Question

To construct a $\Delta PQR$, in which QR= 5.5 cm, $\angle Q={60}^{\circ }$ and PR - PQ = 2.5 cm, the steps of construction are given below. Complete the third step

Here, PR > PQ,
i.e., the side containing base angle is less than the third side.

Steps of construction:
Step 1: Draw the base QR = 5.5 cm
Step 2: At the point Q, make an $\angle XQR={60}^{\circ }$
Step 3 will be

• A. Cut line segment QS = PR - PQ = 2 cm from the line QX extended on opposite side of linesegment QR
• B. Cut line segment QS = PR - PQ = 2.5 cm from the line QX extended on opposite side of line segment QR
• C. Cut line segment QS = PR - PQ = 2.5 cm from the ray QX
• D. Draw the perpendicular bisector LM of SR.

Answer 1

The correct option is B Cut line segment QS = PR - PQ = 2.5 cm from the line QX extended on opposite side of line segment QR


Steps of construction:
Step 1: Draw the base QR = 5.5 cm
Step 2: At the point Q, make an $\angle XQR={60}^{\circ }$
Step 3 : Cut line segment QS = PR - PQ = 2.5 cm from the line QX extended on opposite side of line segment QR.
Join RS and draw the perpendicular bisector of RS.
Step 4 : Let it meet QX at P and join PR.
PQR is the required triangle.