# Area of Parallelograms

• Jul 16th 2009, 07:18 AM
Niceguy78
Area of Parallelograms
find the perimeter and area of each parallelogram Round to the nearest tenth if necessary

http://i25.tinypic.com/x3et6c.png

figure is exactly like that
• Jul 16th 2009, 07:26 AM
skeeter
Quote:

Originally Posted by Niceguy78
find the perimeter and area of each parallelogram Round to the nearest tenth if necessary

http://i25.tinypic.com/x3et6c.png

figure is exactly like that

note the 30-60-90 triangle formed in the upper part of the parallelogram ... the altitude of the parallelogram is the long leg of the 30-60-90 triangle.

ratios of the sides of a 30-60-90 triangle ...

short leg = $\displaystyle x$
long leg = $\displaystyle x\sqrt{3}$
hypotenuse = $\displaystyle 2x$

finally, area of a parallelogram is $\displaystyle A = bh$
• Jul 16th 2009, 07:29 AM
Niceguy78
im looking for the area AND parallelogram and i dont get how you get the x\sqrt{3}
• Jul 16th 2009, 07:40 AM
skeeter
Quote:

Originally Posted by Niceguy78
im looking for the area AND parallelogram and i dont get how you get the x\sqrt{3}

If you have no knowledge of the properties of a 30-60-90 right triangle, then you shouldn't have been assigned this problem.

Go to this site and learn about them ...

30 60 90 Right Triangles - Free Math Help
• Jul 16th 2009, 07:45 AM
Niceguy78
ok ty but now all the problem that i have here are 30 60 90 righ triangles some contain different measurements
• Jul 16th 2009, 08:19 AM
Niceguy78
hi i kind of get it now i got the answer 780 anyone can tell me if i got it right?? sorry for double post
• Jul 16th 2009, 08:21 AM
Plato
It is fairly well known that the area of a parallelogram is $\displaystyle A=a\cdot b\cdot \sin(\theta)$ there $\displaystyle a~\&~b$ are the lengths of two adjacent sides and $\displaystyle \theta$ is the angle between then sides.
• Jul 16th 2009, 08:25 AM
Niceguy78
o wow confusing me.. you give me a different formula?? is my answer right?
• Jul 16th 2009, 09:03 AM
Plato
Quote:

Originally Posted by Niceguy78
o wow confusing me.. you give me a different formula?? is my answer right?

The area of your parallelogram is $\displaystyle (20)(30)\sin\left(60^0\right)$.