Question

The sides of a triangle are $6$ cm, $11$ cm and $15$ cm. Construct incircle and find the radius of the triangle.

• A. $\frac{5\sqrt{2}}{4}$ cm
• B. $3\sqrt{2}$ cm
• C. $6\sqrt{2}$ cm
• D. $\frac{4\sqrt{2}}{5}$ cm

Answer 1

Answer: (A)

$\frac{5\sqrt{2}}{4}$ cm

Let $r$ be the radius of the incircle.

Then, $OF=OE=OD=r$

Given, $a=6$ cm, $b=11$ cm, $c=15$ cm

$\therefore ,s=\frac{a+b+c}{2}$

$\therefore s=16$ cm

By Heron's formula, we have

Area of $△ABC=\sqrt{s}(s-a)(s-b)(s-c)$

$=\sqrt{1}6(16-6)(16-11)(16-15)$

$=20\sqrt{2}c{m}^2$

$OA,OB,OC$ divide the triangle into three parts,

So, Area of $△ABC=$ Area of $△AOC+$ Area of $△AOB+$ Area of $△BOC$

$\Rightarrow 20\sqrt{2}=\frac{1}{2}\times r\times 6+\frac{1}{2}\times r\times 15+\frac{1}{2}\times r\times 11$

$\Rightarrow 20\sqrt{2}=\frac{6+15+11}{2}r$

$\Rightarrow 20\sqrt{2}=16r$

Therefore, radius $=\frac{5\sqrt{2}}{4}$