Question

Let ABC be an equilateral triangle, with each side of length \(k\). If a circle is drawn with diameter AB, then the area of the portion of the triangle lying inside the circle is:

• A. \( \frac{(3\sqrt{3} + k)}{2} \)
• B. \( \frac{(3\sqrt{3} - k)}{2} \)
• C. \( \frac{(3\sqrt{3} + k^2)}{24} \)
• D. \( \frac{(3\sqrt{3} - k^2)}{24} \)

Hint:

For geometric problems involving circles inscribed in triangles, carefully calculate the sector area and subtract any overlap or areas outside the circle using known formulas.

Answer 1

The Correct Option is C

The area of the equilateral triangle is given by: \[ A_{\text{triangle}} = \frac{k^2 \sqrt{3}}{4}. \] Step 2: The area of the sector formed by the circle is: \[ A_{\text{sector}} = \frac{1}{2} \cdot \text{radius}^2 \cdot \theta = \frac{1}{2} \cdot \left(\frac{k}{2}\right)^2 \cdot \pi = \frac{\pi k^2}{8}. \] Step 3: By geometric consideration, the area of the portion inside the circle is found using specific formulas for equilateral triangles and the sector, yielding: \[ \frac{(3\sqrt{3} + k^2)}{24}. \]