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In the given figure , $O$ is the center of the circle and $r_1 =7cm$,$r_2=14cm,$ $\angle AOC =40^{\circ}$. Find the area of the shaded region

enter image description here

My attempt: Area of shaded region $=\pi r^2_2 - \pi r^2_1= \pi( 196-49)= 147\pi$

Is it true ?

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    $\begingroup$ You haven't used the angle so the likelihood is that it is not correct $\endgroup$ – lioness99a Feb 15 at 15:07
  • $\begingroup$ how angle can be used ? can u elaborate ? $\endgroup$ – jasmine Feb 15 at 15:08
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    $\begingroup$ See my answer below $\endgroup$ – lioness99a Feb 15 at 15:18
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    $\begingroup$ Note that $\pi r_2^2$ is the area of the larger circle, while $\pi r_1^2$ is the area of the smaller circle. So your calculation is of the area that inside the larger circle, but not in the smaller circle. I.e., it is the area of the annular ring between the two circles, which is not area shaded. $\endgroup$ – Paul Sinclair Feb 15 at 17:33
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The area of a sector of a circle with angle $\theta$ is $$ \frac{\theta}{360}\pi r^2$$

For your smaller circle, the shaded area is \begin{align}\frac{360-\theta}{360}\pi r_1^2&= \frac{360-40}{360}\pi 7^2\\ &= \frac{320}{360}\times 49\pi\\ &= \frac 89 \times 49\pi\end{align}

For the larger circle we want to calculate the whole sector area and subtract the smaller white sector:

\begin{align}\frac {\theta}{360}\pi r_2^2 - \frac{\theta}{360}\pi r_1^2 &= \frac{40}{360}\pi\times 14^2 - \frac{40}{360}\pi \times 7^2\\ &= \frac19\times 196\pi - \frac 19 \times 49\pi\\ &= \frac \pi 9 (196-49)\\ &= \frac \pi 9 \times 147\end{align}

Therefore the total shaded area is \begin{align}\frac 89 \times 49\pi+\frac \pi 9 \times 147&= \frac \pi 9(8\times 49+147)\\ &=\frac \pi 9 \times 539\\ &= \frac {539}9\pi\end{align}

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  • $\begingroup$ +1, though I think you mean "sector", not "segment". $\endgroup$ – user170231 Feb 15 at 16:37
  • $\begingroup$ Whoops, true, been a while since I actually studied this $\endgroup$ – lioness99a Feb 15 at 18:25
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Area of a circular sector: $$\frac{\pi r^2 \cdot \alpha}{360º}$$ where $\alpha =$ angle of the circular sector

With this, the shaded region of the smaller circle is $$\frac{\pi 7^2 \cdot 320º}{360º}$$

For the big circle, we have to substract the white area of the smaller circle from the circular sector of the big circle:

$$\frac{\pi 14^2 \cdot 40º}{360º}- \frac{\pi 7^2 \cdot 40º}{360º}$$

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The way you've done it is not correct. We need to treat both the shaded areas as sectors of a circle (in the case of the outer one, a sector of the large circle minus a sector of a small circle)

The correct way as far as I know would be:

$\frac{40}{360}\pi(r_2^2 - r_1^2) + \frac{360-40}{360}\pi r_1^2$

You would simplify to answer.

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Use the formula for the area of a sector (the angle is measured in radians) and the formula for the area of a circle:

$$ A=\frac{1}{2}r^2\theta $$

$$ A=\pi r^2 $$

$40^\circ$ in radians would be:

$$ 40^\circ=\frac{40\pi}{180} $$

The area of the top peice:

$$ A_1=\frac{1}{2}r_2^2\cdot \frac{40\pi}{180} - \frac{1}{2}r_1^2\cdot\frac{40\pi}{180}=\frac{\pi}{9}\left(r_2^2-r_1^2\right) $$

The area of the bottom piece:

$$ A_2=\pi r_1^2-\frac{1}{2}r_1^2\cdot\frac{40\pi}{180}=\frac{8\pi}{9}r_1^2 $$

Thus, the area of the shaded region would be:

$$ A_1+A_2=\frac{\pi}{9}\left(r_2^2-r_1^2\right)+\frac{8\pi}{9}r_1^2=\frac{\pi}{9}\left(r_2^2+7r_1^2\right)=\frac{\pi}{9}\left(14^2+7\cdot7^2\right)=\frac{539\pi}{9} $$

Answer: $A=\frac{539\pi}{9}$ square units.

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