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Perimeter and Area of Trapezium

In the follow...

Question

In the following figure, the area of segment ACB is

(a) $(\frac{π}{3} - \frac{\sqrt{3}}{2}) r^{2}$

(b) $(\frac{π}{3} + \frac{\sqrt{3}}{2}) r^{2}$

(c) $(\frac{π}{3} - \frac{\sqrt{2}}{3}) r^{2}$

(d) None of these

Figure

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Solution

We have to find area of segment ACB.

We know that.

Substituting the values we get,

Substituting and we get,

Therefore, area of the segment ACB is.

Hence, the correct answer is option (d).

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Q. In the following figure, If ABC is an equilateral triangle, then shaded area is equal to

(a)

(\frac{π}{3} - \frac{\sqrt{3}}{4}) r^{2}

(\frac{π}{3} - \frac{\sqrt{3}}{2}) r^{2}

(\frac{π}{3} + \frac{\sqrt{3}}{4}) r^{2}

(\frac{π}{3} + \sqrt{3}) r^{2}

\int \frac{1}{\cos x + \sqrt{3} \sin x} d x

is equal to

(a)

\log \tan (\frac{π}{3} + \frac{x}{2}) + C

\log \tan (\frac{x}{2} - \frac{π}{3}) + C

\frac{1}{2} \log \tan (\frac{x}{2} + \frac{π}{3}) + C

(d) none of these

\sqrt{3} \cos θ + \sin θ = \sqrt{2}

, then general value of θ is
(a)

n π + {(- 1)}^{n} \frac{π}{4}, n \in Z

{(- 1)}^{n} \frac{π}{4} - \frac{π}{3}, n \in Z

n π + \frac{π}{4} - \frac{π}{3}, n \in Z

n π + {(- 1)}^{n} \frac{π}{4} - \frac{π}{3}, n \in Z

\sqrt{3} \cos x + \sin x = \sqrt{2}

, then general value of x is
(a)

n π + {(- 1)}^{n} \frac{π}{4}, n \in Z

{(- 1)}^{n} \frac{π}{4} - \frac{π}{3}, n \in Z

n π + \frac{π}{4} - \frac{π}{3}, n \in Z

n π + {(- 1)}^{n} \frac{π}{4} - \frac{π}{3}, n \in Z

\cot x + \cot (\frac{π}{3} + x) + \cot (\frac{π}{3} - x) = 3 \cot 3 x

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Perimeter and Area of Trapezium

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