RD Chapter 10 Congruent Triangles Ex MCQS Solutions
Question - 11 : - In the figure, if AB || CD, then the value of x is
(a) 20°
(b) 30°
(c) 45°
(d) 60°
Answer - 11 : -
In the figure, AB || CD,
and / is transversal
∠1 = x (Vertically opposite angles)
and 120° + x + ∠1 = 180° (Co-interior angles)
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°
Answer - 12 : -
Two lines AB and CD intersect at O
∠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°
Answer - 13 : -
In the figure,
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°
Answer - 14 : - 
∠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
Answer - 15 : -
In the figure, l1 || l2
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
Answer - 16 : - In the figure, l1 || l2 and l3 is the transversal
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 – x /2
(d) 90 – 2x
Answer - 17 : - In the figure,
l1 ||l2 and l3 || l4 and m is theangle 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
(a) 60
(b) 50
(c) 45
d) 30
Answer - 18 : - In the figure, l || m and n is the transversal
⇒ 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°
Answer - 19 : -
In the figure,
AB || HF, DE || FG
∴ 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°
Answer - 20 : -
In the figure, l || m
∴ ∠1 =65° (Corresponding angles)
In ∆BCD,
Ext. ∠1 = x + 20°
⇒ 65° = x + 20°
⇒ x = 65° – 20°
⇒ x = 45° (b)