Question

Two triangles $ABC$ and $DEF$ are similar. The area of $△ABC$ is $9$ sq,cm and the area of $△DEF$ is $16$ sq.cm. If $BC=2.1$cm. Find Length of $EF$.

Answer 1

Finding Length of $EF$:

Given,

Area of $△ABC=$ $9$ sq,cm

Area of $△DEF=16$ sq.cm

$BC=2.1$.

We know that the ratio of area of two similar triangles is equal to the squares of the ratio of corresponding sides.

therefore we have,

$\Rightarrow \frac{\text{Area of}△ABC}{\text{Area of}△DEF}=\frac{B{C}^2}{E{F}^2}$

Substitute the values we get,

$$\begin{gathered} \Rightarrow \frac{9}{16}=\frac{2.1^2}{E{F}^2} \\ \therefore E{F}^2=\frac{4.41\times 16}{9} \\ \Rightarrow E{F}^2=7.84 \\ \Rightarrow EF=\sqrt{7.84} \\ \Rightarrow EF=2.8\text{cm} \end{gathered}$$

Therefore, the value of $EF=2.8\text{cm}$.