GMATPREP1, Circle/Triangle Area

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medea66: Master | Next Rank: 500 Posts; Posts: 186; Joined: Sat Jan 26, 2008 10:05 pm; Thanked: 2 times

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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Source: Beat The GMAT — Problem Solving | Mon Aug 11, 2008 7:52 pm

amitansu: Master | Next Rank: 500 Posts; Posts: 294; Joined: Tue Feb 26, 2008 9:05 pmFollowed by: 1 member; Thanked: 13 times

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)

Amit

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drgmatIL: Master | Next Rank: 500 Posts; Posts: 113; Joined: Sat Aug 02, 2008 12:49 pm; Thanked: 3 times

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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medea66: Master | Next Rank: 500 Posts; Posts: 186; Joined: Sat Jan 26, 2008 10:05 pm; Thanked: 2 times

by medea66 » Tue Aug 12, 2008 5:48 pm

do you just assume that this is a right angle?

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Ian Stewart: GMAT Instructor; Posts: 2623; Joined: Mon Jun 02, 2008 3:17 amFollowed by: 355 members; Location: Montreal; Thanked: 1090 times; GMAT Score:780

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.

For online GMAT math tutoring, or to buy my higher-level Quant books and problem sets, contact me at ianstewartgmat at gmail.com

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sudi760mba: Senior | Next Rank: 100 Posts; Posts: 85; Joined: Wed Aug 13, 2008 2:21 am; Thanked: 1 times

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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