# Finding an area of a triangle:

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• June 27th 2007, 06:39 PM
thumpin_termis
Finding an area of a triangle:
We have a square ABCD with sides = 1, and BE=1, and AE=CE.
What is the area of triangle ABE?

How can I solve this?
• June 27th 2007, 07:05 PM
Jhevon
Quote:

Originally Posted by thumpin_termis
We have a square ABCD with sides = 1, and BE=1, and AE=CE.
What is the area of triangle ABE?

How can I solve this?

I found two ways to solve this and then stop. this is the shorter of the two methods. See the modified diagram below.

from the description we see that $\triangle AEC$ is isosceles. thus if we extend the line BE, it will cut the diagonal at a right angle. from this we can the angles that you see filled in.

now we simply use the formula $\mbox {Area} = \frac {1}{2}(BE)(AE) \sin 135^{ \circ} = \frac { \sqrt {2}}{4}$

if you have any questions say so. i was kind of rushing through this and didn't explain much