Draw $∆$ PQR in which QR = 6cm, PQ + PR = 8.5 cm and $∠$ R - $∠$ Q = 40 $°$

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Solution

Steps of construction:
1. Draw QR = 6 cm.
2. Draw ray RX, such that $∠ QRX = 90 ° + 20 ° (i . e . 90 ° + \frac{∠ R - ∠ Q}{2})$ .
3. With Q as centre and radius 8.5 cm, draw an arc intersecting ray RX at point D.
4. Join Q and D.
5. Draw a ray RY, such that $∠ QRY = 20 ° (i . e . \frac{∠ R - ∠ Q}{2})$ .
6. Draw the perpendicular bisector of seg RE, intersecting seg QD at point P.
7. Join P and R.
8. $△$ PQR is the required triangle.

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