Question

$△$RHP$~$$△$NED, In $△$NED, NE = 7 cm, $\angle$D=30$°$, $\angle$N=20$°$ and $\frac{\mathrm{HP}}{\mathrm{ED}}=\frac{4}{5}$
Construct $△$RHP

Answer 1

$$\begin{gathered} \mathrm{Given}:\frac{\mathrm{HP}}{\mathrm{ED}}=\frac{4}{5} \\ \mathrm{Now},\frac{\mathrm{HP}}{\mathrm{ED}}=\frac{\mathrm{RH}}{\mathrm{NE}}\left(△\mathrm{RHP}~△\mathrm{NED}\right) \\ \Rightarrow \frac{\mathrm{RH}}{\mathrm{NE}}=\frac{4}{5} \\ \Rightarrow \mathrm{RH}=\frac{4}{5}\times 7 \\ =5.6\mathrm{cm} \end{gathered}$$

$$\begin{gathered} \mathrm{Also},\angle \mathrm{N}+\angle \mathrm{E}+\angle \mathrm{D}=180°(\mathrm{Angle}\mathrm{sum}\mathrm{property}) \\ \Rightarrow \angle \mathrm{E}=180°-\angle \mathrm{N}-\angle \mathrm{D}=180°-20°-30°=130° \end{gathered}$$

Steps of construction:
1. Draw a line segment NE = 7 cm.
2. Draw angles $\angle$NEX = 130$°$ and $\angle$ENY = 20$°$.
3. The rays EX and NY intersect at vertex D.
4. Similarly, draw a line segment RH = 5.6 cm.
5. Draw an angle $\angle$HRQ = $\angle$ENY.
6. Draw another angle $\angle$RHS=$\angle$NEX. The rays HS and RQ intersect at point P.

Thus, $∆$RHP is the required triangle, which is similar to $△$NED.