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[ILikePie]

I've got an isosceles trapezoid ABCD (AB || DC and BC = AD) with point E on DC so that AE is perpendicular to DC. AE = 8 and EC = 10.
How do I calculate ABCD's area?

Spoiler: crappy ascii art (click to show/hide)

   a----------b
  / |             \
 /  |              \
d--e------------c

[Phmcw]

AE * EC

[Eugenitor]

AE * EC

This.

If it helps, visualize a point F on the opposite (horizontal) side of point E such that BCF is ADE flipped horizontally. Now flip ADE vertically and put it next to BCF. Voila, you're now calculating the area of a rectangle.

[ILikePie]

Damnit, simplicity will kill me one day, thanks.

[lordnincompoop]

Little did he know that his fate would be to atomize into a conglomeration of simple carbon and hydrogen atoms from a neighbouring nuclear blast.

[ILikePie]

I swear this teacher hates her students.
ABCD is square with point E on AB, and point O somewhere inside, so that EO=BO=CO, and angle AEO=90 degrees. How do I show the value of EO if the square's side=5?

Spoiler: more bad ascii (click to show/hide)

a___e___b
|     |      |
|     |      |
|     o     |
d_______c

[KaminaSquirtle]

I swear this teacher hates her students.
ABCD is square with point E on AB, and point O somewhere inside, so that EO=BO=CO, and angle AEO=90 degrees. How do I show the value of EO if the square's side=5?

Spoiler: more bad ascii (click to show/hide)

a___e___b
|     |      |
|     |      |
|     o     |
d_______c

Where inside?  I don't see any info regarding where o is in the square.  Without that, this is an exercise in futility, as far as I can tell,
Your teacher must really hate her students to give them unsolvable problems.

[ILikePie]

The distances from D, C, and E to O are equal (DO=CO=EO).

[KaminaSquirtle]

Oh, I read the problem wrong, I thought it meant EO=CP=DQ, P and Q shown below:

Spoiler: Ripoff of bad ascii art (click to show/hide)

a___e___b
|     |      |
|     |      |
q     o     p
d_______c

I'm such an idiot.
Ok, let me think for a second then.

Edit:

Spoiler: Solution (click to show/hide)

Well, since CO=DO and AEO=90 degrees, then E is the midpoint of A and B.  Let CO=DO=EO=L.  Then we may form a right triangle using DO, DC/2, and O to the bottom of the square.  We know DC=5, so DC/2=2.5, we also know O to the bottom of the square=BC-L, and since BC=5, we get O to the bottom of the square=5-L, and finally, we know DO=L.  Since DO is the hypotenuse, we may use the Pythagorean Therom.  So we get: sqrt(2.5^2+(5-L)^2)=L, and we solve from there, getting L=EO=3.125.

I'm so rusty at trigonometry....

[penguify]

Point E=B. Otherwise it's not possible, I believe, for that angle to be 90 and the points to be equidistant. Point O is at the midpoint of BC. 2.5. Right?

[KaminaSquirtle]

Just edited my post with what I believe is the solution, hopefully it makes any sense at all without a diagram.

[Jim Groovester]

I swear this teacher hates her students.
ABCD is square with point E on AB, and point O somewhere inside, so that EO=BO=CO, and angle AEO=90 degrees. How do I show the value of EO if the square's side=5?

Spoiler: more bad ascii (click to show/hide)

a___e___b
|     |      |
|     |      |
|     o     |
d_______c

The distances from D, C, and E to O are equal (DO=CO=EO).

Hold on. Which segments are equal to each other? You've contradicted yourself in these two posts and you will get two very different answers depending on which ones are correct.

[KaminaSquirtle]

Good point.  I went with the second case for my solution.

[ILikePie]

It was the second, sorry about that. Thanks.

[eerr]

For DO=CO=EO,(this was what you meant originally, right?)

3.125

For reference, I used DO=X, and assumed point F as the midpoint of DC
FO=EF-EO, therefore FO=5-X
Then, pythagorean theorem,

FO^2+DF^2=DO^2

and solving for X

Same method as KaminaSquirtle.