RD Sharma Class 9 Solutions Chapter 10 Congruent Triangles MCQS

RD Sharma Class 9 Solutions Chapter 10 Congruent Triangles MCQS

These Solutions are part of RD Sharma Class 9 Solutions. Here we have given RD Sharma Class 9 Solutions Chapter 10 Congruent Triangles MCQS

Other Exercises

Mark the correct alternative in each of the following:
Question 1.
One angle is equal to three times its supplement. The measure of the angle is
(a) 130°
(b) 135°
(c) 90°
(d) 120°
Solution:
Let required angle = x
Then its supplement = (180° – x)
x = 3(180° – x) = 540° – 3x
⇒ x + 3x = 540°
⇒ 4x = 540°
⇒ x = \(\frac { { 540 }^{ \circ } }{ 4 }\)  = 135°
∴ Required angle = 135° (b)

Question 2.
Two straight lines AB and CD intersect one another at the point O. If ∠AOC + ∠COB + ∠BOD = 274°, then ∠AOD =
(a) 86°
(b) 90°
(c) 94°
(d) 137°
Solution:
Sum of angles at a point O = 360°
Sum of three angles ∠AOC + ∠COB + ∠BOD = 274°
RD Sharma Class 9 Solutions Chapter 10 Congruent Triangles MCQS Q2.1
∴ Fourth angle ∠AOD = 360° – 274°
= 86° (a)

Question 3.
Two straight lines AB and CD cut each other at O. If ∠BOD = 63°, then ∠BOC =
(a) 63°
(b) 117°
(c) 17°
(d) 153°
Solution:
CD is a line
∴ ∠BOD + ∠BOC = 180° (Linear pair)
RD Sharma Class 9 Solutions Chapter 10 Congruent Triangles MCQS Q3.1
⇒ 63° + ∠BOC = 180°
⇒ ∠BOC = 180° – 63°
∴ ∠BOC =117° (b)

Question 4.
Consider the following statements:
When two straight lines intersect:
(i) adjacent angles are complementary
(ii) adjacent angles are supplementary
(iii) opposite angles are equal
(iv) opposite angles are supplementary Of these statements
(a) (i) and (iii) are correct
(b) (ii) and (iii) are correct
(c) (i) and (iv) are correct
(d) (ii) and (iv) are correct
Solution:
Only (ii) and (iii) arc true. (b)

Question 5.
Given ∠POR = 3x and ∠QOR = 2x + 10°. If POQ is a striaght line, then the value of x is
(a) 30°
(b) 34°
(c) 36°
(d) none of these
Solution:
∵ POQ is a straight line
∴ ∠POR + ∠QOR = 180° (Linear pair)
⇒ 3x + 2x + 10° = 180°
⇒ 5x = 180 – 10° = 170°
∴ x = \(\frac { { 170 }^{ \circ } }{ 5 }\)  = 34° (b)
RD Sharma Class 9 Solutions Chapter 10 Congruent Triangles MCQS Q5.1

Question 6.
In the figure, AOB is a straight line. If ∠AOC + ∠BOD = 85°, then ∠COD =
(a) 85°
(b) 90°
(c) 95°
(d) 100°
RD Sharma Class 9 Solutions Chapter 10 Congruent Triangles MCQS Q6.1
Solution:
AOB is a straight line,
OC and OD are rays on it
and ∠AOC + ∠BOD = 85°
But ∠AOC + ∠BOD + ∠COD = 180°
⇒ 85° + ∠COD = 180°
∠COD = 180° – 85° = 95° (c)

Question 7.
In the figure, the value of y is
(a) 20°
(b) 30°
(c) 45°
(d) 60°
RD Sharma Class 9 Solutions Chapter 10 Congruent Triangles MCQS Q7.1
Solution:
In the figure,
RD Sharma Class 9 Solutions Chapter 10 Congruent Triangles MCQS Q7.2
y = x (Vertically opposite angles)
∠1 = 3x
∠2 = 3x
∴ 2(x + 3x + 2x) = 360° (Angles at a point)
2x + 6x + 4x = 360°
12x = 360° ⇒ x = \(\frac { { 360 }^{ \circ } }{ 12 }\)  = 30°
∴ y = x = 30° (b)

Question 8.
In the figure, the value of x is
(a) 12
(b) 15
(c) 20
(d) 30
RD Sharma Class 9 Solutions Chapter 10 Congruent Triangles MCQS Q8.1
Solution:
∠1 = 3x+ 10 (Vertically opposite angles)
But x + ∠1 + ∠2 = 180°
RD Sharma Class 9 Solutions Chapter 10 Congruent Triangles MCQS Q8.2
⇒ x + 3x + 10° + 90° = 180°
⇒ 4x = 180° – 10° – 90° = 80°
x = \(\frac { { 80 }^{ \circ } }{ 4 }\) = 20   (c)

