Question

In $\Delta ABC,$ if $AB=10,BC=8,CA=12$ and a point $D$ is taken on $AB$ such that $AD:DB$ is $3:2,$ then the length of $CD$ is

• A. $\sqrt{72}$
• B. $\sqrt{36}$
• C. $\sqrt{24}$
• D. $\sqrt{84}$

Answer 1

The correct option is A $\sqrt{72}$

$\frac{\text{AD}}{\text{DB}}=\frac{3}{2}$
And $\frac{\text{AD}}{\text{DB}}=\frac{\text{AC}}{\text{BC}}=\frac{3}{2}$
$\Rightarrow$ $CD$ is angle bisector

$CD=\frac{2AC\cdot BC}{AC+BC}\cdot \cos \frac{C}{2}$ $\dots \left(1\right)$

$\cos C=\frac{{12}^2+8^2-{10}^2}{2\cdot 12\cdot 8}=\frac{9}{16}$
$\Rightarrow 2{\cos }^2\frac{C}{2}-1=\frac{9}{16}$
$\because \cos C=2{\cos }^2\frac{C}{2}-1$
$\therefore {\cos }^2\frac{C}{2}=\frac{1+\cos c}{2}=\frac{1+\frac{9}{16}}{2}=\frac{25}{32}$
$\Rightarrow \cos \frac{C}{2}=\frac{5}{4\sqrt{2}}$

From $\left(1\right)$
$CD=\frac{2\times 12\times 8\times 5}{20\times 4\sqrt{2}}=6\sqrt{2}=\sqrt{72}$
Hence, $CD=\sqrt{72}$