RD Chapter 10 Congruent Triangles Ex 10.2 Solutions
Question - 1 : - In the figure, OA and OB are opposite rays:
(i) If x = 25°, what is the value of y?
(ii) If y = 35°, what is the value of x?
Answer - 1 : -
(i) If x = 25°
∴ 3x = 3 x 25° = 75°
But ∠AOC + ∠BOC = 180° (Linear pair)
⇒ ∠AOC + 75° = 180°
⇒ ∠AOC = 180° – 75°
⇒ ∠AOC = 105°
∴ 2y + 5 = 105° ⇒ 2y= 105° – 5° = 100°
⇒ y =
= 50° ∴ If x = 25° then y = 50°
(ii) If y = 35°, then ∠AOC = 2y + 5
∴ 2y + 5 = 2 x 35° + 5 = 70° + 5 = 75°
But ∠AOC + ∠BOC = 180° (Linear pair)
⇒ 75° + ∠BOC = 180°
⇒ ∠BOC = 180°-75°= 105°
∴ 3x = 105° ⇒ x = 
= 35° ∴ x = 35°
Question - 2 : - In the figure, write all pairs of adjacent angles and all the linear pairs.
Answer - 2 : -
In the given figure,
(i) ∠AOD, ∠COD; ∠BOC, ∠COD; ∠AOD, ∠BOD and ∠AOC, ∠BOC are the pairs of adjacent angles.
(ii) ∠AOD, ∠BOD and ∠AOC, ∠BOC are the pairs of linear pairs.
Question - 3 : - In the figure, find x, further find ∠BOC, ∠COD and ∠AOD.
Answer - 3 : -
In the figure,
AOB is a straight line
∴ ∠AOD + ∠DOB = 180° (Linear pair)
⇒ ∠AOD + ∠DOC + ∠COB = 180°
⇒ x+ 10° + x + x + 20 = 180°
⇒ 3x + 30° = 180°
⇒ 3x= 180° -30°= 150°
⇒ x = 
= 50° ∴ x = 50°
Now ∠BOC =x + 20° = 50° + 20° = 70°
∠COD = x = 50°
and ∠AOD = x + 10° = 50° + 10° = 60°
Question - 4 : - In the figure, rays OA, OB, OC, OD and OE have the common end point O. Show that ∠AOB + ∠BOC + ∠COD + ∠DOE + ∠EOA = 360°.
Answer - 4 : -
Produce AO to F such that AOF is a straight line
Now ∠AOB + ∠BOF = 180° (Linear pair)
⇒ ∠AOB + ∠BOC + ∠COF = 180° …(i)
Similarly, ∠AOE + ∠EOF = 180°
⇒ ∠AOE + ∠EOD + ∠DOF = 180° …(ii)
Adding (i) and (ii)
∠AOB + ∠BOC + ∠COF + ∠DOF + ∠EOD + ∠AOE = 180° + 180°
⇒ ∠AOB + ∠BOC + ∠COD + ∠DOE + ∠EOA = 360°
Hence proved.
Question - 5 : - In the figure, ∠AOC and ∠BOC form a linear pair. If a – 2b = 30°, find a and b.
Answer - 5 : -
∠AOC + ∠BOC = 180° (Linear pair)
⇒ a + 6 = 180° …(i)
and a – 2b = 30° …(ii)
Subtracting (ii) from (i), 3b = 150°
⇒ b = = 50° and a + 50° = 180°
⇒ a= 180°-50°= 130°
∴ a = 130°, b = 50°
Question - 6 : - How many pairs of adjacent angles are formed when two lines intersect in a point?
Answer - 6 : -
If two lines AB and CD intersect each other at a point O, then four pairs of linear pairs are formed i.e.
∠AOC, ∠BOC; ∠BOC, ∠BOD; ∠BOD, ∠AOD and ∠AOD, ∠AOC
Question - 7 : - How many pairs of adjacent angles, in all, can you name in the figure.
Answer - 7 : -
In the given figure 10 pairs of adjacent angles are formed as given below:
∠AOB, ∠BOC; ∠AOB, ∠BOD; ∠AOC, ∠COD; ∠AOD, ∠BOE; ∠AOB, ∠BOE; ∠AOC, ∠COF; ∠BOC, ∠COD; ∠BOC, ∠COE; ∠COD, ∠DOE and ∠BOD, ∠DOE
Question - 8 : - In the figure, determine the value of x.
Answer - 8 : -
In the figure
∠AOC + ∠COB + ∠BOD +∠AOD = 360 (angles at a point )
⇒ 3x + 3x + x + 150° = 360°
⇒ 7x – 360° – 150° = 210°
⇒ x =
= 30° ∴ x = 30°
Question - 9 : - In the figure, AOC is a line, find x.
Answer - 9 : -
In the figure,
∠AOB + ∠BOC = 180° (Linear pair)
⇒ 70° + 2x = 180°
⇒ 2x = 180° – 70°
⇒ 2x = 110°⇒x =
= 55° ∴ x = 55°
Question - 10 : - In the figure, POS is a line, find x.
Answer - 10 : -
In the figure, POS is a line
∴ ∠POQ + ∠QOS = 180° (Linear pair)
⇒ ∠POQ + ∠QOR + ∠ROS = 180°
⇒ 60° + 4x + 40° = 180°
⇒ 4x + 100° – 180°
⇒ 4x = 180° – 100° = 80°
⇒ x = 
=20° ∴ x = 20°