Question 9.
In the figure, which of the following statements must be true?
(i) a + b = d + c
(ii) a + c + e = 180°
(iii) b + f= c + e
(a) (i) only
(b) (ii) only
(c) (iii) only
(d) (ii) and (iii) only
RD Sharma Class 9 Solutions Chapter 10 Congruent Triangles MCQS Q9.1
Solution:
In the figure,
(i) a + b = d + c
a° = d°
b° = e°
c°= f°
(ii) a + b + e = 180°
a + e + c = 180°
⇒ a + c + e = 180°
(iii) b + f= e + c
∴ (ii) and (iii) are true statements (d)

Question 10.
If two interior angles on the same side of a transversal intersecting two parallel lines are in the ratio 2:3, then the measure of the larger angle is
(a) 54°
(b) 120°
(c) 108°
(d) 136°
Solution:
In figure, l || m and p is transversal
RD Sharma Class 9 Solutions Chapter 10 Congruent Triangles MCQS Q10.1
= \(\frac { 3 }{ 5 }\) x 180° = 108° (c)

Question 11.
In the figure, if AB || CD, then the value of x is
(a) 20°
(b) 30°
(c) 45°
(d) 60°
RD Sharma Class 9 Solutions Chapter 10 Congruent Triangles MCQS Q11.1
Solution:
In the figure, AB || CD,
and / is transversal
∠1 = x (Vertically opposite angles)
and 120° + x + ∠1 = 180° (Co-interior angles)
RD Sharma Class 9 Solutions Chapter 10 Congruent Triangles MCQS Q11.2

Question 12.
Two lines AB and CD intersect at O. If ∠AOC + ∠COB + ∠BOD = 270°, then ∠AOC =
(a) 70°
(b) 80°
(c) 90°
(d) 180°
Solution:
Two lines AB and CD intersect at O
RD Sharma Class 9 Solutions Chapter 10 Congruent Triangles MCQS Q12.1
∠AOC + ∠COB + ∠BOD = 270° …(i)
But ∠AOC + ∠COB + ∠BOD + ∠DOA = 360° …(ii)
Subtracting (i) from (ii),
∠DOA = 360° – 270° = 90°
But ∠DOA + ∠AOC = 180°
∴ ∠AOC = 180° – 90° = 90° (c)

Question 13.
In the figure, PQ || RS, ∠AEF = 95°, ∠BHS = 110° and ∠ABC = x°. Then the value of x is
(a) 15°
(b) 25°
(c) 70°
(d) 35°
RD Sharma Class 9 Solutions Chapter 10 Congruent Triangles MCQS Q13.1
Solution:
In the figure,
RD Sharma Class 9 Solutions Chapter 10 Congruent Triangles MCQS Q13.2
PQ || RS, ∠AEF = 95°
∠BHS = 110°, ∠ABC = x
∵ PQ || RS,
∴ ∠AEF = ∠1 = 95° (Corresponding anlges)
But ∠1 + ∠2 = 180° (Linear pair)
⇒ ∠2 = 180° – ∠1 = 180° – 95° = 85°
In ∆AGH,
Ext. ∠BHS = ∠2 +x
⇒ 110° = 85° + x
⇒ x= 110°-85° = 25° (b)

Question 14.
In the figure, if l1 || l2, what is the value of x?
(a) 90°
(b) 85°
(c) 75°
(d) 70°
RD Sharma Class 9 Solutions Chapter 10 Congruent Triangles MCQS Q14.1
Solution:
In the figure,
RD Sharma Class 9 Solutions Chapter 10 Congruent Triangles MCQS Q14.2
∠1 = 58° (Vertically opposite angles)
Similarly, ∠2 = 37°
∵ l1 || l2, EF is transversal
∠GEF + EFD = 180° (Co-interior angles)
⇒ ∠2 + ∠l +x = 180°
⇒ 37° + 58° + x = 180°
⇒ 95° + x= 180°
x = 180°-95° = 85° (b)

Question 15.
In the figure, if l1 || l2, what is x + y in terms of w and z?
(a) 180-w + z
(b) 180° + w- z
(c) 180 -w- z
(d) 180 + w + z
RD Sharma Class 9 Solutions Chapter 10 Congruent Triangles MCQS Q15.1
Solution:
In the figure, l1 || l2
RD Sharma Class 9 Solutions Chapter 10 Congruent Triangles MCQS Q15.2
p and q are transversals
∴ w + x = 180° ⇒ x = 180° – w (Co-interior angle)
z = y (Alternate angles)
∴ x + y = 180° – w + z (a)

