Question:

In the figure, AB and CD are two perpendicular diameters of a circle with center O. OD is the diameter of the smaller circle. If OA = 7 cm, find the ratio of the areas of the smaller and larger circles.

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The ratio of the areas of two circles is the square of the ratio of their radii.
Updated On: Oct 10, 2025
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Solution and Explanation

Let the radius of the larger circle be \( r_1 = OA = 7 \, \text{cm} \). The area of the larger circle is: \[ A_1 = \pi r_1^2 = \pi \times 7^2 = 49\pi \, \text{cm}^2. \]
Step 1: The diameter \( OD \) of the smaller circle is the same as the radius \( OA \) of the larger circle. Therefore, the radius of the smaller circle is: \[ r_2 = \frac{OD}{2} = \frac{7}{2} \, \text{cm}. \] The area of the smaller circle is: \[ A_2 = \pi r_2^2 = \pi \times \left( \frac{7}{2} \right)^2 = \pi \times \frac{49}{4} = \frac{49\pi}{4} \, \text{cm}^2. \]
Step 2: The ratio of the areas of the smaller and larger circles is: \[ \frac{A_2}{A_1} = \frac{\frac{49\pi}{4}}{49\pi} = \frac{1}{4}. \]
Conclusion: The ratio of the areas of the smaller and larger circles is \( \frac{1}{4} \).
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