Question

$△$LTR$~$$△$HYD In $△$HYD, HY = 7.2 cm, YD=6 cm,$\angle$Y=40$°$ and $\frac{\mathrm{LR}}{\mathrm{HD}}=\frac{5}{6}$
Construct $△$LTR

Answer 1

$$\begin{gathered} \mathrm{Given}:\frac{\mathrm{LR}}{\mathrm{HD}}=\frac{5}{6} \\ \mathrm{Now},\frac{\mathrm{LR}}{\mathrm{HD}}=\frac{\mathrm{LT}}{\mathrm{HY}}\left(△\mathrm{LTR}~△\mathrm{HYD}\right) \\ \Rightarrow \frac{\mathrm{LT}}{\mathrm{HY}}=\frac{5}{6} \\ \Rightarrow \mathrm{LT}=\frac{5}{6}\times 7.2 \\ =6\mathrm{cm} \end{gathered}$$

Steps of construction:
1. Draw a line segment HY = 7.2 cm.
2. Draw an angle $\angle$HYX = 40$°$.
3. With Y as the centre, draw an arc of radius 6 cm that cuts the ray YX at vertex D. Join HD.
4. Similarly, Draw a line segment LT = 6 cm.
5. Draw an angle $\angle$LTP = 40$°$.
6. Draw another angle $\angle$TLQ = $\angle$YHD. The rays TP and LQ intersect at point R.

Thus, $∆$LTR is the required triangle, which is similar to $△$HYD.