Question

In Fig., if ABC is an equilateral triangle, then shaded area is equal to___​

Answer 1

Step-by-step explanation:

We have given that ABC is an equilateral triangle. Area of the shaded region = area of the segment BC. Therefore, area of the shaded region is ( π 3 - 3 4 ) r 2

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Answer 2

Given:

ABC is an equilateral triangle

To find:

The area of the shaded region.

Step-by-step explanation:

Since ABC is an equilateral triangle ∠$A$ = $60^\circ$

We know that, $\angle BCA\;=\;\frac12\angle BOC$

⇒ $60^\circ\;=\;\frac12\angle BOC$

⇒ $\angle BOC\;=\;120^\circ$

Area of the shaded region = Area of the segment $BC$

Let us consider $\angle BOC\;=\;\theta$

Area of the segment = $(\mathrm\pi\frac{\mathrm\theta}{360^\circ}\;-\;\sin\;\frac{\mathrm\theta}2\cos\;\frac{\mathrm\theta}2)r^2$

On substituting the values,

∴ Area of the segment = $(\frac{\pi\;\times\;120^\circ}{360^\circ}\;-\;\sin\;60.\cos\;60) \;r^{2}$

⇒ $(\frac{\mathrm\pi}3\;-\;\sin\;60\;.\;\cos\;60)\;r^2$

Substitute $sin\;60\;=\;\frac{\sqrt{3} }{2}$ and $cos\;60\;=\;\frac{1}{2}$

⇒ $(\frac{\mathrm\pi}3\;-\;\frac{\sqrt3}2\;\times\;\frac12)\;r^2$

⇒ $(\frac{\mathrm\pi}3\;-\;\frac{\sqrt3}4)\;r^2$

∴ Area of the shaded region = $(\frac{\mathrm\pi}3\;-\;\frac{\sqrt3}4)\;r^2$

Answer:

Hence, the area of the shaded region in the figure is $(\frac{\mathrm\pi}3\;-\;\frac{\sqrt3}4)\;r^2$