Question

If the radius of the circle is increased by 50% then by how much the area of the circle is increased?
A. 125%
B. 100%
C. 25%
D. 75%

Answer 1

Hint: In this question it is given that the radius of the circle is increased by 50% then we have to find that by how much the area of the circle is increased w.r.t the previous area. So to find the solution we need to find the initial area and area of the area of the increased circle, and after that we have to calculate the increased percentage.
So for this we need to know the formula of area of a circle,
Area = $\pi r^{2}$, where r be the radius.

Complete step-by-step answer:
Let the initial radius of a circle is r.
Therefore, the area$\left( A_{1}\right) =\pi r^{2}$......(1)
Now the radius is increased by 50%,
Therefore the new radius $r_{2}$= r + 50% of r =$r+\dfrac{50}{100} r$=$r+\dfrac{r}{2}$=$\dfrac{3r}{2}$
Therefore, the new area of the circle,
$A_{2}=\pi r^{2}_{2}$
$=\pi \left( \dfrac{3r}{2} \right)^{2}$
$=\pi \dfrac{9r^{2}}{4}$
$=\dfrac{9}{4} \pi r^{2}$

Now the as we know that,
$\text{Increase percentage} =\dfrac{\text{Increase value} }{\text{Original value} } \times 100\%$
Therefore, increase area = $\dfrac{\left( A_{2}-A_{1}\right) }{A_{1}} \times 100\%$
$=\dfrac{\left( \dfrac{9}{4} \pi r^{2}-\pi r^{2}\right) }{\pi r^{2}} \times 100\%$
$=\dfrac{\left( \dfrac{9}{4} -1\right) \pi r^{2}}{\pi r^{2}} \times 100\%$
$=\left( \dfrac{9}{4} -1\right) \times 100\%$
$=\left( \dfrac{5}{4} \right) \times 100\%$
$=125\%$
Hence, the correct option is option A.

Note: To solve this type of question you need to know that if you have given that the radius increased by x% , then the increased radius is $\left( \dfrac{x}{100} \times \text{radius} \right)$ and the new radius will be $(\text{initial radius} +\text{increased radius} )$.