10m b)8m,20m c)16m,20m d)8m, 10m

3. A rhombus of area 216 sq cm has one of its diagonals of 24 cm. Find the other diagonal and side of the rhombus. a)18cm,30cm b) 18cm, 15cm c) 9 cm, 15 cm d) Data inadequate

Answers l .c 2.a 3.b

Rule 40 Theorem: If one of the diagonals of a rhombus of side 'x' units measures'd' units, then the area of the rhombus is

r \ r n 2 \x - \ 1 UJ J

given by d

other diagonal is 2 x

sq units and the length of the

( 1 \ 2 2

i f * -V

units.

Illustrative Example Ex.: One of the diagonals of a rhombus of side 5 cm mea-

sures 8 cm. Find the area of the rhombus. Soln: Detail Method: We know that the diagonals of a rhom-

bus bisect each other at right angle. From the figure we can write for right angled triangle,

5 cm r

d 2 \ > ' 8 cm

D

= V25-16 =3 cm

.-. d2 = 3x2 = 6 cm

Area of the rhombus

= xd, xd1 =x8x6 = 24 cm 2 2

Quicker Method: Applying the above theorem,

f i - - \

Area= 8 = 8 x ^ 9 =8x3 = 24 sq cm

Note: Expression of the above theorem can be written as follows, (i) Area of the rhombus

'Perimeter^ ( d^2 4 J {2

(ii) Length of the other diagonal

sq units.

= 2 ( Perimeter^ ( d

I * J ~{~2) units. " I f perimeter and one of the diagonals of a rhombus are given, then the area and the length of the other diagonal can be calculated."

Exercise 1. Find the area of a rhombus one side of which measures

20 cm and one of whose diagonals is 24 cin. a) 384 sqcm b) 348 sqcm c) 484 sq cm d) Can't be determined

2. One side of a rhombus is 10 cm and one of its diagonals is 12 cm. The area of the rhombus is:

a) 120 cm1

c) 80 cm2

Answers

b)60 cm1

d)96 cm2

24V T j

= 16x24 1. a; Hint: Required answer = 24 ^|20 -

= 384 sq cm 2. d

Rule 41 To find the area of a rhombus if its diagonals are given.

A

Area of a rhombus

1 x D, x D2 = ^-(Product of diagonals)

Illustrative Example Ex.: Find the area of a rhombus one of whose diagonals

measures 8 cm and the other 10 cm.

1 8x10 Soln: Area = (product of diagonals) = r = 40 sq cm.

Exercise 1. I f the perimeter of a rhombus is 4a and lengths of the

diagonals are x and y, then its area is: 111

d ) - x y a)a(x + y) b)x2+yr c)xy

(NDA1990) 2. In a rhombus whose area is 144 sq cm one of its diago-

nals is twice as long as the other. The lengths of its diagonals are:

a) 24 cm, 48 cm b) 12 cm, 24 cm

c) c m > 12V2 c m d)6cm, 12 cm (CDS 1989)

3. Find the area of a rhombus one of whose diagonals mea-sures 6 cm and the other 12 cm. a) 36 sq cm b) 24 sq cm c) 20 sq cm d) None of these

4. Find the area of a rhombus one of whose diagonals mea-sures 8 cm and the other 18 cm. a) 42 sq cm b) 72 sq cm c) 52 sq cm d) Data inadequate

Answers 1. d; 2. b; Hint: Let its diagonals be x cm and 2x cm. Then

1 , - x x x 2 x = 144=>x 2 =144 or,x=12 2

3.a Lengths of diagonals are 12 cm, 24 cm

4.b

Rule 42 To find the sides of the rhombus if its two diagonals are given.

Side of rhombus = ^x^D2 +D22 ; Where, D, and D2

are the two diagonals.

Illustrative Example Ex.: Find the side of a rhombus one of whose diagonals

measures 6 cm and the other 8 cm. Soln: Applying the above formula, we have

side= X- x V(6)2 + (8) 2 = ~ x ^(36 + 64) = 5 cm.

Exercise 1. Find the side of a rhombus one of whose diagonals mea-

sures 4 cm and the other 3 cm.

