Question

$△LTR\sim$ $△HYD$, $△HYD$ is such that $HY=7.2cm,YD=6cm,\angle Y={40}^{\circ }$ and $\frac{LR}{HD}=\frac{5}{6}$ and construct $△LTR.$

Answer 1

The sides of $△LTR$ will be $\frac{5}{6}$ times the sides of $△HYD$

$\therefore LT=\frac{5}{6}\times HY=\frac{5}{6}\times 7.2=6cm$

$TR=\frac{5}{6}\times YD=\frac{5}{6}\times 6=5cm$

$\angle T=\angle Y={40}^{\circ }$

Construction:

Step $1$: Draw base $LT=6cm$ with a ruler.

Step $2$: Draw a line at an angle of ${40}^{\circ }$ from point $T$.

Step $3$: Place the compass on point $T$ and draw an arc of length of $5cm$.

Step $4$: Mark the intersection point of the line and arc in steps $2$ and $3$ as point $R$ and draw lines joining $R-L$ and $R-T$.

$△LTR$ is the required triangle.