Show that area enclosed by a parallelogram is a rectangle ?
Answers
Answered by
0
They are of the same area,,,
PROOF:
The quadrilateral shape bounded by DAFC is common to both the parallelogram ABCD and rectangle DEFC, so only need to prove that the area of triangles ADE and BFC are the same, and to do that, we proof that those two triangles are congruent.
First, both triangles are right angle triangles, due to the rectangle (all of its inner angles are right angles), so Pythagoras Theorem can prove useful.
|EF| = |AB| (opposing sides of a parallelogram have the same length, and both ABCD and DEFC share the same base |DC|)
|EF| = |EA| + |AF|
|AB| = |AF| + |FB|
=> |EA| = |FB|
|BC| = |AD| (as per above for parallelogram)
the remaining side |ED| = |FC|, per Pythagoras theorem.
Since the 2 triangles have sides that are exactly the same, the triangles EAD and FBC are congruent via the SSS rule (side-side-side).
Since the two triangles are congruent, they also have the same area.
Thus, adding the triangle to the part that the Parallelogram and the Rectangle have in common, it concludes that the area of the Rectangle is equal to the Parallelogram.
PROOF:
The quadrilateral shape bounded by DAFC is common to both the parallelogram ABCD and rectangle DEFC, so only need to prove that the area of triangles ADE and BFC are the same, and to do that, we proof that those two triangles are congruent.
First, both triangles are right angle triangles, due to the rectangle (all of its inner angles are right angles), so Pythagoras Theorem can prove useful.
|EF| = |AB| (opposing sides of a parallelogram have the same length, and both ABCD and DEFC share the same base |DC|)
|EF| = |EA| + |AF|
|AB| = |AF| + |FB|
=> |EA| = |FB|
|BC| = |AD| (as per above for parallelogram)
the remaining side |ED| = |FC|, per Pythagoras theorem.
Since the 2 triangles have sides that are exactly the same, the triangles EAD and FBC are congruent via the SSS rule (side-side-side).
Since the two triangles are congruent, they also have the same area.
Thus, adding the triangle to the part that the Parallelogram and the Rectangle have in common, it concludes that the area of the Rectangle is equal to the Parallelogram.
Similar questions