Question

Find the area of the shaded region in the figure given below

Given that AC = 24cm, BC = 10cm, and O is the centre of the circle (Use $\pi =3.14$)

Answer 1

- Hint: Find the area of the triangle ABC. Use the property that angle in a semicircle is $90{}^\circ $. The area of the shaded region is the area of the triangle subtracted from the area of the semicircle—use area of the circle $=\pi {{r}^{2}}$. Use Pythagora's theorem.

Complete step-by-step solution -

Since AOB is a diameter of the circle we have $\angle ACB=90{}^\circ $ because “the angle in a semicircle is $90{}^\circ $”
So, $\Delta ABC$ is a right-angled triangle right angled at C.
We know that in a right-angle triangle ABC right angled at B $A{{B}^{2}}+B{{C}^{2}}=A{{C}^{2}}\text{ (Pythagora }\!\!'\!\!\text{ s Theorem)}$
So we have in $\Delta ABC$ $A{{C}^{2}}+B{{C}^{2}}=A{{B}^{2}}$
Given AC = 24cm and BC = 10cm
Substituting we get
$\begin{align}
  & A{{B}^{2}}={{24}^{2}}+{{10}^{2}} \\
 & =576+100=676 \\
\end{align}$
So $AB=\sqrt{676}=26$.
Since AB is the diameter of the circle we have 2r = AB =26, i.e. r = 13cm.
Now we know that if the lengths of legs of a right-angled triangle are a and b, then the area of the triangle is $\dfrac{1}{2}ab$.
Using the above formula, we get
$ar\left( \Delta ABC \right)=\dfrac{1}{2}\times 24\times 10=120c{{m}^{2}}$
Hence area of triangle ABC = 120 sq-cm
Also, the area of a circle $=\pi {{r}^{2}}$
Hence the area of a semicircle $=\dfrac{\pi {{r}^{2}}}{2}$
Using the above formula, we have the area of the semicircle $=\dfrac{\pi \times {{13}^{2}}}{2}=\dfrac{3.14\times 169}{2}=265.33c{{m}^{2}}$.
Hence the area of the semicircle = 265.33 sq-cm
Hence the area of the shaded region = area of the semicircle – the area of the triangle = 265.33-120 = 145.33 sq-cm
Hence the area of the shaded region = 145.33 sq-cm.

Note: [1] The area of the shaded regions cannot be calculated individually easily.
[2] However, the combined area of the two segments was calculated easily by noting the fact that the sum of areas of segments is the area of the semicircle – the area of the triangle. Both the area of the triangle and the area of the semicircle can be easily calculated.
[3] Although we have used the property that the area of a right angle triangle is $\dfrac{1}{2}ab$ where a and b are the lengths of the legs of the right-angled triangle, we can calculate the area of the triangle using heron's formula also.
Here we have a = 24 , b = 10 and c = 26
So $s=\dfrac{a+b+c}{2}=\dfrac{24+10+26}{2}=\dfrac{60}{2}=30$
We know the area of a triangle $=\sqrt{s\left( s-a \right)\left( s-b \right)\left( s-c \right)}$
Using, we get
$\sqrt{30\left( 30-24 \right)\left( 30-10 \right)\left( 30-26 \right)}=\sqrt{30\times 6\times 20\times 4}=\sqrt{14400}=120$
Hence the area of the triangle ABC = 120 sq-cm.