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Proof of the area of a parallelogramStart your proof of the area of a parallelogram by drawing a parallelogram ABCD
  Call the rectangle that is formed by drawing the heights from vertices D and C, EDCF  The proof is not complicated. We already know that the area of the rectangle is equal to base times height Therefore, if we can show that we can play with parallelogram ABCD to make rectangle EDCF, we are done. In this case, area of parallelogram ABCD must equal area of rectangle EDCF What we are trying to do is illustrated below:  If this is really the case, you coud just cut triangle ADE and place it over triangle FCB until they coincide and here we go! You had just made your rectangle by cutting triangle ADE from the parallelogram ABCD And again, if you can do that, both the paralelogram and the rectangle must have the same area All we have to do now is to show that triangle ADE is the same as triangle FCB and we are done! Both triangles are right triangles, so if we can show that two sides are the same, the last side must be the same according to the Pythagorean Theorem and both triangles will therefore be the same In parallelogram ABCD, segment AD = segment BC because in a parallelogram opposite sides are equal In rectangle, EDCF, segment ED = segment FC because in a rectangle opposite sides are equal We have found two sides that are equal! We are done with the whole proof Proof of the area of a parallelogram has come to completion. Any questions? Contact me
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