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Class-IX CBSE
Latest Pattern Sample Paper {Mathematics}
Term-I Examination (SA I)
Time: 3hours Max. Marks: 90
Section-A
1. Let x be a rational number and y be an irrational number. It x + y necessarily an
irrational number?
Solution:
Yes, it is necessarily that sum of rational and irrational numbers is an irrational.
2. If an isosceles right angled triangle has an area of 12 cm2. Then find the length of its
base.
Solution:
Let equal sides of an isosceles triangle be x cm i.e. AB-BC= x cm.
Given, area of an isosceles triangle is 12 cm2.
General Instructions
(i) All questions are compulsory.
(ii) The question paper consists of 31 questions divided into 4 sections A, B, C
and D. Section A comprises of 4 questions of 1 mark each, Section B comprises of
6 questions of 2 marks each, Section C comprises of 10 questions of 3 marks
each and Section D comprises of 11 questions of 4 marks each.
(iii) There is no overall choice. However, internal choice has been provided in 1
question of 2 marks, 3 questions of 3 marks each and 2 questions of 4 marks
each. You have to attempt only one of the alternatives in all such questions.
(iv)Use of calculator is not permitted.
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Area = - x Base × Altitude
12
12 x x2
224 x
x 24 [taking positive square root]
Hence, length of the base of a triangle is 24 cm.
3. Find value of the polynomial p{x) = 5x + 3 – 4x2 at x = 2.
Solution:
Given, p(x) = 5x + 3 – 4x2
At x=2, then p(2) = 5 × 2 + 3 – 4 × (2)2
= 10 + 3– 4 ×4
= 13 – 16 = – 3
Hence, the value of p(x) is – 3 at x = 2.
4. Euclid divided his book 'Elements' into how many chapters?
Solution:
Euclid divided his book 'Elements' into 13 chapters.
Section-B
5. The polynomials kx3 + 3x2 – 8 and 3x3 – 5x + k are divided by x + 2. If the remainder in
each case is the same, then find the value of k.
Solution:
Let p(x) = Kx3+ 3x2 – 8
and q(x) = 3x3 – 5x + k.
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When we divided p(x) and q(x) by x + 2, we get the remainder p(–2) and q(–2).
But according to the question, p(–2) = q(–2)
(–2)3 + 3(–2)2 – 8 = 3(–2)3 – 5(–2) + k
– 8k +12 – 8 = – 24 +10 + k
8k + 4 = –14 + k
–9k = – 18
k =2
6. The angles of a triangle are (x – 40)°), (x – 20)° and 0
1x 10
2
, find the angles of
triangle.
Or
In the given figure, AOC and BOC form a linear pair and a – b = 70°, find the values of
a and b.
Solution:
Let angles of a triangle be
A = (x – 40)°, S = (x – 20)°
and 0
1C x 10
2
We know that, A + B + C = 180°
[by angle sum property of a triangle]
x – 40° + x – 20° + – x –10° = 180°
0 012x x 70 180
2
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012x x 250
2
04x x250
2
0250 2
x5
x = 1000
A = x – 40° = 100° - 40° = 60°
B = x – 20° = 100° – 20° = 80°
and C = 1
2 x – 10° =
1
2 × 100° – 10°= 50°-10° = 40°
Or
According to the question,
AOC + BOC = 180° [linear pair axiom]
a +b = 180° ...(i)
and a – b = 70° ...(ii) [given]
On adding Eqs. (i) and (ii), we get
a + b =180°
0
0
a b 70
2a 250
0
0250a 125
2
On putting the value of a in Eq. (i), we get
125° + b = 180°
b = 180° – 125°
b = 55°
7. Express 15.712 in the form p/q.
Solution:
Let x = 15.712
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x = 15.71212...
On multiplying both sides by 10, we get
10x = 157.1212 ...(i)
On multiplying both sides by 100, we get
1000x = 157121212 ...(ii)
On subtracting Eq (i) from Eq (ii), we get
1000x – 10x = 15712.1212...– 1571212...
