Question

In the given figure, ∆ABC is right-angled at A. Find the area of the shaded region if AB = 6 cm, BC = 10 cm and O is the centre of the incircle of ∆ABC.

Answer 1

Using Pythagoras' theorem for triangle ABC, we have:

$C{A}^2+A{B}^2=B{C}^2$

$$\begin{gathered} CA=\sqrt{B{C}^2-A{B}^2} \\ =\sqrt{100-36} \\ =\sqrt{64} \\ =8\mathrm{cm} \end{gathered}$$

Now, we must find the radius of the incircle. Draw OE, OD and OF perpendicular to AC, AB and BC, respectively.



Consider quadrilateral AEOD.
Here,
$EO=OD(\mathrm{Both}\mathrm{are}\mathrm{radii}.)$

Because the circle is an incircle, AE and AD are tangents to the circle.
$\angle AEO=\angle ADO=90°$

Also,
$\angle A=90°$
Therefore, AEOD is a square.
Thus, we can say that $AE=EO=OD=AD=r$.

$$\begin{gathered} CE=CF=8-r \\ BF=BD=6-r \\ CF+BF=10 \\ \Rightarrow \left(8-r\right)+\left(6-r\right)=10 \\ \Rightarrow 14-2r=10 \\ \Rightarrow r=2\mathrm{cm} \end{gathered}$$

Area of the shaded part = Area of the triangle $-$ Area of the circle
$$\begin{gathered} =\left\{\frac{1}{2}\times 6\times 8\right\}-\left\{\mathrm{\pi }\times 2\times 2\right\} \\ =24-12.56 \\ =11.44{\mathrm{cm}}^2 \end{gathered}$$