Selina Concise Class 6 Math Chapter 22 Angles (with Their Types And Properties) Exercise 22A Solution
EXERCISE 22A
(1) For each angle given below, write the name of the vertex, the names of the arms and the name of the angle.
(i) Vertex = 0; arms are OA and OB; angle AOB.
(ii) Vertex = Q, arms are QP and QR; angle PQR.
(iii) Vertex = M; arms are MN and ML; angle NML.
(2) Name the points:
(i) a, b and x
(ii) d, m, n, s and t
(3) ∠AOB, ∠AOC, ∠AOD, ∠AOE, ∠BOC, ∠BOD, ∠BOE, ∠COD, ∠COE and ∠DOE.
(4) Add:
(5) In the figure given alongside, name:
(i) ∠AOB and ∠BOC; ∠BOC and ∠COD, ∠COD and ∠DOA
(ii) ∠AOB and ∠AOD
(iii) ∠BOC and ∠COD
(iv) Reflex ∠AOB and reflex ∠COD
(6) (i) 180o – 75o = 105o
(ii) 180o – 110o = 70o
(7) In the given figure, AOC is a straight line. If angle AOB = 50o angle = AOE = 90o and angle COD = 25o, find the measure of:
(i) ∠BOC = 180o – 50o = 130o
(ii) ∠EOD = 180o – (90o + 25o) = 180o – 115o = 65o
(iii) Obtuse ∠BOD = ∠BOC + ∠COD = (130+25)o = 155o
(iv) Reflex ∠BOD
= ∠BOA + ∠AOE + ∠DOE
= (50o + 90o + 65o)
= 205o
(v) Reflex ∠COE
= ∠COB + ∠BOA + ∠AOE
= 130o + 50o + 90o
= 270o
(8) In the given figure, if:
(i) b = 360o – 130o = 230o
(ii) a = 360o – 200o = 160o
(9) In the figure given alongside, ABC is a straight line.
(i) y = 180o – 53o = 127o
(10) In the figure given alongside, AOB is a straight line. Find the value of x and also answer each of the following:
∠AOP + ∠BOP = 180o
⇒ x + 30o + x – 30o = 180o
⇒ 2x = 180o
⇒ x = 90o
(i) ∠AOP = x + 30o = 90o + 30o = 120o
(ii) ∠BOP = x – 30o = 90o – 30o = 60o
(iii) ∠AOP
(iv) ∠BOP
(11) In the figure given alongside, PQR is a straight line. Find x. Then complete the following:
∠PQA + ∠AQB + ∠BQR = 180o
⇒ x + 20o + 2x + 10o + x – 10o = 180o
⇒ 4x = 180o – 20o
⇒ 4x = 160o
⇒ x = 40o
(i) ∠AQB = 2x + 10 = (2 × 40) + 10 = 90o
(ii) ∠BQP
= ∠PQA + ∠AQB
= x + 20 + 2x + 10
= 3x + 30
= (3 × 40) + 30
= 120 + 30 = 150o
(iii) ∠AQR
= ∠AQB + ∠BQR
= 2x + 10 x – 10
= 3x = (3 × 40) = 120o
(12) In the figure given alongside, lines AB and CD intersect at point O.
(i) Here we see that AOB is a straight line which angle is 180o.
Then, ∠a = 180o – 68o = 112o
(ii) ∠AOC and ∠BOD; ∠AOD and ∠BOC
(iii) ∠AOC and ∠BOC; ∠BOC and ∠BOD; ∠BOD and ∠DOA; ∠DOA and ∠AOC
(iv) Reflex ∠BOC =180o + 68o = 248o;
Reflex ∠BOC = 180o + 68o = 248o
Reflex ∠AOC = 180o + 112o = 292o
Reflex = ∠AOD = 180o + 112o = 292o