Question

∆PQR~∆PTS in ∆ PQR PQ= 6.3 cm, angle RQP= 50° , PR= 5.6 cm PQ|PT 7|5 construct ∆ AHE ​

Answer 1

Answer:

1) Draw the base QR = 6 cm.

At point Q draw a ray QX making an ∠ QXR = 60o.

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.