Question

The radius of the incircle of the triangle PQR with the three sides as $6\sqrt 2 ,6\sqrt 2 $ and $4\sqrt 2 $ is,
$\left( a \right)$ 4
$\left( b \right)$ 3
$\left( c \right)$ 8/3
$\left( d \right)$ 2

Answer 1

Hint: In this question use the concept that the radius of the incircle is the ratio of the area of the triangle to the half of the perimeter of the triangle, and use the concept that if all the sides of the triangle is given then the area of the triangle is calculated as $\sqrt {s\left( {s - a} \right)\left( {s - b} \right)\left( {s - c} \right)} $ so use these concepts to reach the solution of the question.

Complete step-by-step answer:
Given data:
Sides of the triangle is $6\sqrt 2 ,6\sqrt 2 $ and $4\sqrt 2 $
Consider the triangle PQR as shown above in the diagram.
Let, PQ = $6\sqrt 2 $, PR = $6\sqrt 2 $ and QR = $4\sqrt 2 $
Consider an incircle as shown in the above figure with center o and radius r units.
Now as we know that are area of the triangle if all three sides are given is calculated as,
$ \Rightarrow {\left( {{\text{Area}}} \right)_\Delta } = \sqrt {s\left( {s - a} \right)\left( {s - b} \right)\left( {s - c} \right)} $........................ (1)
Where, s = half of the perimeter and a, b, c are the lengths of the triangle.
Let, a = PQ, b = PR, and c = QR
Now as we know that the perimeter of any shape is the sum of all the sides of the triangle.
So the perimeter of the triangle is the sum of all the side lengths of the triangle, so the perimeter (P) of the triangle is,
$ \Rightarrow P = PQ + QR + RP$
$ \Rightarrow P = 6\sqrt 2 + 4\sqrt 2 + 6\sqrt 2 = 16\sqrt 2 $
Therefore, s = $\dfrac{{{\text{perimeter of triangle}}}}{2}$
$ \Rightarrow s = \dfrac{{16\sqrt 2 }}{2} = 8\sqrt 2 $
So the area of the triangle from equation (1) is,
$ \Rightarrow {\left( {{\text{Area}}} \right)_\Delta } = \sqrt {8\sqrt 2 \left( {8\sqrt 2 - 6\sqrt 2 } \right)\left( {8\sqrt 2 - 6\sqrt 2 } \right)\left( {8\sqrt 2 - 4\sqrt 2 } \right)} $
Now simplify this we have,
$ \Rightarrow {\left( {{\text{Area}}} \right)_\Delta } = \sqrt {8\sqrt 2 \left( {2\sqrt 2 } \right)\left( {2\sqrt 2 } \right)\left( {4\sqrt 2 } \right)} = \sqrt {256\left( 2 \right)} = 16\sqrt 2 $
Now as we know that the radius of the incircle is the ratio of the area of the triangle to the half of the perimeter of the triangle.
So the radius of the incircle is, r = $\dfrac{{{{\left( {{\text{Area}}} \right)}_\Delta }}}{s}$
Now substitute the values we have,
So the radius of the incircle is, r = $\dfrac{{16\sqrt 2 }}{{8\sqrt 2 }} = 2$
So this is the required radius of the incircle.
Hence option (d) is the correct answer.

Note: Whenever we face such types of questions first draw the pictorial representation of the problem it will help us to understand what we have to calculate, and always recall the formula of the area of the triangle if all the sides are given also the formula of the radius of the incircle which is stated above.