Given: m∠PQR = 105° and m∠QRP = 40°
We know that sum of angles of a triangle is 180°.
⇒ m∠PQR + m∠QRP + m∠QPR = 180°
⇒ 105° + 40° + m∠QPR = 180°
⇒ 145° + m∠QPR = 180°
⇒ m∠QPR = 180° – 145°
⇒ m∠QPR = 35°
To construct: ∆PQR where m∠P = 35° , m∠Q = 105° and PQ = 5 cm.
Steps of construction:
(a) Draw a line segment PQ = 5 cm.
(b) At point P, draw ∠XPQ = 35° with the help of protractor.
(c) At point Q, draw ∠YQP = 105° with the help of protractor.
(d) XP and YQ intersect at point R.
It is the required triangle PQR.

Class 7 Maths Chapter 10 Exercise 10.4

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