PRACTICE BOOK ON QUICKER MATHS

a) 6 cm b) 3.5 cm c) 2.5 cm d)5cm Find the side of a rhombus one of whose diagonals mea-sures 12 cmjand the other 16 cm. a) 10cm I b)20cm c)12cm d)15cm Find the side of a rhombus one of whose diagonals mea-sures 18 cm and the other 24 cm. a) 20 cm b)30cm c)15cm d) Data inadequate

Answers l .c Trapezium

2. a 3.c

Rule 43 To find the area of a trapezium, when length of parallel sides and the perpendicular distance between them is given.

1 Area of a trapezium = (sum ofparallel sides x perpen-

1 dicular distance between the parallel sides) = (a + b)h; where a and b are the parallel sides of the trapezium and h is the perpendicular distance between the sides a and b.

Illustrative Example Ex.: A trapezium has the perpendicular distance between

the two parallel sides 60 m. I f the lengths of the paral-lel sides be 40 m and 130 m, then find the area of the trapezium. ^

Soln: Applying the above formula,

Area of a trapezium = ^ (l 30 + 40)60 = 85 x 60 = 5100

sqm.

Exercise 1. The cost of ploughing trapezoid field at the rate of Rs

1.35 per square metre is Rs 421.20. The difference be-tween the parallel sides is 8 metres and the perpendicu-lar distance between them is 24 metres. Find the length of parallel sides. a)17m,9m b)28m,20m c) 34 m, 26 m d) Can't be determined

2. The two parallel sides of a trapezium are 1 m and 2 m respectively. The perpendicular distance between them is 6 m. The area of trapezium is:

a ) 9 m 2 b)\2m2 c)6 m2 d)18/w 2 3. The cross section of a canal is a trapezium in shape. I f

the canal is 10 m wide at the top and 6 m wide at the bottom and the area of cross section is 640 m2, the length of canal is: a)40m b)80m c)160ri1 d)384m

4. The area of a trapezium is 384 sq cm. I f its parallel sides are in the ratio 3 : 5 and the perpendicular distance be-tween them be 12 cm, the smaller of parallel sides is: a) 16 cm b)24cm c)32cm c)40cm


Answers 1. a; Hint: Let the length of parallel sides be x and y.

421.20 Area = = 312 sq m

Now, applying the given rule,

312 = - ( x + v)x24 2

.-. x + y = 2 6 (i)

and x - y = 8 (given) (ii) By solving equ (i) and equ (ii), we get

x = 17mandy = 9m. 2. a 3. b; Hint: let the length of canal be x m.

l . . , . n 640x2 Then, - (10 + 6 )x* = 640=>;r = = 80m.

2 16

1 3 84 x 2 4. b;Hint: - ( 3 s + 5*)xl2 = 384 =>x = =s

2. 8x12 .-. Smaller side = 24 cm.

Rule 44 Theorem: To find the area of a trapezium, when the lengths of parallel sides and non-parallel sides are given.

Area of a trapezium = ^(s-kfe-cfc-d)

where, k = (a - b) ie the difference between the parallel sides and c and d are the two non-parallel sidec of the

k + c + d trapezium. And s =

Illustrative Example Ex.: In a trapezium, parallel sides are 60 and 90 cms re-

spectively and non-parallel sides are 40 and 50 cms respectively. Find its area.

Soln: k = difference between the parallel sides = 90-60 = 30 cm

Let c be 40 cm then d = 50 cm

k + c + d 30 + 40 + 50. 120 , Now,s= = - = = 60 ciri

Area a + b

k

60 + 90

s(s - k\s - c)[s - d)

^60(60 - 30X60 - 40X60 - 50) 30

= 5V60x30x20xl0 = 5 x 600 = 3000 sq cm.

Exercise 1. A field is in the form of a trapezium whose parallel sides

are 120 metres and 75 metres and the non-parallel sides

are 105 metres and 72 metres. Find the cost of ploughing the field at the rate of 60 paise per square metres. a)Rs3404 b)Rs3440 c)Rs3574 d)Rs3414

2 The two parallel sides of a trapezium measure 58 metres and 42 metres respectively. The other two sides are equal, each being 17 metres. Find its area. a) 570 sqm b) 750 sqm c) 740 sq m d) 760 sq m

3. Find the area of a trapezium whose parallel sides are 11 metres and 25 metres long and the non-parallel sides are 15 metres and 13 metres long respectively. a)216sqm b)316sqm c)215sqm d)

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