Question

The perimeter of a sector of a circle of area $25\pi \text{ c}{{\text{m}}^{2}}$ is 20 cm. Find the area of the sector.

Answer 1

Hint: We first find the radius of the circle from the given area of the circle. We use that radius to find the perimeter of the circle. Then assuming sector value, we find the ratio of the sector to the main circle and use that to find the area of the sector.

Complete step-by-step solution:
It is given that the area of the circle is $25\pi \text{ c}{{\text{m}}^{2}}$. Let’s assume that the radius of the circle is r cm.
That’s why the area of the circle with radius r cm will be $\pi {{r}^{2}}\text{ c}{{\text{m}}^{2}}$. We equate this value with the value of $25\pi \text{ c}{{\text{m}}^{2}}$ and get $\pi {{r}^{2}}=25\pi $.
Now we solve the equation and get ${{r}^{2}}=25$ which gives $r=5$. Value of r can’t be negative as that’s a radius value.
We now find the perimeter of the circle. The formula of perimeter for a circle with radius r is $2\pi r$ cm.
We place the value $r=5$ in that and get the perimeter as $2\pi r=2\pi \times 5=10\pi $ cm.
Now the area and the perimeter of a sector or a part of a circle changes with the same ratio.
Let the sector be $\dfrac{1}{x}$ of the main circle. This means the perimeter of that sector will also be $\dfrac{1}{x}$ of the main circle’s perimeter.
The perimeter is $\dfrac{10\pi }{x}$ cm which is equal to 20. So, $\dfrac{10\pi }{x}=20$.
This gives $x=\dfrac{\pi }{2}$.
Now the area of that sector will also be $\dfrac{1}{x}$ of the main circle’s area. The area of the full circle is $25\pi \text{ c}{{\text{m}}^{2}}$.
The area of the sector of the circle is $\dfrac{25\pi }{x}$. We place the value of x and get the area as
$\dfrac{25\pi }{x}=\dfrac{25\pi }{\dfrac{\pi }{2}}=50$.
The area of the sector is $50\text{ c}{{\text{m}}^{2}}$.

Note: We also can use the angular sector instead of partition but that will only complicate the solution as all the relations are the derivation of the sector partition ratio. The radius, area, perimeter is in linear relation with the ratio.