Construct $△$ PQR where base QR = 4.1 cm, $∠$ Q = 60 $°$ and PQ + QR = 8.5 cm.

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Solution

Steps of construction:
1. Draw QR = 4.1 cm.
2. Draw ray QX, such that $∠ RQX = 60 °$ .
3. With Q as centre and radius 8.5 cm, draw an arc, intersecting ray QX at point A.
4. Join A and R.
5. Draw the perpendicular bisector of seg AR, intersecting seg AQ at point P.
6. Join P and R.
7. Hence, $△$ PQR is the required triangle.

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