Question

Construct a triangle $PQR$ in which $QR=5cm,\angle R={40}^o$ and $PR-PQ=1cm$.

Answer 1

It is given that $QR=5$ cm, the difference between other two sides $PR-PQ=1$ cm and $\angle R={40}^{\circ }$.
Steps of construction.
1. Draw a line segment $QR=5cm$.
2. At $R$, construct at angle of ${40}^{\circ }$ and produce it.
3. By taking $R$ as centre, draw an arc of radius $1$ cm cutting $RT$ at $S$.
4. Join $QS$.
5. Also, draw $\perp$ bisector of $QS$ which meets $RT$ at $P$
6. Join $PQ$.
Hence, $\Delta PQR$ is the required triangle.