Question

$△$ABC$~$$△$DEF, In $△$ABC, AB = 5.2 cm, BC=4.6 cm,$\angle$B=45$°$ and $\frac{\mathrm{BC}}{\mathrm{EF}}=\frac{2}{3}$
Construct $△$DEF

Answer 1

$$\begin{gathered} \mathrm{Given}:\frac{\mathrm{BC}}{\mathrm{EF}}=\frac{2}{3} \\ \Rightarrow \mathrm{EF}=\frac{3}{2}\times 4.6 \\ =6.9\mathrm{cm} \end{gathered}$$

Steps of construction:
1. Draw a line segment BC = 4.6 cm.
2. Draw an angle $\angle$CBX = 45$°$.
3. With B as the centre, draw an arc of radius 5.2 cm that cuts the ray BX at vertex A. Join AC.
4. Similarly, draw a line segment FE = 6.9 cm.
5. Draw an angle $\angle$FEY = 45$°$.
6. Draw another angle $\angle$BCA=$\angle$EFP. The rays FP and EY intersect at point D.

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