Question

Find the area of the smaller segment cut-off from the circle \( x^2 + y^2 = 25 \) by \( x = 3 \):

• A. \( 75 \cos^{-1}\left( \frac{3}{5} \right) - 12 \)
• B. \( 25 \cos^{-1}\left( \frac{3}{5} \right) - 24 \)
• C. \( 25 \cos^{-1}\left( \frac{3}{5} \right) - 12 \)
• D. \( 25 \cos^{-1}\left( \frac{3}{5} \right) - 6 \)
• E. \( 50 \cos^{-1}\left( \frac{3}{5} \right) - 12 \)

Hint:

For segment area problems, use the formula involving the central angle \( \theta \) and the radius \( r \). Remember to simplify the expression carefully for the correct result.

Answer 1

The Correct Option is C

The area of the segment of a circle is given by the formula: \[ A = \frac{1}{2} r^2 \left( \theta - \sin \theta \right) \] where: - \( r = 5 \) (the radius of the circle), - \( \theta = \cos^{-1}\left( \frac{3}{5} \right) \), which is the central angle corresponding to the segment. 
First, calculate the angle \( \theta \). The cosine inverse of \( \frac{3}{5} \) gives \( \theta \). 
Now, substitute the values of \( r \) and \( \theta \) into the formula for the area: \[ A = \frac{1}{2} \times 5^2 \left( \cos^{-1}\left( \frac{3}{5} \right) - \sin \left( \cos^{-1}\left( \frac{3}{5} \right) \right) \right) \] This simplifies to: \[ A = 25 \cos^{-1}\left( \frac{3}{5} \right) - 12 \] Thus, the area of the segment is \( 25 \cos^{-1}\left( \frac{3}{5} \right) - 12 \).