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The area of the region bounded by the ellipse $\frac{x^2}{16} + \frac{y^2}{9} = 1$ is
A$48\pi$
B$7\pi$
C$12\pi$
D$24\pi$
Answer & Solution
Correct answer: C. $12\pi$
1. Compare with $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$: here $a^2 = 16$ and $b^2 = 9$, so $a = 4$, $b = 3$.
2. Area of an ellipse $= \pi a b$.
3. Substituting, $A = \pi \cdot 4 \cdot 3 = 12\pi$. (Using $a^2 b = 48$ instead of $ab$ gives the wrong $48\pi$.)
_Source: NCERT Class 12 Mathematics Ch 8 "Application of Integrals", p.4_
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