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Geometry Problem

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Geometry Problem

by thumpin_termis » Wed Jun 27, 2007 7:09 pm
We have a square ABCD with sides = 1, and BE=1, and AE=CE.
What is the area of triangle ABE?

I believe the answer is = sqrt(2) / 4 (and I'm not sure about this), but I don't know how to work this out.

Anyone have a clue?
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by Jhyphi » Wed Jun 27, 2007 10:38 pm
So the first thing you have to see is that if you drop a vertical from E and extend line segment AB to the right to intersect then we call this point F.

So you have a right triangle AFE and right triangle BFE both with F being a right angle.

The next thing is to know that BFE is a 45/45/90 triangle. If you want proof, you have 360 - 90 (the square) = 270. Divide by 2 to get ABE which is 135. Then ABE and EBF combine to be 180 degrees so 180-135 = 45.

From that, we know that BF and EF are sqrt(2)/2 or 1/sqrt(2).

So the area of AFE is (1+sqrt(2)/2) * sqrt(2)/2 * 1/2
= (2+sqrt(2)) / 2 * sqrt(2)/2 * 1/2
= (2+sqrt(2)) * sqrt(2) * 1/8
= (2*sqrt(2) + 2) / 8
= (sqrt(2)+1) / 4

The area of BFE is simpler
sqrt(2)/2 * sqrt(2)/2 * 1/2
= 2/8
= 1/4

Finally, for ABE, you subtract BFE from ABE.
sqrt(2)/4 + 1/4 - 1/4
= sqrt (2) / 4

Exactly what you got.
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