Question

Question 2
It is given that $\Delta ABC\sim \Delta EDF$ such that AB = 5cm, AC = 7cm, DF = 15cm and DE = 12cm. Find the lengths of the remaining sides of the triangles.

Answer 1

Given, $\Delta ABC\sim \Delta EDF$, so the corresponding sides of $\Delta ABC$ and $\Delta EDF$ are in the same ratio.

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


Also, AB = 5 cm, AC = 7 cm
DF = 15 cm and DE = 12 cm
On putting these values in Eq. (i), we get,
$\frac{5}{12}=\frac{7}{EF}=\frac{BC}{15}...\left(ii\right)$

On taking first and second terms from eq(ii), we get
$\frac{5}{12}=\frac{7}{EF}$
$\Rightarrow EF=\frac{7\times 12}{5}=16.8cm$

On taking first and third terms from eq(ii), we get
$\frac{5}{12}=\frac{BC}{15}$
$\Rightarrow BC=\frac{5\times 15}{12}=6.25cm$

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