Question

In a $△ABC,$ right angled at $A$. The radius of the inscribed circle is $2$cm.Radius of the circle touching the side $BC$ and also sides $AB$ and $AC$ produced is $15$cm.The length of the side $BC$ measured in cm is

• A. $11$
• B. $12$
• C. $13$
• D. $14$

Answer 1

The correct option is C $13$

$\because$Radius of the inscribed circle $r=\frac{△}{s}=2$
and Radius of the touching circle$r_1=\frac{△}{s-a}=15$
Thus, $\frac{s-a}{s}=\frac{r}{r_1}=\frac{2}{15}$
or $1-\frac{a}{s}=\frac{2}{15}$
or $\frac{a}{s}=1-\frac{2}{15}=\frac{13}{15}$
$\therefore s=\frac{15}{13}a$
So, $\frac{△}{s}=2$
$\Rightarrow △=2s$
$\Rightarrow △=2\times \frac{15}{13}a=\frac{30}{13}a$
$\Rightarrow \frac{1}{2}bc=\frac{30}{13}a$
$\Rightarrow bc=\frac{60}{13}a$ .............$\left(1\right)$
$\because A=\frac{\pi }{2}$(given)
Also, $b+c=2s-a=\frac{30}{13}a-a=\frac{17}{13}a$ ..............$\left(2\right)$
Solving eqns$\left(1\right)$ and $\left(2\right)$, we get
$b+c=\frac{17}{13}a$
since $12\times 5=60$ and $12+5=17$ we have
$b=\frac{12}{13}a$ and $c=\frac{5}{13}a$
Substituting for $a,b,c$ we get
Also, $\frac{△}{s}=\frac{\frac{bc}{2}}{\frac{a+b+c}{2}}=\frac{bc}{a+b+c}$
$\Rightarrow 2=\frac{2}{13}a$
$\Rightarrow a=13$