Question

. Construct a triangle PQR, given QR=3cm, and PQ+QR=6cm. (not steps)​

Answer 1

Answer:

Step-by-step explanation:

Given, inv ΔPQR, QR = 3 cm, ∠PQR = 45°

and QP – PR = 2 cm

Since, C lies on the perpendicular bisector RS of AY.

To construct ΔPQR, use the following steps.

1.Draw the base QR of length 3 cm.

2.Make an angle XQR = 45° at point Q of base QR.

3.Cut the line segment QS =QP- PR = 2 cm from the ray QX.

4.Join SR and draw the perpendicular bisector of SR say AB.

5.Let bisector AB intersect QX at P. Join PR Thus, ΔPQR is the required triangle.

Justification

Base QR and ∠PQR are drawn as given.

Since, the point P lies on the perpendicular bisector of SR.

PS = PR

Now, QS = PQ – PS

= PQ -PR

Thus, our construction is justified