Question

$△RHP\sim$ $△NED$ in $△NED,NE=7cm,\angle E={30}^{\circ },$ $\angle N={20}^{\circ }$ and $\frac{HP}{ED}=\frac{4}{5}$; Construct $△RHP$.

Answer 1

The sides of $△RHP$ will be $\frac{5}{4}$ times the sides of $△NED$

$\therefore RH=\frac{5}{4}\times NE=\frac{5}{4}\times 7=8.75cm$

$\angle H=\angle E={30}^{\circ }$

$\angle R=\angle N={20}^{\circ }$

Construction:

Step $1$: Draw base $RH=8.75cm$ with a ruler.

Step $2$: Draw a line at an angle of ${30}^{\circ }$ from point $H$.

Step $3$: Draw a line at an angle of ${20}^{\circ }$ from point $R$.

Step $4$: Mark the intersection point of the lines in steps $2$ and $3$ as point $P$ and draw lines joining $P-R$ and $P-H$.

$△RHP$ is the required triangle.