Q. 144.4( 7 Votes )

Construct <span l

Answer :

Steps of Construction:

(i) Draw the base, QR = 3 cm.

At point B, draw a ray QX, which makes RQX = 45°

(ii) Here, QP – QR = 2 cm


Hence, cut the line segment QD is equal to QP – QR = 2 cm from the ray QX.

(iii) Now, join RD and draw its perpendicular bisector AB, which bisects RD at M (say).

Let P be the intersection point of perpendicular bisector AB and ray QX. Join PR.

Thus, ΔPQR is the required triangle.


Base QR and Q are drawn as given. Since, the point P lies on the perpendicular bisector of DR.


QD = PQ – PD

QD = QP – QR

Thus, the construction is justified.

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