Question

In the given figure, ABC is a quadrant of a circle of radius 28 cm and a semicircle is drawn with BC as diameter. Find the area of the shaded region in the figure.

Answer 1

Hint: Use the formula: Area of shaded region = Area of triangle ABC + Area of semicircle with diameter BC − Area of quadrant ABC. Use $\dfrac{1}{2}\times $ base $\times $ height to find area of triangle ABC, \[\dfrac{\pi {{r}^{2}}}{2}\] to find area of semi-circle, and \[\dfrac{\pi {{R}^{2}}}{4}\] to find the area of the quadrant ABC. Find these values and substitute in the first equation to get the final answer.

Complete step by step answer:
In this question, we are given that ABC is a quadrant of a circle of radius 28 cm and a semicircle is drawn with BC as diameter.
Using this information, we need to find the area of the shaded region in the figure.

We can clearly see from the figure that:
Area of shaded region = Area of triangle ABC + Area of semicircle with diameter BC − Area of quadrant ABC …(1)
We will calculate the value of each term separately and then substitute the values in the above equation.
We will first find the area of triangle ABC.
We can see that triangle ABC is a right angled triangle.
Also, both of its sides are equal to the radius of the quadrant = 28 cm.
We know that Area of a triangle = $\dfrac{1}{2}\times $ base $\times $ height.
For triangle ABC, we have base = height = 28 cm.
So, area of triangle ABC = $\dfrac{1}{2}\times 28\times 28=392c{{m}^{2}}$ …..…(2)
Now, we will use Pythagoras theorem on triangle ABC to find the length of BC.
$A{{B}^{2}}+A{{C}^{2}}=B{{C}^{2}}$
${{28}^{2}}+{{28}^{2}}=B{{C}^{2}}$
$2\times {{28}^{2}}=B{{C}^{2}}$
$BC=28\sqrt{2}cm$
So, the radius of the semi-circle, $r=\dfrac{28\sqrt{2}}{2}=14\sqrt{2}cm$
Area of semi-circle with diameter BC = \[\dfrac{\pi {{r}^{2}}}{2}=\dfrac{\pi \times 14\sqrt{2}\times 14\sqrt{2}}{2}=616c{{m}^{2}}\] …(3)
Now, area of quadrant ABC = \[\dfrac{\pi {{R}^{2}}}{4}\]
Here, R is the radius of the quadrant ABC = 28 cm
So, area of quadrant ABC = \[\dfrac{\pi {{R}^{2}}}{4}=\dfrac{\pi \times 28\times 28}{4}=616c{{m}^{2}}\] …(4)
Now, we will substitute equations (2), (3), and (4) in equation (1).
Area of shaded region = Area of triangle ABC + Area of semicircle with diameter BC − Area of quadrant ABC
Area of shaded region = \[392+616-616=392c{{m}^{2}}\]
This is the final answer.

Note: In this question, it is very important to note that we took triangle ABC as a right angled triangle. This is because it is a part of the quadrant ABC and a quadrant is one fourth of a circle which means its centre is a right angle.