Question

What is the area of a circle that is inscribed in a right angled triangle with sides of length $8m,{\rm{ }}15m {\rm{ \ and\ \ }}17m$?
a) $9\pi {m^2}$
b) $11\dfrac{1}{9}\pi {m^2}$
c) $14\pi {m^2}$
d) $10\pi {m^2}$

Answer 1

Hint: In this question first draw the figure. This will give you a clear picture of what we have to find out. Here, we will divide the given triangle into three triangles to find out the radius of the incircle. So, draw a perpendicular line from the centre of the circle to sides of the given triangle to get three triangles. And, the sum of the area of all three triangles is equal to the given triangle. Hence, radius can be found consequently find the area of incircle.



Complete step-by-step answer:
In the question it is given that a right angle triangle with sides of length $8m,{\rm{ }}15m {\rm{ \ and\ \ }}17m$.
Let,
 $\begin{array}{l}
{\rm{base, }}b = 8m\\
{\rm{height, }}a = 15m\\
{\rm{hypotenuse, }}c = 17m
\end{array}$
Here, the longest side must be considered as a hypotenuse.
It is also given that circle is inscribed in a right angled triangle as shown below,

In the above figure, draw perpendicular from the centre ‘O’ to all the three sides of the triangle and name them as E, F and D. Also join all the vertices to all the centre of the circle as shown below,

Now we will find out area of triangle using Heron’s formula,
${\rm{Area\ \ of\ \ triangle}} = \sqrt {(s)(s - a)(s - b)(s - c)} $
Where, ‘$a$’, ‘$b$’, and ‘$c$’ are the sides of a triangle and ‘$s$’ is the semi perimeter. We will find ‘$s$’ using formula $s = \dfrac{{a + b + c}}{2}$
Substituting ‘$a$’, ‘$b$’, and ‘$c$’ as sides of triangle,
$ \Rightarrow s = \dfrac{{15 + 8 + 17}}{2}$
$ \Rightarrow s = \dfrac{{40}}{2}$
$ \Rightarrow s = 20m$
Then equation (1) becomes,
$ \Rightarrow {\rm{Area\ \ of\ \ triangle}} = \sqrt {(20)(20 - 15)(20 - 8)(20 - 17)} $
Simplifying the above equation,
$ \Rightarrow {\rm{Area \ \ of\ \ triangle}} = \sqrt {(20)(5)(12)(3)} $
$ \Rightarrow {\rm{Area \ \ of\ \ triangle}} = \sqrt {3600} $
$ \Rightarrow {\rm{Area \ \ of\ \ triangle}} = 60{m^2}$
In the figure we can clearly see that,
$ \Rightarrow {\rm{Area of }}\Delta {\rm{AOB}} + {\rm{Area of }}\Delta {\rm{BOC}} + {\rm{Area of }}\Delta {\rm{AOC}} = {\rm{Area of }}\Delta {\rm{ABC}}$
We will use this concept to find the radius of the given incircle.
We know that, ${\rm{Area of triangle}} = \dfrac{1}{2} \times {\rm{base}} \times {\rm{height}}$
So equation (2) becomes
$ \Rightarrow \left( {\dfrac{1}{2} \times {\rm{AB}} \times r} \right) + \left( {\dfrac{1}{2} \times {\rm{BC}} \times r} \right) + \left( {\dfrac{1}{2} \times {\rm{AC}} \times r} \right) = 60$
For all the triangles onto LHS, height is the same i.e., ‘$r$’.
Solving equation (3) by substituting,
${\rm{AB}} = 15m$, ${\rm{BC}} = 8m$, ${\rm{AC}} = 17m$.
$ \Rightarrow \left( {\dfrac{1}{2} \times 15 \times r} \right) + \left( {\dfrac{1}{2} \times 8 \times r} \right) + \left( {\dfrac{1}{2} \times 17 \times r} \right) = 60$
Multiplying above equation by ‘2’ we get,
$ \Rightarrow 15r + 8r + 17r = 120$
$ \Rightarrow 40r = 120$
$ \Rightarrow r = \frac{{120}}{{40}}$
$ \Rightarrow r = 3m$
$\therefore $ radius of the incircle is $3m$
Now, we need to find out the area of the incircle using formula.
${\rm{Area of incircle}} = \pi {r^2}$
Substituting $r = 3$ we get,
$ \Rightarrow {\rm{Area \ \ of\ \ incircle}} = \pi {(3)^2}$
$ \Rightarrow {\rm{Area\ \ of\ \ incircle}} = 9\pi {m^2}$
Thus, option (a) is correct.

Note: Whenever we face such a type of problem, first draw the pictorial representation of the given problem. Here, we have used both. Heron’s formula and basic ${\rm{Area = }}\dfrac{1}{2} \times {\rm{base}} \times {\rm{height}}$ formula to find area of triangle. When all the sides of a triangle are given then use Heron’s formula to find the area of the triangle.