Question

Construct $∆$ PQR in base QR = 6.7 cm, $\angle$PQR = 45$°$ and PR $-$ PQ = 2.8 cm.

Answer 1

Steps of construction:

1. Draw QR = 6.7 cm.
2. Draw ray QX, such that $\angle \mathrm{RQX}=45°$.
3. With Q as centre and radius 2.8 cm, draw an arc intersecting ray QX(produced in downward direction) at point D.
4. Join D and R.
5. Draw the perpendicular bisector of seg DR, intersecting seg QX at point P.
6. Join P and R.
7. Hence, $△$PQR is the required triangle.