Question

Question 3
Construct a triangle PQR in which $QR=6cm,\angle Q={60}^{\circ }andPR-PQ=2cm$

Answer 1

Steps to construct the required triangle are as follows :
Step 1: Draw a line segment QR of 6cm. At point Q, draw an angle of ${60}^{\circ }$, say $\angle XQR={60}^{\circ }$.
Step 2: Cut a line segment QS = PR-PQ = 2cm from the line QX extended on the opposite side of line segment QR and join SR.
Step 3: Draw perpendicular bisector AB of line segment SR. Let it intersect QX at point P.
Step 4: Join PR, then $\Delta PQR$ is the required triangle.