construct a PQR in which QR=6cm and angle Q=60 or PR
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Step-by-step explanation:
[ fig. is in the attachment]
Given:
In ∆PQR, QR= 6 cm, PR-PQ=2 cm & ∠Q= 60°
Steps of construction:
1. Draw the base QR=6 cm.
At Point Q draw a ray QX making an ∠XQR=60°
Here , PR -PQ= 2cm
PR>PQ
The side containing the base angle Q is less than third side.
2. Cut the line segment QS equal to PR-PQ=2 cm , from the ray QX extended on opposite side of base QR.
3. Join SR and draw its perpendicular bisector ray AB which intersect SR at M.
4. Let P be the intersection point of SX and perpendicular bisector AB. Then join PR.
Thus ∆PQR is the required Triangle.
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