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# The circles shown are tangent at point B. Point A is the ce

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The circles shown are tangent at point B. Point A is the ce [#permalink]  07 Mar 2018, 14:36
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Question Stats:

88% (00:41) correct 11% (02:37) wrong based on 9 sessions
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circles.jpg [ 12.06 KiB | Viewed 828 times ]

The circles shown are tangent at point B. Point A is the center of the larger circle, and line segment AB (not shown) is a diameter of the smaller circle. The area of the smaller circle is what fraction of the area of the larger circle?

[Reveal] Spoiler: OA
$$\frac{1}{4}$$

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Re: The circles shown are tangent at point B. Point A is the ce [#permalink]  07 Mar 2018, 19:22
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The fastest way to solve this problem is to know that for any two similar shapes, the ratio of their areas is equal to the square of the ratio of their lengths. For example, a square with a side of 5 and an area of 25, and a square with a side of 10 and an area of 100. The ratio of their side lengths is 1:2, but the ratio of their areas is 1:4.

In this problem, we can quickly see that the diameter of the small circle is half that of the larger circle. Squaring 1/2 gets us 1/4, so the small circle has 1/4 the area of the larger.
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Re: The circles shown are tangent at point B. Point A is the ce [#permalink]  02 May 2018, 07:49
Expert's post
Carcass wrote:
Attachment:
circles.jpg

The circles shown are tangent at point B. Point A is the center of the larger circle, and line segment AB (not shown) is a diameter of the smaller circle. The area of the smaller circle is what fraction of the area of the larger circle?

[Reveal] Spoiler: OA
$$\frac{1}{4}$$

Let's assign some values that meet the given conditions.

So, let's say that AB = 2
This means the RADIUS of the LARGE circle = 2

It also means that the DIAMETER of the SMALL circle = 2
If the DIAMETER of the SMALL circle = 2, then the RADIUS of the SMALL circle = 1

The area of the smaller circle is what fraction of the area of the larger circle?

So, area of BIG circle = π(2²) = 4π
And the area of SMALL circle = π(1²) = 1π

From here, we can see that the area of the smaller circle is 1/4 of the area of the larger circle (i.e., 1π is 1/4 of 4π)

Cheers,
Brent
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Re: The circles shown are tangent at point B. Point A is the ce [#permalink]  02 May 2018, 11:11
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Even though I like both the above methods, the technique I prefer is this. Area of big circle = pi*r^2. Area of small circle = pi*(r/2)^2= pi*r^2/4. i took less times using this. I guess the way everyone practices and has learned their basics matters with such techniques. So till our techniques are right and quick, its all good.
Re: The circles shown are tangent at point B. Point A is the ce   [#permalink] 02 May 2018, 11:11
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