RD Chapter 10 Congruent Triangles Ex MCQS Solutions
Question - 21 : - In the figure, if AB || CD, then x =
(a) 100°
(b) 105°
(c) 110°
(d) 115°
Answer - 21 : -
In the figure, AB || CD
Through P, draw PQ || AB or CD
∠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°
Answer - 22 : -
In the figure, l || m
∠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°
Answer - 23 : -
In the figure,
l || m and n is the transversal
∠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°
Answer - 24 : -
In the figure, l || m
and PQ is the transversal
∠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
Answer - 25 : -
Let first angle = x
Then its complementary angle = 90° – x
∴ 2x = 3(90° – x)
⇒ 2x = 270° – 3x
⇒ 2x + 3x = 270°
⇒ 5x = 270°
⇒ x =
= 54° ∴ second angle = 90° – 54° = 36°
∴ smaller angle = 36° (c)
Question - 26 : - 
Answer - 26 : -
Question - 27 : - In the figure, AB || CD || EF and GH || KL.
The measure of ∠HKL is
(a) 85°
(b) 135°
(c) 145°
(d) 215°
Answer - 27 : -
In the figure, AB || CD || EF and GH || KL and GH is product to meet AB in L.
∵ 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°
Answer - 28 : -
AB || CD and PQ is the transversal EL is the bisector of ∠FEB and ∠LEB = 35°
∴ ∠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
Answer - 29 : -
In the figure, AB || CD
BD is transversal
∴ ∠ABD + ∠BDC = 180° (Co-interior angles)
⇒y + 2y+y + 5y = 180°
⇒ 9y = 180° ⇒ y =
= 20° (d)
Question - 30 : - In the figure, if CP || DQ, then the measure of x is
(a) 130°
(b) 105°
(c) 175°
(d) 125°
Answer - 30 : -
In the figure, CP || DQ
BA is transversal
Produce PC to meet BA at D
∵ QB || PD
∴ ∠D = 105° (Corresponding angles)
In ∆ADC,
Ext. ∠ACP = ∠CDA + ∠DAC
⇒ x = ∠1 + 25°
= 105° + 25° = 130° (a)