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Asked by Sriomprakash Last Modified
Answer 
Tasneem
The steps mentioned below will be followed to construct the required triangle.
Step I: Draw line segment QR of 6 cm. At point Q, draw an angle of 60°, say ∠XQR.
Step II: Cut a line segment QS of 2 cm from the line segment QT extended in the opposite side of line segment XQ. (As PR > PQ and PR − PQ = 2 cm). Join SR.
Step III: Draw perpendicular bisector AB of line segment SR. Let it intersect QX at point P. Join PQ, PR.
ΔPQR is the required triangle.
XQR.
PQR is the required triangle.
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