Q. 144.4( 7 Votes )
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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
∴ QP > QR
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.
∴ PD = PR
QD = PQ – PD
⇒ QD = QP – QR
Thus, the construction is justified.
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