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MathRevolution
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circle.jpg
A smaller circle is inscribed in a larger circle shown as above figure. If the smaller circle passes through a center of the larger circle,what is the ratio of the region shaded to the larger circle’s area?

A. 1/2
B. 2/3
C. 3/4
D. 4/5
E. 5/6


* A solution will be posted in two days.

From Figure we see that

radius of bigger circle = Diameter of smaller circle

Let, Radius of Bigger circle = 2 = Diameter of smaller circle

i.e. Radius of Smaller circle = 2/2 = 1

Area of Bigger circle = Pi 2^2 = 4 Pi
Area of Smaller circle = Pi 1^2 = Pi

Area of Shaded Region = Big circle - Small Circle = 4 Pi - Pi = 3Pi

Shaded Area/Area of Big circle = 3Pi/4 Pi = 3/4

Answer: Option C
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Nice Question..
here we just need to figure out that radius of the larger circle is the diameter of the smaller circle.
Hence C as Shaded area => 3/4 of the larger area
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As we know that the smaller circle will have radius half to that of main circle
So area of smaller circle = 1/4* area of larger circle
Now as per question it asks us the ratio of shaded region to that of whole
So shaded region = area of larger - area of smaller = 1- 1/4 = 3/4

So answer is C

Sent from my Mi 4i using Tapatalk
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A smaller circle is inscribed in a larger circle shown as above figure. If the smaller circle passes through a center of the larger circle,what is the ratio of the region shaded to the larger circle’s area?

A. 1/2
B. 2/3
C. 3/4
D. 4/5
E. 5/6


-> As the ratio of the radius is 1:2, the ratio of the area is 1:4.
If the area of the small circle is a, the shaded area is 3a.
Thus, 3a/(a+3a)=3/4 and the answer is C.
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I have a small doubt in this question. If you consider the radius of the smaller circle to be 3 and the radius of the larger circle to be 5 then the area of the smaller is pi and the larger area is 25pi . Therefore the shaded part is 25pi-pi=16pi
16pi/25pi

So here i am missing out on the fact that the smaller circle passes through the center of the larger circle. Is that so? and in what circumstance does this method hold valid. Thank you
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