21)

In \(\triangle ABC\), \(\angle C = 90^\circ\), AC = 5 cm and BC = 12 cm. The bisector of \( \angle A \) meets BC at D. What is the length of AD ?

SSC CGL 2020

A)

\(\frac{2}{3}\sqrt{13} \) cm

B)

\(2\sqrt{13}\) \(\)cm

C)

\(\frac{4}{3}\sqrt{13}\) cm

D)

\(\frac{5\sqrt{13}}{3}\) cm

Correct Option: D

\(\frac{5\sqrt{13}}{3}\) cm

---

By the Pythagoras theorem,
\((AB)^2 = (AC)^2 + (BC)^2\); ⇒ \((AB)^2 = (5)^2 + (12)^2\);
⇒  AB = 13 cm;
By angle bisector theorem,
\(\frac{AB}{BD} = \frac{AC}{CD}\);
Let CD be x cm.
\(\frac{13}{12 - x} = \frac{5}{x};\)
⇒  x = 60/18 = 10/3;
In \(\triangle ACD\),
\((AD)^2 = (AC)^2 + (CD)^2;\)
⇒ \((AD)^2 = (5)^2 + (\frac{10}{3})^2\);  ⇒  AD = \(\frac{5\sqrt13}{3}\)