Question

63. If the radius of a circle is increasod by 50%, then its area will increase by which
of the following percentages?
(1) 50%
(2) 75%
(3) 100%
(4) 125%

Answer 1

✪ᴀɴsᴡᴇʀ✪

• $\boxed{\bf{Area\:increases\:by\:125\%.}}$

☆ɢɪᴠᴇɴ☆

• Raduis of a circle is increased by 50%.

★ᴛᴏ ғɪɴᴅ★

• Percentage increase in area.

⍟ᴄᴀʟᴄᴜʟᴀᴛɪᴏɴ⍟

Let

• $R_1, A_1$ be original radius and area of the circle.

• $R_2, A_2$ be new radius and area of the circle.

Since radius in increased by 50%, we have

$\:\:\:\:\: R_2 = R_1 + 50\%\:of\:R_1$

$\implies R_2 = R_1 + \frac{50}{100}R_1$

$\implies R_2 = R_1 + \frac{R_1}{2 }$

$\implies R_2 = \frac{3R_1}{2 }$

$\implies \frac{R_2}{R_1} = \frac{3}{2}$

Now,

$\:\:\:\:\: \frac{A_2}{A_1} = \frac{\pi R_2^2}{\pi R_1^2}$

$\implies \frac{A_2}{A_1} = \left[\frac{R_2}{R_1}\right]^2$

$\implies \frac{A_2}{A_1} = \left[\frac{3}{2}\right]^2$

$\implies \frac{A_2}{A_1} = \frac{9}{4}$

$\implies \frac{A_2}{A_1} - 1 = \frac{9}{4} - 1$

$\implies \frac{A_2-A_1}{A_1} = \frac{9-4}{4}$

$\implies \frac{A_2-A_1}{A_1} = \frac{5}{4}$

$\implies \frac{A_2-A_1}{A_1} \times 100 = \frac{5}{4}\times 100$

$\implies \boxed{ \bf{\frac{A_2-A_1}{A_1} \times 100 = 125\%}}$

Thus, area increases by 125%.