RD Sharma Class 9 Solutions Chapter 10 Congruent Triangles MCQs

In this chapter, we provide RD Sharma Solutions for Chapter 10 Congruent Triangles MCQs for English medium students, Which will very helpful for every student in their exams. Students can download the latest RD Sharma Solutions for Chapter 10 Congruent Triangles MCQs Maths pdf, free RD Sharma Solutions for Chapter 10 Congruent Triangles MCQs Maths book pdf download. Now you will get step by step solution to each question.

TextbookNCERT
ClassClass 9
SubjectMaths
ChapterChapter 10
Chapter NameCongruent Triangles
ExerciseMCQs

RD Sharma Solutions for Class 9 Chapter 10 Congruent Triangles MCQs Download PDF

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 = 540∘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°
Class 9 RD Sharma Solutions Chapter 10 Congruent Triangles
∴ 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)
Class 9 Maths Chapter 10 Congruent Triangles RD Sharma Solutions
⇒ 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 = 170∘5  = 34° (b)
RD Sharma Book Class 9 PDF Free Download Chapter 10 Congruent Triangles

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 Book Chapter 10 Congruent Triangles
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°
Congruent Triangles With Solutions PDF RD Sharma Class 9 Solutions
Solution:
In the figure,
RD Sharma Class 9 Maths Book Questions Chapter 10 Congruent Triangles
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 = 360∘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 Mathematics Class 9 Solutions Chapter 10 Congruent Triangles
Solution:
∠1 = 3x+ 10 (Vertically opposite angles)
But x + ∠1 + ∠2 = 180°
Solution Of Rd Sharma Class 9 Chapter 10 Congruent Triangles
⇒ x + 3x + 10° + 90° = 180°
⇒ 4x = 180° – 10° – 90° = 80°
x = 80∘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 Math Solution Class 9 Chapter 10 Congruent Triangles
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 Questions Chapter 10 Congruent Triangles
= 35 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°
Maths RD Sharma Class 9 Chapter 10 Congruent Triangles
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 Chapter 10 Congruent Triangles

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
∠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 Solutions Class 9 Chapter 10 Congruent Triangles
Solution:
In the figure,
RD Sharma Class 9 PDF Chapter 10 Congruent Triangles
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°
Congruent Triangles Class 9 RD Sharma Solutions
Solution:
In the figure,
RD Sharma Class 9 Solution Chapter 10 Congruent Triangles
∠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
Class 9 RD Sharma Solutions Chapter 10 Congruent Triangles width=
Solution:
In the figure, l1 || l2
Class 9 Maths Chapter 10 Congruent Triangles RD Sharma Solutions
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 Book Class 9 PDF Free Download Chapter 10 Congruent Triangles
Solution:
In the figure, l1 || l2 and l3 is the transversal
RD Sharma Class 9 Book Chapter 10 Congruent Triangles

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 – x2
(d) 90 – 2x
Congruent Triangles With Solutions PDF RD Sharma Class 9 Solutions
Solution:
In the figure,
RD Sharma Class 9 Maths Book Questions Chapter 10 Congruent Triangles
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 = 540∘−x4
= 90° – x2 (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 Mathematics Class 9 Solutions Chapter 10 Congruent Triangles
Solution:
In the figure, l || m and n is the transversal
Solution Of Rd Sharma Class 9 Chapter 10 Congruent Triangles
⇒ 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°
Maths RD Sharma Class 9 Chapter 10 Congruent Triangles
Solution:
In the figure,
AB || HF, DE || FG
Class 9 RD Sharma Solutions Chapter 10 Congruent Triangles
∴ 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°
Class 9 Maths Chapter 10 Congruent Triangles RD Sharma Solutions
Solution:
In the figure, l || m
RD Sharma Book Class 9 PDF Free Download Chapter 10 Congruent Triangles
∴ ∠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 Book Chapter 10 Congruent Triangles
Solution:
In the figure, AB || CD
Through P, draw PQ || AB or CD
Congruent Triangles With Solutions PDF RD Sharma Class 9 Solutions
∠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 Maths Book Questions Chapter 10 Congruent Triangles
Solution:
In the figure, l || m
RD Sharma Mathematics Class 9 Solutions Chapter 10 Congruent Triangles
∠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°
Solution Of Rd Sharma Class 9 Chapter 10 Congruent Triangles
Solution:
In the figure,
l || m and n is the transversal
RD Sharma Math Solution Class 9 Chapter 10 Congruent Triangles
∠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 Questions Chapter 10 Congruent Triangles
Solution:
In the figure, l || m
and PQ is the transversal
Maths RD Sharma Class 9 Chapter 10 Congruent Triangles
∠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 = 270∘5  = 54°
∴ second angle = 90° – 54° = 36°
∴ smaller angle = 36° (c)

Question 26.
RD Sharma Class 9 Chapter 10 Congruent Triangles
Solution:
RD Sharma Class 9 Solutions Chapter 10 Congruent Triangles

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 Solutions Class 9 Chapter 10 Congruent Triangles
Solution:
In the figure, AB || CD || EF and GH || KL and GH is product to meet AB in L.
RD Sharma Class 9 PDF Chapter 10 Congruent Triangles
∵ 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°
Congruent Triangles Class 9 RD Sharma Solutions
∴ ∠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 Solution Chapter 10 Congruent Triangles
Solution:
In the figure, AB || CD
BD is transversal
∴ ∠ABD + ∠BDC = 180° (Co-interior angles)
⇒y + 2y+y + 5y = 180°
⇒ 9y = 180° ⇒ y = 180∘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°
Class 9 RD Sharma Solutions Chapter 10 Congruent Triangles
Solution:
In the figure, CP || DQ
BA is transversal
Produce PC to meet BA at D
Class 9 Maths Chapter 10 Congruent Triangles RD Sharma Solutions
∵ QB || PD
∴ ∠D = 105° (Corresponding angles)
In ∆ADC,
Ext. ∠ACP = ∠CDA + ∠DAC
⇒ x = ∠1 + 25°
= 105° + 25° = 130° (a)

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