Question

In Fig. 7.246, ∠ABC = 90°, AD = 15 cm and DC = 20 cm. If BD is the bisector of ∠ABC, then perimeter of ∆ABC is ________.

Answer 1

In ∆ABC, BD is the bisector of ∠ABC.

$\therefore \frac{\mathrm{AB}}{\mathrm{BC}}=\frac{\mathrm{AD}}{\mathrm{CD}}$ (An angle bisector of a triangle divides the opposite side in the ratio of the sides containing the angle)

$\Rightarrow \frac{\mathrm{AB}}{\mathrm{BC}}=\frac{15\mathrm{cm}}{20\mathrm{cm}}=\frac{3}{4}$

Let AB = 3k cm and BC = 4k cm.

In right ∆ABC,

$$\begin{gathered} {\mathrm{AB}}^2+{\mathrm{BC}}^2={\mathrm{AC}}^2\left(\mathrm{Pythagoras}\mathrm{theorem}\right) \\ \Rightarrow {\left(3k\right)}^2+{\left(4k\right)}^2={\left(15+20\right)}^2={\left(35\right)}^2 \\ \Rightarrow 9k^{\mathit{2}}+16k^2=1225 \\ \Rightarrow 25k^2=1225 \end{gathered}$$
$$\begin{gathered} \Rightarrow k^2=49 \\ \Rightarrow k=7 \end{gathered}$$

∴ AB = 3 × 7 = 21 cm and BC = 4 × 7 = 28 cm

Perimeter of ∆ABC = AB + BC + CA = 21 + 28 + 35 = 84 cm

In Fig. 7.246, ∠ABC = 90°, AD = 15 cm and DC = 20 cm. If BD is the bisector of ∠ABC, then perimeter of ∆ABC is ___84 cm____.