# The Triangle and Its Properties NCERT Solution for Class 7 Maths Chapter 6 Exercise 6.3

## The Triangle and Its Properties Exercise 6.3 NCERT Solution for Class 7 Maths Chapter 6

Exercise 6.3

Q.1 :- Find the value of the unknown x in the following diagrams:

Solution 1:-

Q.1 :-  Find the values of the unknowns x and y in the following diagrams:

Solution 1:-  By The  Angle Sum Property Of A Triangle

Formula:- The total measure of the three angles of a triangle is 180°.

Solution 1.(i) :- In Triangle ABC,
BAC + ACB + ABC = 180º,     [ By the angle sum property of triangle ]

x + 60º+ 50º = 180º,
x + 110º = 180º,
x = 180º – 110º,
x= 70º.

Solution 1.(ii) :- In Triangle PQR
PRQ + PQR + RPQ = 180º,     [ By the angle sum property of triangle ]

x + 30º+ 90º = 180º,
x + 120º = 180º,
x = 180º – 120º,
x= 60º.

Solution 1.(iii) :- In Triangle XYZ
XZY + XYZ + ZXY = 180º,     [ By the angle sum property of triangle]

x + 110º+ 30º = 180º,
x + 140º = 180º,
x = 180º – 140º,
x= 40º.

Solution 1.(iv) :- In The Isosceles Triangle
∠1 + ∠2 + ∠3 = 180º,     [ By the angle sum property of triangle]

x + x+ 50º = 180º,
2x + 50º = 180º,
2x = 180º – 50º,
x = 130º/2 ,
x = 65º.

Solution 1.(v) :- In The Equilateral Triangle
∠1 + ∠2 + ∠3 = 180º,     [ By the angle sum property of triangle]

x + x+ x = 180º,
3x = 180º,
x = 180º/3 ,
x = 60º.

Solution 1.(vi) :- In The Given Right Angle Triangle
∠1 + ∠2 + ∠3 = 180º,     [ By the angle sum property of triangle]

x + 2x+ 90º = 180º,
3x + 90º = 180º,
3x = 180º – 90º,
x = 90º/3,
x = 30º.

Q.2:- Find the values of the unknowns x and y in the following diagrams:

Solution 2:-

Solution 2.(i) :- In The Given Triangle
x + 50º = 120º,     [ By the Exterior angle property of a triangle ]
x = 120º – 50º,
x = 70º

Now, For The Value of Y, using Angle sum property of a triangle,
x + y + 50º = 180º,
70º + y + 50º = 180º,
y + 120º = 180º,
y = 180º – 120º,
y = 60º.

Hence, x=70º and y=60º.

Solution 2.(ii) :- In The Given Triangle
y = 80º,     [ By the Vertically Opposite angle property of a triangle ]

Now, For The Value of X, using Angle sum property of a triangle,
x + y + 50º = 180º,
x + 80º + 50º = 180º,
x + 130º = 180º,
x = 180º – 130º,
x = 50º.

Hence, x=50º and y=80º.

Solution 2.(iii) :-In The Given Triangle
x = 50º + 6,     [ By the Exterior angle property of a triangle ]
x = 110º ,

Now, For The Value of Y, using Angle sum property of a triangle,
y + 60º + 50º = 180º,
y + 110º = 180º,
y = 180º – 110º,
y = 70º.

Hence, x=110º and y=70º.

Solution 2.(iv) :-In The Given Triangle
x = 60º,                         [ By the Vertically Opposite angle property of a triangle ]

Now, For The Value of Y, using Angle sum property of a triangle,
x + y + 30º = 180º,
60º + y + 30º = 180º,
y + 90º = 180º,
y = 180º – 90º,
y = 90º.

Hence, x=60º and y=90º.

Solution 2.(v) :-In The Given Triangle
y = 90º,                    [ By the Vertically Opposite angle property of a triangle ]

Now, For The Value of X, using Angle sum property of a triangle,
x + x + y = 180º,
2x + 90º  = 180º,
2x = 180º – 90º,
2x = 90º,
x = 90º/2,
x = 45º.

Hence, x=45º and y=90º.

Solution 2.(vi) :- In The Given Triangle
x = y,                    [ By the Vertically Opposite angle property of a triangle ]

Now, For The Value of Y, using Angle sum property of a triangle,
x + x + y = 180º,
x + x + x  = 180º,
3x = 180º,
x = 180º/3,
x = 60º.

Hence, x=60º and y=60º.

NCERT Solutions for Class 7 Math, Chapter 6 – The Triangle and Its Properties, Exercise-6.3