Question

construct∆PQR in which QR= 6cm angle Q=45° ,PR - PQ =3cm. Also write the measure of PR and PQ.​

Answer 1

Answer:

Given:  

In ∆PQR, QR= 6 cm, PR-PQ=2 cm & ∠Q= 60°

Steps of construction:

1. Draw the base QR=6 cm.

At Point Q draw a ray QX making an  ∠XQR=60°

Here , PR -PQ= 2cm

PR>PQ

The side containing the base angle Q is less than third side.

 

2. Cut the line segment QS equal to PR-PQ=2 cm , from the ray QX extended on  opposite side of base QR.

3. Join SR and draw its perpendicular bisector ray AB which intersect SR at M.

4. Let P be the intersection point of SX and perpendicular bisector AB. Then join PR.

Thus ∆PQR is the required Triangle.

Step-by-step explanation: