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

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- Jul 16th 2009, 07:18 AMNiceguy78Area 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 AMskeeter
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 AMNiceguy78
im looking for the area AND parallelogram and i dont get how you get the x\sqrt{3}

- Jul 16th 2009, 07:40 AMskeeter
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 AMNiceguy78
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 AMNiceguy78
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 AMPlato
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 AMNiceguy78
o wow confusing me.. you give me a different formula?? is my answer right?

- Jul 16th 2009, 09:03 AMPlato