Math, asked by 9110111968, 1 year ago

prove the area of parallelogram

Answers

Answered by pragya2785
3
Proof: Since a rectangle is also a parallelogram so, the result is a direct consequence of the above theorem. Theorem: The area of a parallelogram is the product of its base and the corresponding altitude. Given: In a parallelogram ABCD, AB is the base.
Answered by Anonymous
0
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:

Carefully look at the figure again. Do you think that triangle ADE is the same as triangle FCB?

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 parallelogram 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
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