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M23-27

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M23-27  [#permalink]

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New post 16 Sep 2014, 01:19
1
3
00:00
A
B
C
D
E

Difficulty:

  45% (medium)

Question Stats:

69% (01:25) correct 31% (01:23) wrong based on 152 sessions

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Re M23-27  [#permalink]

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New post 16 Sep 2014, 01:19
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Re: M23-27  [#permalink]

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New post 26 Dec 2014, 03:25
Or even more simple:

Identify that you have a 3-4-5 (6-8-10) triangle and without calculation you know that the hypothenuse is 10 => radius is 10.
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Re: M23-27  [#permalink]

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New post 28 Jan 2016, 11:10
I took the wrong approach with this question, would someone kindly clarify?

I had done a quick sketch of 0,6 on a graph and assumed that since the distance between the first point and where x meets the number line (x, 0) then the hypotenuse would be 16, and the radius would therefore be 16.

Where am I going wrong? I recognize that a 6-8-10 triangle can be formed from the side with a value of 6 however how does length 16 factor into this? Thanks
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Re: M23-27  [#permalink]

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New post 28 Jan 2016, 23:38
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Anonamy wrote:
I took the wrong approach with this question, would someone kindly clarify?

I had done a quick sketch of 0,6 on a graph and assumed that since the distance between the first point and where x meets the number line (x, 0) then the hypotenuse would be 16, and the radius would therefore be 16.

Where am I going wrong? I recognize that a 6-8-10 triangle can be formed from the side with a value of 6 however how does length 16 factor into this? Thanks


Image

The center of the circle is on y-axis, so it's symmetric around it. We know that the the distance between the two points where the circle intersects the x-axis is 16, hence half of it (8 units) must be to the left of 0 and the remaining half (another 8 units) to the right of 0.

As you can see the radius of the circle is a hypotenuse of a right triangle with the sides equal to 6 and 16/2=8 (6-8-10 right triangle), so radius=hypotenuse=10. The area is \(\pi{r^2}=100\pi\).

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Circle.png

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Re M23-27  [#permalink]

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New post 25 Jul 2016, 06:54
I think this is a high-quality question and I agree with explanation. Wow. This is a 700 level question certainly not 600 as classified by you Bunuel. Anyway terrific question in my view! Great work!
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New post 22 Mar 2018, 10:13
+1 for option E. Draw the graph , the answer will stand out almost immediately !
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New post 22 Mar 2018, 13:12
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New post 26 Mar 2018, 10:59

Solution:



Given:

    • Centre of the circle is at (6,0)

    • The distance between points where the circle intersects the x-axis = 16 units

Working out:

We need to find out the area of the circle.

To solve this problem, let us first draw a proper diagram.

Image
The distance AB is 16 units (given in the question)

Since the diagram is symmetrical, AC=CB= AC/2 = 8 units.

OC = 6 units

Thus, AO (Radius of the circle) = \(\sqrt{( 6^2 + 8^2)}\) = \(\sqrt{100}\) = 10 units

Hence the area of the square = \(pie * 10^2 = 100 pie\)

Answer: Option E
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Re: M23-27   [#permalink] 26 Mar 2018, 10:59
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