Question

$∆$XYZ $~$ $∆$DEF; In $∆$DEF; DE = 5.5 cm, $\angle$E = 40$°$, EF = 4.0 cm and $\frac{XY}{DE}=\frac{6}{5}$ then construct $∆$XYZ.

Answer 1

$$\begin{gathered} \mathrm{Given}:\hspace{0.17em}\frac{\mathrm{XY}}{\mathrm{DE}}=\frac{6}{5} \\ \Rightarrow \mathrm{XY}=\frac{6}{5}\times 5.5 \\ =6.6\mathrm{cm} \end{gathered}$$

Steps of construction:
1. Draw a line segment DE = 5.5 cm.
2. Draw $\angle$DEP = 40$°$.
3. With E as the centre, draw an arc of radius 4 cm that cuts the ray EP at F. Join DF.
4. Similarly, draw a line segment XY = 6.6 cm.
5. Draw $\angle$XYQ = 40$°$.
6. Draw another angle $\angle$YXR = $\angle$EDF. The rays XR and YQ intersect at point Z.

Thus, $∆$XYZ is the required triangle, which is similar to $△$DEF.