Question

construct a triangle PQR where Angle P=70°, Angle R=50° and PQ=5cm​

Answer 1

Diagram

Construct Δ PQR such that ∠P = 70° , ∠R = 50° , QR = 7.3 cm and constructs its circumcircle.

SOLUTION

By angle sum property of triangles,

In Δ PQR,

∠P + ∠Q +∠R = 180°

⇒ 70° + ∠Q + 50° = 180°

⇒ ∠Q = 60°

Steps of construction:

1. Draw a line QR = 7.3 cm.

2. With Q as centre, draw an angle of 60° using the protractor. Similarly, draw an angle of 50° with R as centre.

Where the rays RY and XQ meet, name the point as P.

Thus, Δ PQR is obtained.

3. Draw the perpendicular bisectors of the lines PR and QR. Let these perpendicular bisectors meet at point O.

4. With O as centre and OP as radius, construct a circle touching all the vertices of the Δ PQR.

This circle is thus the required circumcircle.