990x =15555
15555
x990
3111
x198
[dividing numerator and denominator by 5]
8. Two points with coordinates (4, 3) and (4, – 2) lie on a line, parallel to which axis?
Solution:
As x-coordinate of both points is 4.
So, both points lie on the line x = 4 which is parallel to y-axis.
9. Evaluate 103 × 107, without multiplying directly.
Solution:
103 × 107=(100 + 3)(100 + 7)
= (100)2 +100 × 7 + 3 ×100 + 3 × 7
= 10000+ 100(7+3)+21
= 10000 + 1000 + 21
= 11021
10. Find the factors of polynomial
4x2 + y2+4xy + 8x + 4y + 4
Solution:
4x2 + y2 + 4xy + 8x+4y+4
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= (2x)2 + (y)2 + 2(2x)(y) +2(2x)(2) + 2(2)(y) + 22
= (2x)2+(y)2+(2)2 + 2(2x)(y) + 2(2)(y) + 2(2x)(2)
= (2x + y + 2)2
[ (a + b + c)2 =a2 + b2 +c2 + 2ab + 2bc + 2ca]
Section-C
11. Evaluate
(i) (104)3 (ii) (999)3
Solution:
(i) We have, (104)3 =(100 + 4)3
= (100)3 + (4)3 + 3 × 100 × 4(100 + 4)
[ (a + B)3 = a3 + b3 +3ab (a + b)]
= 1000000 + 64 + 1200 × 104
= 1000000 + 64 + 124800
= 1124864
(ii) We have, (999)3 = (1000 – 1)3
= (1000)3 – (1)3 – 3 × 1000 × 1 × (1000–1)
[ (a – b)3 = a3 – (a – b)3 = a3 – b3 (a – b)]
= 1000000000 –1 – 3000 × 999
= 1000000000 – 1 – 2997000
= 997002999
12. Find the remainder when p(x) = x3 + 3x2 + 3x +1 is divided by 5 + 2x.
Or
Simplify
(x + y + z)2 – (x – y + z)2.
Solution:
By remainder theorem, when p(x) is divided by
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5 + 2x = 25
x2
, then remainder is given by 5
p2
.
Now, p(x) = x3+3x2+3x+1
3 2
5 5 5 5p 3 3 1
2 2 2 2
= 125 25 15
3 18 4 2
= 125 75 15
18 4 2
= 125 150 60 8 27
8 8
Hence, the required remainder is 27
8 .
Or
(x + y + z)2 – (x – y + z)2
= [(x + y +z)2 – {x + (–y) + z}2]
= (x2 +y2 +z2 + 2xy + 2yz + 2zx) – [x2 + (–y)2 + (z)2 + 2(x) – (–y) + 2(–y) (z) + 2zx]
[ (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca]
= (x2 + y2 + z2 + 2xy + 2yz + 2zx)
– (x2 + y2 + z2 – 2xy – 2yz + 2zx)
= x2 + y2 + z2 + 2xy + 2yz + 2zx – x2 – y2 – z2 + 2xy + 2yz – 2zx
= 4xy + 4yz
= 4y(x + z)
13. Evaluate
15
10 20 40 5 80 .
take
5 = 2.236 and 10 = 3.162)
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Or
If x =9 4 5 , then find the value of 1
xx
.
Solution:
Now, 10 20 40 5 80
= 10 2 5 2 10 5 4 5
= 3 10 3 5 3( 10 5)
15
10 20 40 5 80
= 15 5
3( 10 5) 10 5
= 15 10 5
10 5 10 5
[multiplying numerator and denominator by 10 50 ]
= 2 2
5( 10 5) 5( 10 5)
10 5( 10) ( 5)
[ a2 – b2 = (a – b) (a + b)]
= 5( 10 5)
5
= 10 5 3.162 2.236 10 3.162
and 5 2.236
= 5.398
Or
Given x = 9 – 4 5
1 1
x 9 4 5
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9
= 1 9 4 5
9 4 5 9 4 5
[multiplying numerator and denominator by 9 + 4 5 ]
= 9 4 5
9 4 581 80
[ 2 2a b a b a b ]
and 1
x 9 4 5 9 4 5 18x
..(i)
Now,
21 1 1
x x 2 x.xx x
[ (a – b)2 = a2 + b2 – 2ab]
= 1
x 2x
= 18 – 2 = 16 [from Eq. (i)]
1
x x 4x
14. Simplify
7/2 5/21 2 2 3
2 4 3 5
2 3 2 3
2 3 2 3
.
