Geometry

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by newera » Sat Aug 02, 2008 4:21 pm
I got B. Whats the OA?

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by gabriel » Sat Aug 02, 2008 9:26 pm
newera wrote:I got B. Whats the OA?
Give your solution too please. Remember this is not a test :P but a way to share your knowledge.

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by Stuart@KaplanGMAT » Sat Aug 02, 2008 11:28 pm
We know that the perimiter of the shape is the 2 identical sides of the triangle + 3/4 the circumference of the circle.

As always, when a question has a confusing part and a simple part, start with the simple part.

The radius of the circle is 2, so the circumference is 2pi(2) = 4pi. However, we only want 3/4 of the circumference, so the correct answer is "3(pi) + the 2 equal sides of the triangle"... eliminate (d) and (e).

Since we're using 3/4 of the circle, the remaining sector of the circle is 1/4 of the circle, so the angle formed by the centre of the circle and the two top points of the triangle is 90 degrees (1/4 of 360 degrees). Therefore, the triangle inside the circle is a 45/45/90 (since it's isosceles) and the two equal sides have length 2 (since each one is a radius).

A 45/45/90 triangle is in the ratio of x:x:xroot2. Since each of the x sides is 2, the hypotenuse is 2(root2).

So, we have a triangle with base of 2(root2) and height 5. We need to find each of the other two sides of the triangle - let's call each side "y".

We can turn the triangle into two right angle triangles, each of which has one side of root2 and one side of 5. So, to solve for y:

(root2)^2 + 5^2 = y^2

2 + 25 = y^2
27 = y^2
root27 = y
3(root3) = y

So, y=3(root3)

Therefore, the final answer is 3(pi) + 2(3(root3)) = 3pi + 6(root3)... choose (b).
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by Nycgrl » Sun Aug 03, 2008 6:59 pm
OA is B

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by beeparoo » Mon Aug 04, 2008 11:00 am
This is a good problem. Worthy of the real GMAT, I think.
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