Question 16.
In the figure, if l1 || l2, what is the value of y?
(a) 100
(b) 120
(c) 135
(d) 150
RD Sharma Class 9 Solutions Chapter 10 Congruent Triangles MCQS Q16.1
Solution:
In the figure, l1 || l2 and l3 is the transversal
RD Sharma Class 9 Solutions Chapter 10 Congruent Triangles MCQS Q16.2

Question 17.
In the figure, if l1 || l2 and l3 || l4 what is y in terms of x?
(a) 90 + x
(b) 90 + 2x
(c) 90 – \(\frac { x }{ 2 }\)
(d) 90 – 2x
RD Sharma Class 9 Solutions Chapter 10 Congruent Triangles MCQS Q17.1
Solution:
In the figure,
RD Sharma Class 9 Solutions Chapter 10 Congruent Triangles MCQS Q17.2
l1 || l2 and l3 || l4 and m is the angle bisector
∴ ∠2 = ∠3 = y
∵ l1 || l2
∠1 = x (Corresponding angles)
∵ l3 || l4
∴ ∠1 + (∠2 + ∠3) = 180° (Co-interior angles)
⇒ x + 2y= 180°
⇒ 2y= 180°-x
⇒ y = \(\frac { { 540 }^{ \circ }-x }{ 4 }\)
= 90° – \(\frac { x }{ 2 }\) (c)

Question 18.
In the figure, if 11| m, what is the value of x?
(a) 60
(b) 50
(c) 45
d) 30
RD Sharma Class 9 Solutions Chapter 10 Congruent Triangles MCQS Q18.1
Solution:
In the figure, l || m and n is the transversal
RD Sharma Class 9 Solutions Chapter 10 Congruent Triangles MCQS Q18.2
⇒ y = 25°
But 2y + 25° = x+ 15°
(Vertically opposite angles) ⇒ x = 2y + 25° – 15° = 2y+ 10°
= 2 x 25°+10° = 50°+10° = 60° (a)

Question 19.
In the figure, if AB || HF and DE || FG, then the measure of ∠FDE is
(a) 108°
(b) 80°
(c) 100°
(d) 90°
RD Sharma Class 9 Solutions Chapter 10 Congruent Triangles MCQS Q19.1
Solution:
In the figure,
AB || HF, DE || FG
RD Sharma Class 9 Solutions Chapter 10 Congruent Triangles MCQS Q19.2
∴ HF || AB
∠1 =28° (Corresponding angles)
But ∠1 + ∠FDE + 72° – 180° (Angles of a straight line)
⇒ 28° + ∠FDE + 72° = 180°
⇒ ∠FDE + 100° = 180°
⇒ ∠FDE = 180° – 100 = 80° (b)

Question 20.
In the figure, if lines l and m are parallel, then x =
(a) 20°
(b) 45°
(c) 65°
(d) 85°
RD Sharma Class 9 Solutions Chapter 10 Congruent Triangles MCQS Q20.1
Solution:
In the figure, l || m
RD Sharma Class 9 Solutions Chapter 10 Congruent Triangles MCQS Q20.2
∴ ∠1 =65° (Corresponding angles)
In ∆BCD,
Ext. ∠1 = x + 20°
⇒ 65° = x + 20°
⇒ x = 65° – 20°
⇒ x = 45° (b)

Question 21.
In the figure, if AB || CD, then x =
(a) 100°
(b) 105°
(c) 110°
(d) 115°
RD Sharma Class 9 Solutions Chapter 10 Congruent Triangles MCQS Q21.1
Solution:
In the figure, AB || CD
Through P, draw PQ || AB or CD
RD Sharma Class 9 Solutions Chapter 10 Congruent Triangles MCQS Q21.2
∠A + ∠1 = 180° (Co-interior angles)
⇒ 132° + ∠1 = 180°
⇒ ∠1 = 180°- 132° = 48°
∴ ∠2 = 148° – ∠1 = 148° – 48° = 100°
∵ DQ || CP
∴ ∠2 = x (Corresponding angles)
∴ x = 100° (a)

Question 22.
In tlie figure, if lines l and in are parallel lines, then x =
(a) 70°
(b) 100°
(c) 40°
(d) 30°
RD Sharma Class 9 Solutions Chapter 10 Congruent Triangles MCQS Q22.1
Solution:
In the figure, l || m
RD Sharma Class 9 Solutions Chapter 10 Congruent Triangles MCQS Q22.2
∠l =70° (Corresponding angles)
In ∆DEF,
Ext. ∠l = x + 30°
⇒ 70° = x + 30°
⇒ x = 70° – 30° = 40° (c)

Question 23.
In the figure, if l || m, then x =
(a) 105°
(b) 65°
(c) 40°
(d) 25°
RD Sharma Class 9 Solutions Chapter 10 Congruent Triangles MCQS Q23.1
Solution:
In the figure,
l || m and n is the transversal
RD Sharma Class 9 Solutions Chapter 10 Congruent Triangles MCQS Q23.2
∠1 = 65° (Alternate angles)
In ∆GHF,
Ext. x = ∠1 + 40° = 65° + 40°
⇒ x = 105°
∴ x = 105° (a)

