Question

Construct $∆$ PQR, such that QR = 7.4 cm, $\angle$PQR= ${60}^{°}$ and PQ $-$ PR = 2.5 cm.

Answer 1

​Steps of construction:

(1) Draw seg QR of length 6.5 cm.
(2) Draw ray QA such that ∠RQA = 60°
(3) Take point B on ray QA such that QB = 2.5 cm.
(4) Construct the perpendicular bisector of seg BR.
(5) Name the point of intersection of ray QA and the perpendicular bisector of seg BR as P.
(6) Draw seg PR.

Therefore, △PQR is required triangle.