Draw $△$ ABC in which BC = 5 cm, $∠$ A = 50 $°$ and AB +AC = 7.5cm.

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Solution

Steps of construction:
1. Draw BD = 7.5 cm.
2. Draw ray DX, such that $∠ BDX = 25 ° (i . e . \frac{1}{2} ∠ A)$ .
3. With B as centre and radius 5 cm, draw an arc, intersecting ray DX at point C.
4. Join B and C.
5. Draw the perpendicular bisector of seg DC, intersecting BD at point A.
6. Hence, $△$ ABC is the required triangle.

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