1950 AHSME Problems/Problem 8: Difference between revisions
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==Solution== | ==Solution== | ||
Increasing by < | Increasing by <imath>100\%</imath> is the same as doubling the radius. If we let <imath>r</imath> be the radius of the old circle, then the radius of the new circle is <imath>2r.</imath> | ||
Since the area of the circle is given by the formula < | Since the area of the circle is given by the formula <imath>\pi r^2,</imath> the area of the new circle is <imath>\pi (2r)^2 = 4\pi r^2.</imath> The area is quadrupled, or increased by <imath>\boxed{\mathrm{(C) }300\%.}</imath> | ||
==Solution== | |||
\[ | |||
\textbf{Alternate Solution:} | |||
\] | |||
Let the original radius be \( r \) and the original area be | |||
\[ | |||
A_1 = \pi r^2. | |||
\] | |||
A \(100\%\) increase in radius means the new radius is | |||
\[ | |||
r_2 = r + 100\% \times r = 2r. | |||
\] | |||
The new area is | |||
\[ | |||
A_2 = \pi (r_2)^2 = \pi (2r)^2 = 4\pi r^2. | |||
\] | |||
Now, compute the percent increase in area: | |||
\[ | |||
\text{Percent Increase} = \frac{A_2 - A_1}{A_1} \times 100\% | |||
= \frac{4\pi r^2 - \pi r^2}{\pi r^2} \times 100\% | |||
= 3 \times 100\% = 300\%. | |||
\] | |||
\[ | |||
\boxed{\text{The area is increased by } 300\%.} | |||
\] | |||
==See Also== | ==See Also== | ||
Revision as of 17:44, 9 November 2025
Problem
If the radius of a circle is increased 
, the area is increased:
Solution
Increasing by is the same as doubling the radius. If we let 
be the radius of the old circle, then the radius of the new circle is 
Since the area of the circle is given by the formula 
the area of the new circle is 
The area is quadrupled, or increased by 
Solution
\[ \textbf{Alternate Solution:} \]
Let the original radius be \( r \) and the original area be \[ A_1 = \pi r^2. \]
A \(100\%\) increase in radius means the new radius is \[ r_2 = r + 100\% \times r = 2r. \]
The new area is \[ A_2 = \pi (r_2)^2 = \pi (2r)^2 = 4\pi r^2. \]
Now, compute the percent increase in area: \[ \text{Percent Increase} = \frac{A_2 - A_1}{A_1} \times 100\% = \frac{4\pi r^2 - \pi r^2}{\pi r^2} \times 100\% = 3 \times 100\% = 300\%. \]
\[ \boxed{\text{The area is increased by } 300\%.} \]
See Also
| 1950 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 7 |
Followed by Problem 9 | |
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