Question

It is given that ∆ ABC ~ ∆ EDF such that $AB=5cm,AC=7cm,DF=15cm$ and $DE=12cm$. Find the lengths of the remaining sides of the triangles.

Answer 1

Find the lengths of the remaining sides of the triangles :

Given, $∆ABC~∆EDF$

According to the property of a similar triangle, if two triangles are similar then the ratio of their corresponding sides is also equal.

So, we know that the corresponding sides of $∆ABC$ and $∆EDF$ are in the same ratio.

$\frac{AB}{ED}=\frac{AC}{EF}=\frac{BC}{DF}$ …………..….$\left(i\right)$

It is given that

$AB=5cm,AC=7cm,DF=15cm$ and $DE=12cm$

Substituting these values in $\left(i\right)$, we get,

$\frac{5}{12}=\frac{7}{EF}=\frac{BC}{15}$

Now, considering the first two terms, we get

$\frac{5}{12}=\frac{7}{EF}$

$EF=\frac{12\times 7}{5}=16.8cm$

On considering the first and last terms, we get

$\frac{5}{12}=\frac{BC}{15}$

$BC=\frac{5\times 15}{12}=6.25cm$

Hence, the lengths of the remaining sides of the triangles are $EF=16.8cm$ and $BC=6.25cm$