Or
If 2x = 3y = 6–z, then prove that 1 1 1
0.x y z
Solution:
7/2 5/21 2 2 3
2 4 3 5
2 3 2 3
2 3 2 3
=
7/22 4 3 5
2 1 3 2
3 3 3 3
2 2 2 2
m
m
1a
a
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=
7/2 5/26 8
3 5
3 3
2 2
[ m n m na a a ]
=
7/2 5/26 5
7/2 83
3 2
32
[ m ma 1/a ]
= 6 7/2 5 5/2
3 7/2 8 5/2
3 2
2 3
nm mn[ a a ]
= 21 25/2
21/2 20
3 2
2 3
= 21 20 25/2 21/2
3 2
m
m n
n
aa
a
= 1 2
3 2 3 4 12
To prove, 1 1 1
0x y z
Let 2x = 3y = 6-z = k
Thus, 2= k1/x, 3 = k1/y = and 6 = k–1/z
Now, 2 × 3 = 6
1/x 1/y 1/zk k k
[put 2 = 1/x 1/y 1/zk , 3 k , 6 k ]
1/x 1/y 1/z m n m nk k [ a a a ]
On comparing the exponent, we get
1 1 1
x y z
1 1 1
0x y z
15. In the given figure, O is the mid-point of each of the line segments AB and CD. Prove that
AC = BD and AC BD.
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Solution:
Given In the figure, AO = OB, OC = OD
To prove AC = BD and AC BD
Proof In AOC and BOD, we have AO = BO
[ O is the mid-point of AB]
AOC = BOD [vertical opposite angles]
and CO = DO [ O is the mid-point of CD]
AOC BOD [by SAS congruence rule]
Then, AC = BD [by CPCT]
and CAO = DBO [by CPCT]
CAB = DBA ...(i)
But CAB and DBA are alternate interior angles formed when transversal AB
intersects CA at A and DB at B.
AC BD
Hence, AC = BD and AC BD. Hence proved.
16. Write down Euclid's five postulates.
Solution:
First A straight line may be drawn from any one point to any other point.
Second A terminated line can be produced indefinitely.
Third A circle can be drawn with any centre and any radius.
Fourth All right angles are equal to one another.
Fifth For every line L and for every point P not lying on L, there exists a unique line M
passing through P and parallel to L.
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17. In the given figure, if ,AB CD, EFCD and GED = 126°, then find AGE, GEF and
FGE.
Solution:
Given, AB CD, GE is a transversal line, then
AGE = GED
AGE =126° [alternate interior angles]
Now, GEF = GED – FED
GEF = 126° – 90° [ FED = 90°]
GEF =36°
Also, FGE + AGE = 180° [linear pair axiom]
FGE +126° =180°
FGE = 180°– 126° = 54°
Hence, AGE =126°, GEF =36°
and FGE = 54°.
18. Prove that two distinct lines cannot have more than one point in common.
Solution:
Given Two distinct lines l1 and l2
To prove Lines l1 and l2 have only one point in common.
Proof Suppose, lines l1 and 12 intersects at two distinct points, say P and Q.
Then, line l1 contains points P and Q.
Also, line 12 contains points P and Q.
So, two lines l1 and l2 pass through two distinct points P and Q.
But only one (unique) line can pass through two distinct points.
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So, our assumption that two lines can pass through two distinct points is wrong.
Hence, two distinct lines cannot have more than one point in common.
19. If the sides of a triangle are produced in order, then prove that the sum of the exterior
angles so formed is equal to four right angles.
