# Thread: Area of a Parallelogram

1. ## Area of a Parallelogram

I have to find the area of this parallelogram, but the only three things I'm told are that AD = 5, EF = 11, and EFC = 90º. So far I've tried using the pythagoras theorem, and the special 30-60-90 triangle theorem (longer cathet = √3 * shorter cather OR (√3/2) * Hypotenuse), but they all lead to a dead end.

2. ## Re: Area of a Parallelogram

$A = bh = 5 \cdot 11$

3. ## Re: Area of a Parallelogram

But 11 is not the length, DC is the length, 11 is EF.

4. ## Re: Area of a Parallelogram

Originally Posted by kaperdomo
But 11 is not the length, DC is the length, 11 is EF.
base & height are perpendicular ...

EF is the height

AD = BC is the base

5. ## Re: Area of a Parallelogram

Originally Posted by skeeter
base & height are perpendicular ...

EF is the height
How is EF the height? DC is the height and nothing says they are congruent.

Originally Posted by skeeter
AD = BC is the base
I get this, it's the EF part I have a problem with.

6. ## Re: Area of a Parallelogram

DC is not the height.

One more time ... base and height of a parallelogram are perpendicular.

turn the attached diagram 90 degrees ...