Construct a ∆PQR, in which QR = 6.5 cm, Q = 60° and PR - PQ = 1.5 cm and justify it.

Given, in ∆PQR,

QR = 6.5 cm, ZQ = 60° and

PR - PQ =1.5 cm

**Steps of construction**

(i)Draw the base, QR = 6.5 cm

i.e. the side containing the base angle 0 is less than third side, so it is the case II.

(ii)Cut line segment OS equal to PR-PQ, i.e.

OS =1.5 cm from the ray OX extended on opposite side of base QR.

(iii)Join SR and draw its perpendicular bisector ray AB, which intersects SR at M (say).

(iv)Let P be the intersection point of SX and perpendicular bisector AB. Then, join PR.

Thus, ∆PQR is the required triangle.

Justification

Base QR and ∠Q are drawn as given.

Since, AB is the perpendicular bisector of SR and P lies on it.

∴ PS = PR

Now, QS = PS - PQ

=> QS= PR-PQ

Thus, construction justified.