# Math Help - Area of Parallelograms

1. ## Area of Parallelograms

find the perimeter and area of each parallelogram Round to the nearest tenth if necessary

figure is exactly like that

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

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

finally, area of a parallelogram is $A = bh$

3. im looking for the area AND parallelogram and i dont get how you get the x\sqrt{3}

4. 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

5. ok ty but now all the problem that i have here are 30 60 90 righ triangles some contain different measurements

6. 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

7. It is fairly well known that the area of a parallelogram is $A=a\cdot b\cdot \sin(\theta)$ there $a~\&~b$ are the lengths of two adjacent sides and $\theta$ is the angle between then sides.

8. o wow confusing me.. you give me a different formula?? is my answer right?

9. Originally Posted by Niceguy78
o wow confusing me.. you give me a different formula?? is my answer right?
The area of your parallelogram is $(20)(30)\sin\left(60^0\right)$.