Ncert Solutions For Class 7 Maths Ex 6.3

Ncert Solutions For Class 7 Maths Chapter 6 Ex 6.3

Question 1:

Find the value of the unknown a in the following given diagrams.

8

Answers:

(a) In Δ PQR

QPR + PRQ + PQR = 180o [ By angle sum property of a triangle]

=> a + 50o + 60o = 180o

=> a + 110o = 180o

=> a = 180o – 110 = 70o

(b) In Δ ABC,

CAB + ABC + CAB = 90o [ By angle sum property of a triangle]

=> 90o + 30o + a = 180o

=> a + 120= 180o

=> a = 180o – 120o = 60o

(c ) In Δ ABC ,

CAB + ABC + BCA = 180o [ By angle of sum property of a triangle]

=> 30 + 110+ a  = 180o

=> a + 140o = 180o

=> a = 180– 140O = 40O

(d) In the given isosceles triangle,

a + a + 50o = 180o

=> 2a + 50 = 180o

=> 2a = 180o – 50o

=> 2a = 130o

=> a=1302 = 65o

(e) In the given equilateral triangle,

a + a + a = 180o                                              [ By angle sum property of a triangle]

=> 3 a = 180o

=> a=1803 = 60o

(f) In the given right angled triangle,

a + 2a + 90o = 180o      [ By the angle sum property of a triangle]

=> 3a + 90o = 180o

=> 3a = 180o – 90o

=> 3a = 90o

=> a=903 = 30o

 

Question 2

Find the values a and b in the following diagrams that are given below:

9

Answers:

(a) 50+ a = 120o                                 [Exterior angle property of a Δ ]

=> a = 120 o – 50 o = 70 o

Now, 50 o + a + b = 180 o                               [Angle sum property of a Δ ]

=> 50 o + 70 o + b = 180 o

=> 120 o + b = 180 o

=> b = 180 o – 120 o = 60 o

(b) b = 80 o – – – – – – – – – – – (1)                         [ Vertically opposite angle]

Now, 50 o + a + b = 180 o                     [ Angle sum property of a Δ ]

=> 50 o + 80 o + b = 180 o                      [From equation (1) ]

=> 130 o + b = 180 o

=> b = 180 o – 130 o = 50 o

( c) 50 o + 60 o = a                               [Exterior angle property of a Δ ]

=> a = 110 o

Now, 50 o + 60 o + b = 180 o                  [Angle sum property of a Δ ]

=> 110 o + b = 180 o

=> b = 180 o – 110 o

=> b = 70 o

(d) a = 60 o   – – – – – – – – – – – – – – – (1)             [ Vertically opposite angle]

Now, 30 o + a + b = 180 o                                 [ Angle sum property of a Δ ]

=> 50 o + 60 o + b = 180 o                                  [From equation (1) ]

=> 90 o + b = 180 o

=> b = 180 o – 90 o = 90 o

(e) b = 90 o – – – – – – – – – – – – – (1)                    [ Vertically opposite angle]

Now, b + a + a = 180 o                                                 [ Angle sum property of a Δ ]

=> 90 o + 2a = 180                                          [From equation (1) ]

=> 2a = 180 o – 90 o

=> 2a = 90 o

=> a=902 = 45o

(f) a = b – – – – – – – – – – – (1)                             [ Vertically opposite angle]

Now, a + a + b = 180 o                                                 [ Angle sum property of a Δ ]

=> 2a + a = 180 o                                             [From equation (1) ]

=> 3a = 180 o

=> a=1803 = 60o

 

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