Question 24.
In the figure, if lines l and m are parallel, then the value of x is
(a) 35°
(b) 55°
(c) 65°
(d) 75°
RD Sharma Class 9 Solutions Chapter 10 Congruent Triangles MCQS Q24.1
Solution:
In the figure, l || m
and PQ is the transversal
RD Sharma Class 9 Solutions Chapter 10 Congruent Triangles MCQS Q24.2
∠1 = 90°
In ∆EFG,
Ext. ∠G = ∠E + ∠F
⇒ 125° = x + ∠1 = x + 90°
⇒ x = 125° – 90° = 35° (a)

Question 25.
Two complementary angles are such that two times the measure of one is equal to three times the measure of the other. The measure of the smaller angle is
(a) 45°
(b) 30°
(c) 36°
(d) none of these
Solution:
Let first angle = x
Then its complementary angle = 90° – x
∴ 2x = 3(90° – x)
⇒ 2x = 270° – 3x
⇒ 2x + 3x = 270°
⇒ 5x = 270°
⇒ x = \(\frac { { 270 }^{ \circ } }{ 5 }\)  = 54°
∴ second angle = 90° – 54° = 36°
∴ smaller angle = 36° (c)

Question 26.
RD Sharma Class 9 Solutions Chapter 10 Congruent Triangles MCQS Q26.1
Solution:
RD Sharma Class 9 Solutions Chapter 10 Congruent Triangles MCQS Q26.2

Question 27.
In the figure, AB || CD || EF and GH || KL.
The measure of ∠HKL is
(a) 85°
(b) 135°
(c) 145°
(d) 215°
RD Sharma Class 9 Solutions Chapter 10 Congruent Triangles MCQS Q27.1
Solution:
In the figure, AB || CD || EF and GH || KL and GH is product to meet AB in L.
RD Sharma Class 9 Solutions Chapter 10 Congruent Triangles MCQS Q27.2
∵ AB || CD
∴ ∠1 = 25° (Alternate angle)
and GH || KL
∴ ∠4 = 60° (Corresponding angles)
∠5 = ∠4 = 60° (Vertically opposite angle)
∠5 + ∠2 = 180° (Co-interior anlges)
∴ ⇒ 60° + ∠2 = 180°
∠2 = 180° – 60° = 120°
Now ∠HKL = ∠1 + ∠2 = 25° + 120°
= 145° (c)

Question 28.
AB and CD are two parallel lines. PQ cuts AB and CD at E and F respectively. EL is the bisector of ∠FEB. If ∠LEB = 35°, then ∠CFQ will be
(a) 55°
(b) 70°
(c) 110°
(d) 130°
Solution:
AB || CD and PQ is the transversal EL is the bisector of ∠FEB and ∠LEB = 35°
RD Sharma Class 9 Solutions Chapter 10 Congruent Triangles MCQS Q28.1
∴ ∠FEB = 2 x 35° = 70°
∵ AB || CD
∴ ∠FEB + ∠EFD = 180°
(Co-interior angles)
70° + ∠EFD = 180°
∴ ∠EFD = 180°-70°= 110°
But ∠CFQ = ∠EFD
(Vertically opposite angles)
∴ ∠CFQ =110° (c)

Question 29.
In the figure, if line segment AB is parallel to the line segment CD, what is the value of y?
(a) 12
(b) 15
(c) 18
(d) 20
RD Sharma Class 9 Solutions Chapter 10 Congruent Triangles MCQS Q29.1
Solution:
In the figure, AB || CD
BD is transversal
∴ ∠ABD + ∠BDC = 180° (Co-interior angles)
⇒y + 2y+y + 5y = 180°
⇒ 9y = 180° ⇒ y = \(\frac { { 180 }^{ \circ } }{ 9 }\)  = 20° (d)

Question 30.
In the figure, if CP || DQ, then the measure of x is
(a) 130°
(b) 105°
(c) 175°
(d) 125°
RD Sharma Class 9 Solutions Chapter 10 Congruent Triangles MCQS Q30.1
Solution:
In the figure, CP || DQ
BA is transversal
Produce PC to meet BA at D
RD Sharma Class 9 Solutions Chapter 10 Congruent Triangles MCQS Q30.2
∵ QB || PD
∴ ∠D = 105° (Corresponding angles)
In ∆ADC,
Ext. ∠ACP = ∠CDA + ∠DAC
⇒ x = ∠1 + 25°
= 105° + 25° = 130° (a)

Hope given RD Sharma Class 9 Solutions Chapter 10 Congruent Triangles MCQS are helpful to complete your math homework.

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