Question

28. Construct a triangle PQR in which QR = 6cm, 20 = 60°and PR PQ = 2cm. with steps of construction​

Answer 1

Answer:

Draw the base QR = 6 cm. ...

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

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

Let P be the intersection point of SX and perpendicular bisector AB.

Answer 2

Answer:

1) Draw the base QR = 6 cm.

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

o

.

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.