Let ABC be a triangle whose sides AB, BC and CA are produced in order, forming
exterior CBF, ACD and BAE.
In ABC, we have
CBF =1 + 3 ...(i)
[exterior angle is equal to the sum of opposite interior angles]
Similarly, ACD = 1 + 2 ...(ii)
and BAE = 2 + 3 ...(iii)
On adding Eqs. (i), (ii) and (iii), we get CBF + ACD + BAE =2[1 + 2 +3]
= 2 × 180° = 4 × 90°
[by angle sum property of a triangle is 180°]
CBF + ACD + BAE = 4 right angles
Thus, if the sides of a triangle are produced in order, then the sum of exterior angles so
formed is equal to four right angles.
20. Suppose E and F are the mid-points of the sides AB and AC of ABC. CE and BF are
produced to X and Y respectively, so that EX = CE and FY = BF.AX and AY are joined. Find
in figure, a triangle congruent to AEX and demonstrate the congruency. Prove that XAY
is a straight Line.
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Given E and F are the mid-points of the sides AB and AC of AABC, CE and BF are
produced to X and Y, respectively such that EX = CE and FY = BF.
To prove (i) AEX = BFC
(ii) XAY is a straight line.
Construction Join AX and AY.
Proof In AEX and BEC, we have
AE = BE
[ E is the mid-point of AB]
AEX = BEC [vertically opposite angles]
and EX = EC [given]
AEX BEC [by SAS congruence rule]
XAE = CBE [by CPCT]
or XAB = CBA
[ XAE = XAB and CBE = CBA]
But XAB and CBA are alternate interior angles formed when a transversal AB meets
XA at A and BC at B.
XA BC ...(i)
Similarly, it can be proved that
AFY CFB and AY BC ...(ii)
From Eqs. (i) and (ii), we get
BC XA and BC Ay
Hence, XAY is a straight line. Hence proved.
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Section-D
21. In the given figure, if AB = AC, then prove that AF > AE.
In the given figure, B < A and C < D. Prove that AD< BC.
Solution:
Given in the Figure, AB = AC
To prove AF > AE
Proof hi ABC, we have
AC = AB [given]
1 = 2 ...(i)
[angles opposite to equal sides are equal]
In DBE, whose side BE is extended to A, we have
5 > 1 ...(ii)
[exterior angle > each opposite interior angle]
From Eqs. (i) and (ii), we get
5 > 2 ...(iii)
Now, 3 = 4 ...(iv)
[vertical opposite angles]
Considering FDC, whose side DC is extended to B, we have
2 > 3 ...(v)
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From Eqs. (iv) and (v), we get
2 >4 ...(vi)
From Eqs. (iii) and (vi), we get
5 > 4
AF > AE
Or
[sides opposite to greater angle is longer]
Hence, AF > AE
Or
Given In the figure B < A and C < D
To prove AD < BC
Proof In ABO, we have
B < A [given]
AO < BO . ..(i)
[sides opposite to smaller angle is shorter]
In COD, we have
C < D
OD < OC ...(ii)
[sides opposite to smaller angle is shorter]
On adding Eqs. (i) and (ii), we get
AO + OD < BO + OC
AD < BC
Hence, AD < BC
22. Draw the quadrilateral with vertices (– 4,4), (–6,0), (–4, –4)and (–2, 0). Name the type of
quadrilateral and find its area.
Solution:
Firstly, plot the points A(–4,4), B(–6,0),C(–4, –4) and D(–2,0) on a graph paper and join
all these points.
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We obtained quadrilateral is a rhombus because its all sides are equal.
i.e., AB = BC = CD = DA
and diagonals are not equal.
Now, area of a rhombus = 1 2
1d d
2 1
2
d 4
and d 8
= 1
4 82
= 16 sq units
Hence, are a of quadrilateral is 16 sq units.
23. If x = 1 1
2 22 5 2 5 and 1
2y 2 5 2 5 , then evaluate x2 + y2.
