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What is the area of the circular region

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by melguy » Fri Aug 09, 2013 12:30 am
I am confused with the language of the question. The question is asking what is the area of the circular region centered at the origin although in the answer explanation it is telling us to find area of the triangle!

Its just the language of the question that is bothering me. I thought after we find the hypotenuse we need to find the area of the circle. Although in the explanation only the hypotenuse is found and that is the final answer.

OA is D.
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Source: — Problem Solving |

by GMATGuruNY » Fri Aug 09, 2013 1:41 am
What is the area of the circle shown below?
Image
When a shape is drawn on the coordinate plane, look for what the shape and the coordinate plane have IN COMMON.
Point (-10, 24) is both a point the circle and the vertex of the triangle shown below:
Image

The red triangle is a multiple of a 5-12-13 triangle:
10-24-26.
When shapes overlap, look for what the shapes have in common.
The hypotenuse of the red triangle is also the RADIUS of the circle.
Thus, the radius of the circle = 26.
Since the area of a circle = πr², we get:
A = π26² = 676π.

The correct answer is D.
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by macattack » Mon Aug 12, 2013 12:38 am
For those who are not familiar with the 5-12-13 triangle or frequently forget that this family exists here is another way: Area of a circle= pi*r^2
In the above figure r=sqrt(10^2+24^2)=sqrt(676)
so r^2 =676--->Area=676*pi
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by Brent@GMATPrepNow » Mon Aug 12, 2013 6:34 am
As Mitch has pointed out, we can often solve geometry questions quickly if we're aware of some popular right triangles that feature integer lengths.
The two most popular right triangles are the 3-4-5 and the 5-12-13 right triangles (and, of course, multiples of these triangles such as 6-8-10, and 10-24-26 right triangles)

There are 2 more right triangles you might want to memorize: 8-15-17 and 7-24-25. These right triangles don't appear frequently, but if they do, you can save yourself the hassle (and time) of applying the Pythagorean Theorem.

Cheers,
Brent
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