The sides of a triangle are $6$ cm, $11$ cm and $15$ cm. Construct incircle and find the radius of the triangle.
• A
$\dfrac{5\sqrt{2}}{4}$ cm
• B
$3\sqrt{2}$ cm
• C
$6\sqrt{2}$ cm
• D
$\dfrac{4\sqrt{2}}{5}$ cm

#### Solution

Let $r$ be the radius of the incircle.
Then, $OF = OE = OD = r$
Given, $a = 6$ cm, $b = 11$ cm, $c = 15$ cm
$\therefore, s = \dfrac{a+b+c}{2}$
$\therefore s = 16$ cm
By Heron's formula, we have
Area of $\triangle ABC = \sqrt s(s-a)(s-b)(s-c)$
$= \sqrt 16(16 - 6)(16 - 11)(16 - 15)$
$= 20\sqrt 2 cm^2$
$OA, OB, OC$ divide the triangle into three parts,
So, Area of $\triangle ABC =$ Area of $\triangle AOC +$ Area of $\triangle AOB +$ Area of $\triangle BOC$
$\Rightarrow 20\sqrt 2 = \dfrac {1}{2}\times r\times 6 + \dfrac {1}{2}\times r\times 15 + \dfrac {1}{2}\times r\times 11$
$\Rightarrow 20\sqrt 2 = \dfrac{6+15+11}{2} r$
$\Rightarrow 20\sqrt 2 = 16r$
Therefore, radius $= \dfrac {5\sqrt 2}{4}$