Solution:
Given, 1/2 1/2
x 2 5 2 5
and 1/2 1/2
y 2 5 2 5
Now, 2 1/2 1/2x y [(2x 5) (2 5) + 1/2 2(2 5) (2 5)]
= 1/2 2[2(2 5) ]
= 1
224[2 5]
n
m mn[ a a ]
= 4(2 5) 8 4 5
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and 1/2 1/22xy 2[(2 5) (2 5) ] × 1/2 1/2[(2 5) (2 5) ]
= 1/2 2 1/2 22[{(2 5) } {(2 5) } ]
[ 2 2a b a b a b ]
= 2[2 5 2 5] 2 4
= 8
2
x y 2xy 8 4 5 8
2 2x y 2xy 2xy 4 5
[ (a + b)2 = a2 + b2 + 2ab]
x2 + y2 = 4 5
24. In a class, a teacher conducted a small quiz to solve a question on blackboard. She needs
two students and a prize will be given to the students who solve the question first. For
this purpose she choose a boy and a girl. The problem is that in the given figure, AB CD.
Find the values of x, y and z.
Which of these values is depicted by the teacher in this question?
(i) Social value
(ii) Freedom
(iii) Truth value
(iv) Gender equality
Solution:
In ACO, AO AC [given]
Then, ACO = AOC
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[angles opposite to equal sides are equal]
In ACO + AOC + CAO =180° [ by angle sum property of a triangle is 180° ]
ACO + ACO + 74°=180°
[ AOC = ACO]
ACO =180° – 74°
ACO = 106°
0
0106ACO 53
2
i.e., ACO = AOC = 53°
Again, AOC + COP = 180°
[linear pair axiom]
53° + x = 180° [ AOC = 53°]
x = 180° – 53°
x = 127° = POC
In A= POC, POC + OCP + CPO = 1800
[by angle sum property of a triangle is 1800]
127°+15°+ y =180°
142° + y =180°
y = 180° – 142°
= 380
Since, AB CD and AP is a transversal line.
Then, y=z [alternate interior angles]
z = 38°
Hence, x = 127°, y = 38° and z = 38°
Value depicted by the teacher is gender equality.
25. If in two right triangles, the hypotenuse and one side of one triangle are equal to the
hypotenuse and one side of the other triangle, then prove that the two triangles are
congruent.
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Solution:
Given Two right angled ABC and DEF in which B = E = 90°, AC = DF and
BC = EF
To prove ABC DEF
Construction Produce DE to G such that GE = AB.
Join GF.
Proof In ABC and GEF, we have
BC = EF [given]
ABC = GEF = 90°
[by construction]
AB = GE [by construction]
ABC GEF
[by SAS congruence rule]
Then, A = G [by CPCT ]
and AC = OF ...(i) [by CPCT]
But AC = DP [given]
GF = DF
D = G
[angles opposite to equal sides are equal]
A = D
Now, in ABC and DEF,
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A = D [ from Eq. (i)]
and B = E [given]
Remaining C = Remaining F
Now, in ABC and DEF, we have
BC = EF [given]
C = F
and AC = DE [given]
ABC DEF
[by SAS congruence rule] Hence proved.
26. Prove that (a + b + c)3 – a3 – b3 – c3 = 3(a + b)(b + c)(c + a)
Solution:
To prove,
(a + b + c)3 –a3 –b3 –c3 = 3(a + b)(b + c)(c + a)
LHS = [(a + b + c)– a3]– [b3 + c3]
= [(a + b + c) – a][(a + b + c)2 + a(a + b + c) + a2] – (b + c) (b2 + c2 – bc]
[ x3 – y3 = (x – y) (x2 + y2 + xy)
and x3 + y3 =(x + y)(x2 +y 2–xy]
= (b + c)[a2 + b2 +c2 + 2ab + 2bc +2ca + a2 + ab + ac + a2] –[(b + c) (b2 + c2 – bc)
= (b + c)[3a2 + b2 +c2 + 3ab +2bc +3ca – b2 – c2 + bc]
= (b + c) [3a2 + 3ab + 3bc + 3ca]
= 3(b + c) [a2 + ab + bc + ca]
= 3(b + c) [a(a + b) + c(b + a)]
= 3(a + b) (b + c) (c + a) = RHS Hence proved.
