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 $$= \dfrac {5\sqrt 2}{4}$$