Question

Construct a right triangle whose hypotenuse and one side measures $10cm$ and $8cm$ respectively. Then construct another triangle whose sides are $\frac{4}{5}$ times the corresponding sides of this triangle.

Answer 1

1. Draw a line segment $RQ=8c.m$.

2. Construct $RL$ perpendicular $RQ$.

3. Taking $Q$ as centre and whose radius as $10c.m.$ drawn an arc that cuts the ray $RE$ are point $P$.

4. Join $PQ$ to obtain $△PQR$.

5. Below $RQ$, make an acute angle $QRX$

6. Along $RX$ make $5$ points $R_1,R_2,R_3,R_4,R_5$ such that $R{R}_1=R_1{R}_2=R_2{R}_3=\dots \dots ..=R_4{R}_5$

7. Join $R_{5}Q$.

8. From $R_4$. Drawn $R_{4}C\parallel R_{5}C$

9. From $Q$ draw $R^{\prime }{P}^{\prime },RP$ meeting $RP$ at point $P$ then $P^{\prime }{Q}^{\prime }R$ as the required triangle.