Question

$△$LMN$~$$△$XYZ, In $△$LMN, LM = 6 cm, MN = 6.8 cm, LN = 7.6 cm and $\frac{\mathrm{LM}}{\mathrm{XY}}=\frac{4}{3}$
Construct $△$XYZ

Answer 1

$$\begin{gathered} \mathrm{Given}:\frac{\mathrm{LM}}{\mathrm{XY}}=\frac{4}{3} \\ \Rightarrow \mathrm{XY}=\frac{3}{4}\times 6 \\ =4.5\mathrm{cm} \end{gathered}$$

Steps of construction:-
1. Draw a line segment LM = 6 cm.
2. With L as the centre, draw an arc of radius 7.6 cm.
3. With M as the centre, draw an arc of radius 6.8 cm that cuts the previously drawn arc at N. Join LN and MN.
4. Similarly, draw a line segment XY = 4.5 cm.
5. Draw another angle $\angle$YXQ =$\angle$MLN.
6. Draw another angle $\angle$XYP=$\angle$LMN. The rays YP and XQ intersect at point Z.

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