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jn0r

Two large circles (each of radius 0.5 cm) and two smaller circles are inscribed into a larger circle of radius 1cm. Find area of shaded part of the figure.

A) \(\frac{5\pi}{18}\)
B) \(2\pi\)
C) \(5\pi\)
D) \(4\pi\)
E) \(2\pi/3\)

Attachment:
Screenshot_7.png


Of course we can find the radius by joining the Center of each circle.

But the choices give us answer immediately. A 20 seconds analysis.

The area of the entire figure = \(\pi r^2=\pi * 1=\pi\)
B, C and D can be eliminated immediately.

Clearly, the unshaded part is less than the unshaded part, thus the area of shaded part has to be less than \(\frac{\pi }{2}\)

Only A left.

I don't see how the radius of the smaller circle is calculated. Could you explain it? Thanks a lot.
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jn0r

Two large circles (each of radius 0.5 cm) and two smaller circles are inscribed into a larger circle of radius 1cm. Find area of shaded part of the figure.

A) \(\frac{5\pi}{18}\)
B) \(2\pi\)
C) \(5\pi\)
D) \(4\pi\)
E) \(2\pi/3\)

Let us join the centre A and C to get hypotenuse AC.
Draw a perpendicular on the diameter to meet at B

So ABC is a right angled triangle, where AC=0.5+r, BC=0.5 and AB=larger radius-r=1-r

Pythagorus theorem tells us \((1-r)^2+0.5^2=(0.5+r)^2........1+r^2-2r+0.5^2=0.5^2+r+r^2......r=\frac{1}{3}\)

Area of all the 4 circles =\(2*\pi *\frac{1}{2}^2+2*\pi *\frac{1}{3}^2=\pi (\frac{1}{2}+\frac{2}{9})=\frac{13\pi }{18}\)

Thus area of shaded region = \(\pi *1^2-\frac{13\pi }{18}=\frac{5\pi }{18}\)

A

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541E5B9A-8DA3-486B-BFE2-C690B49A08FE.jpeg
541E5B9A-8DA3-486B-BFE2-C690B49A08FE.jpeg [ 27.7 KiB | Viewed 1884 times ]

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chetan2u
jn0r

Two large circles (each of radius 0.5 cm) and two smaller circles are inscribed into a larger circle of radius 1cm. Find area of shaded part of the figure.

A) \(\frac{5\pi}{18}\)
B) \(2\pi\)
C) \(5\pi\)
D) \(4\pi\)
E) \(2\pi/3\)

Attachment:
The attachment Screenshot_7.png is no longer available


Of course we can find the radius by joining the Center of each circle.

But the choices give us answer immediately. A 20 seconds analysis.

The area of the entire figure = \(\pi r^2=\pi * 1=\pi\)
B, C and D can be eliminated immediately.

Clearly, the unshaded part is less than the unshaded part, thus the area of shaded part has to be less than \(\frac{\pi }{2}\)

Only A left.

I don't see how the radius of the smaller circle is calculated. Could you explain it? Thanks a lot.

Hi, I have attached an image below highlighting the method to solve for the radius of smaller circle. I am not a pro at this, so pls do with my bad hand writing.
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WhatsApp Image 2021-06-19 at 7.46.01 PM.jpeg
WhatsApp Image 2021-06-19 at 7.46.01 PM.jpeg [ 86.1 KiB | Viewed 1856 times ]

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