Question

Question 2
Construct the following and give justification :

A $\Delta PQR$, given that QR = 3cm, $\angle PQR={45}^{\circ }$ and QP – PR = 2cm.

Answer 1

Given, in $\Delta PQR,QR=3cm$
$\angle PQR={45}^{\circ }$
QP – PR = 2cm

Steps to construct $\Delta PQR$
(i) Draw the base QR of length 3cm.
(ii) Make an angle XQR = ${45}^{\circ }$ at a point Q of the base QR.
(iii) Cut the line segment QS = QP – PR = 2cm on the ray QX.
(iv) Join SR and draw the perpendicular bisector of SR say AB.
(v) Let bisector AB intersect QX at P, join RP.
Thus, $\Delta PQR$ is the required triangle.

Justification
Base QR and $\angle PQR$ are drawn as given,
Since the point, P lies on the perpendicular bisector of SR
$\therefore$ PS = PR
Now, QS = PQ - PS
= PQ - PR
Thus, our construction is justified.