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प्रश्न
In the following figure, If ABC is an equilateral triangle, then shaded area is equal to
पर्याय
\[\left( \frac{\pi}{3} - \frac{\sqrt{3}}{4} \right) r^2\]
\[\left( \frac{\pi}{3} - \frac{\sqrt{3}}{2} \right) r^2\]
\[\left( \frac{\pi}{3} + \frac{\sqrt{3}}{4} \right) r^2\]
\[\left( \frac{\pi}{3} + \sqrt{3} \right) r^2\]
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उत्तर
We have given that ABC is an equilateral triangle.
`∴ ∠A=60°`
As we know that,`∠BCA=1/2 m (∠BOC)`
`∴ 60=1/2 m (BOC)`
`m(∠BOC)=120°`
Area of the shaded region = area of the segment BC.
Let `∠BOC=θ`
∴ Area of the segment= `(piθ/360-sin θ/2 cos θ/2)`
Substituting the values we get,
Area of the segment= `((pixx120)/360-sin60cos60)r^2`
∴ Area of the segment=`(pi/3-sin60cos60)r^2`
Substituting `sin 60=sqrt3/2` and `60=1/2`we get,
∴ Area of the segment=`(pi/3-1/2xxsqrt3/2)r^2`
∴ Area of the segment=`(pi/3-sqrt3/4)r^2`
Therefore, area of the shaded region is` (pi/3-sqrt3/4)r^2`.
