Question

Draw $∆$PQR in which QR = 5 cm, $\angle$P = 50$°$ and PQ $-$ PR = 2 cm.

Answer 1

Steps of construction:
1. Draw ray QY. Mark the point D, such that QD = 2 cm.
2. Draw ray DX, such that $\angle Q\mathrm{DX}=115°$ $\left(90°+\frac{50°}{2}\right)$.
3. With Q as centre and radius 5 cm, draw an arc, intersecting ray QX in point R.
4. Join R and Q.
5. Draw the perpendicular bisector of seg RD, intersecting QY at point P.
6. Hence, $△$PQR is the required triangle.