Proof of the area of a parallelogram
Start your proof of the area of a parallelogram by drawing a parallelogram ABCD.
Then, draw heights from vertices D and C to Segment AB. You may have to extend segment AB as you draw the height from C
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:
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
Proof of the area of a parallelogram has come to completion. Any questions? Contact me
Feb 17, 19 12:04 PM
There is no rational number whose square is 2. An easy to follow proof by contraction.
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