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GMATPREP1, Circle/Triangle Area

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by medea66 » Mon Aug 11, 2008 7:52 pm
This seemed like such an easy question w/ base and height given. All I had to do was a= 1/2(bh)......so why is the answer what it is?
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by amitansu » Mon Aug 11, 2008 8:37 pm
You are right, but i think you forgot to square 2 and that makes sqrt (4-3) as sqrt(3) not 1 !! (The height is AB here so, AC^2-BC^2=AB^2)


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by drgmatIL » Tue Aug 12, 2008 3:15 am
ABC is right angle
BC = 1
AC = 2 so AB is root of 3

the area is (3root * 1 )\2
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by medea66 » Tue Aug 12, 2008 5:48 pm
do you just assume that this is a right angle?
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by Ian Stewart » Tue Aug 12, 2008 8:01 pm
medea66 wrote:do you just assume that this is a right angle?
No, that's not an assumption. If you ever connect a diameter AB of a circle to a different point C on a circle, the angle at C will be a right angle. That's not hard to prove regardless, but it can be very useful to know for the GMAT.
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Re: GMATPREP1, Circle/Triangle Area

by sudi760mba » Fri Mar 06, 2009 3:25 am
medea66 wrote:This seemed like such an easy question w/ base and height given. All I had to do was a= 1/2(bh)......so why is the answer what it is?

Here's what we do know:

1) AC=2 because 2*radius = 2
2) BC=1
3) ABC is a right angle

With this we can state that we need to find AB as the base and we're given BC as the height of 1. To find AB, we can use the Pythagorean theorem
of
a^2 + b^2 = c^2

or AB^2 + BC^2= AC^2
AB^2 + 1^2 = 2^2
AB^2 = 4 -1
AB = square root of (3)

Base * height /2 =Area of triangle
(sqrt 3)*(1) /2 = (sqrt 3)/2
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