A Parallelogram has area 85 $\displaystyle cm^2$. The side lengths are 10 cm and 9 cm. What are the measures of the interior angles

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- Apr 16th 2009, 11:42 AMsupersaiyanArea of Parallelogram
A Parallelogram has area 85 $\displaystyle cm^2$. The side lengths are 10 cm and 9 cm. What are the measures of the interior angles

- Apr 16th 2009, 11:58 AMTwig
Given the area is 85cm^2. Use the base as 10 cm, so $\displaystyle 10 \cdot h = 85 \, \Rightarrow h = 8.5 $

This gives one angle $\displaystyle \alpha = arcsin(\frac{8.5}{9}) $

So the other angle: $\displaystyle \beta = 180 - arcsin(\frac{8.5}{9}) $ - Apr 16th 2009, 12:01 PMmylestone
Let's stand the parallelogram on the table so that the long edge--10cm--is flat (horizontal). You are given the area of the parallelogram (85 sq. cm). You recall how that area was found? It's the base times the vertical height of the parallelogram. The vertical height is different from the other side (which is why the area isn't 10*9, Twig--if we were playing with squares or rectangles that would be the case though). Use the long edge and the area to find the height (Area=base*height, and you know two of these already). Then you have a right triangle and you can use what you know about right triangles to find the measure of the acute angle (there are two of them, but they're of equal measure). And you know that all four angles in any 4-sided polygon add to 360 degrees. That should get you through.

Incidentally, you can think of a square or a rectangle as special kinds of parallelograms since they do fit the definition of parallelogram. If you do this then, like I mentioned above, the height turns out to be the length of the vertical side. You may have done this already in class (I remember doing it in my geometry class), but if you didn't know the formula for the area of a parallelogram, just cut one out of a piece of paper. Then draw a vertical line from a top corner to the base (through the inside of the parallelogram). Cut along this line to get that right triangle. Lastly, slide that right triangle over to the other side of the parallelogram...SCORE! You just made a square and you can see why the area of a parallelogram must come from the base times the VERTICAL HEIGHT, and not just the product of two sides. Happy mathing! - Apr 16th 2009, 12:11 PMsupersaiyan
Thanks, is this correct?

http://i39.tinypic.com/2z7lj4z.jpg - Apr 16th 2009, 01:58 PMmylestone
Looks good from my house!