GMATPREP1, Circle/Triangle Area

[medea66]

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?

[amitansu]

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

[drgmatIL]

ABC is right angle
BC = 1
AC = 2 so AB is root of 3

the area is (3root * 1 )\2

[medea66]

do you just assume that this is a right angle?

[Ian Stewart]

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.

[sudi760mba]

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