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
Verified
485.7k+ views
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 $$(\text{initial radius} +\text{increased radius} )$$.
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 $$(\text{initial radius} +\text{increased radius} )$$.
Recently Updated Pages
How to find how many moles are in an ion I am given class 11 chemistry CBSE
Class 11 Question and Answer - Your Ultimate Solutions Guide
Identify how many lines of symmetry drawn are there class 8 maths CBSE
State true or false If two lines intersect and if one class 8 maths CBSE
Tina had 20m 5cm long cloth She cuts 4m 50cm lengt-class-8-maths-CBSE
Which sentence is punctuated correctly A Always ask class 8 english CBSE
Trending doubts
The reservoir of dam is called Govind Sagar A Jayakwadi class 11 social science CBSE
10 examples of friction in our daily life
What problem did Carter face when he reached the mummy class 11 english CBSE
Difference Between Prokaryotic Cells and Eukaryotic Cells
State and prove Bernoullis theorem class 11 physics CBSE
What organs are located on the left side of your body class 11 biology CBSE