27. Simplify
3/7 1/4 1/3
2
4 5 2.
2187 256 1331
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Solution:
3/7 1/4 1/3
2
4 5 2
2187 256 1331
=
3/7 1/4 37 4
4 5 2
[ 11 ]2 1/33 4
=
3 1 2
4 5 2
3 4 11 [
nm mna a ]
= 3 1 2
4 3 5 4 2 11 m
m
1a
a
= 4 × 27 – 5 × 4 + 2 × 121
= 108 – 20 + 242
= 350 – 20 = 330
28. Factroise 3 2 2 31 1a a b ab b
3 27 .
Factorise 3
3
1x 14.
x
Solution:
3 2 2 31 1a a b ab b
3 27
= 2 3
3 2 1 1 1a 3 a b 3 a b b
3 3 3
= 3
1a b
3
[ (a – b)3 = a3 – b3 – 3ab (a – b)]
= 1 1 1
a b a b a b3 3 3
Hence, 3 2 2 31 1a a b ab b
3 27 =
1 1 1a b a b a b
3 3 3
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OR
3 3
3 3
1 1x 14 x 8 6
x x =
333 1 1
x 2 3 x 2x x
= 2
221 1 1 1x 2 x 2 x 2 2 x
x x x x
[ x3 + y3 + z3 – 3xyz = (x + y + z)
(x2 + y2 + z2 – xy – yz – zx)]
= 2
2
1 1 2x 2 x 4 1 2x
x x x
= 2
2
1 1 2x 2 x 5 2x
x x x
29. Find the area of the cyclic quadrilateral AB CD by using Brahmagupta's formula, in
which AB = 9cm, BC = 12 cm, CD = 12 cm and DA = 15 cm.
Solution:
According to the given informations, the rough sketch of the cyclic quadrilateral ABCD
will be as shown alongside.
Let a = AB = 9cm, t = BC = 12cm,
C = CD = 12cm and d = AD=15cm
a b c d
s2
= 9 12 12 15
2
=
48
2 = 24 cm
Area of cyclic quadrilateral
= s a s b s c s d [by Brahmagupta’s formula]
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= 24 9 24 12 24 12 24 15
= 15 12 12 9 = 3 5 12 12 3 3
= 12 3 3 5 = 236 15 cm
Hence, area of cyclic quadrilateral is 236 15 cm .
30. What is the maximum number of digits in the repeating block of digits in the decimal
expansion of 1
17? Perform the division to determine your answer.
Solution:
By long division method, we get
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0.0588235294117647058
17 100
85
150
136
140
136
40
34
60
51
90
85
50
34
160
153
70
68
20
17
30
17
130
119
110
102
80
68
120
119
100
85
150
136
14
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The remainders start repeating after 16 divisions.
10.0588235294117647
17
Hence, the maximum number of digits in the repeating block of digits in the decimal
expansion of 1
17 is 16.
31. A point O is taken inside an equilateral four sided figure ABCD such that its distances
from the angular points D and B are equal. Show that AO and OC are in the same straight
line.
Solution:
Let AB CD and transversal XY cuts AB and CD at P and Q, respectively. Let PR and QR be
the bisectors of BPQ and DQP respectively, meet at R.
AB CD and XPQY is the transversal. Then, BPQ + DQP = 180°
[cointerior angles]
01BPQ DQP 180
2
[dividing both sides by 2]
01 2 90 ...(i)
1BPQ 1
2
1DQP 2
2
In PRQ , we have
RPQ + PQR + PRQ = 180°
[ by angle sum property of a triangle is 180°]
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1 + 2 + 3 = 180°
90° + 3 = 180°
[from Eq. (i), 1 + 2 = 90°]
3 = 180° – 90° = 90°
Hence, PRQ = 90°