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प्रश्न
Find the area of the smaller part of the circle x2 + y2 = a2 cut off by the line `x = a/sqrt2`
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उत्तर
The area of the smaller part of the circle, x2 + y2 = a2, cut off by the line, `x = a/sqrt2`, is the area ABCDA.
It can be observed that the area ABCD is symmetrical about x-axis.
∴ Area ABCD = 2 × Area ABC
Therefore, the area of smaller part of the circle, x2 + y2 = a2, cut off by the line, x = `a/sqrt2` is `a^2/2 (pi/2 - 1)` units
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