# parallelogram

• Nov 18th 2009, 05:31 AM
thereddevils
parallelogram
Prove that parallelograms on the same base and betweem the same parallels are equal in area .
• Nov 18th 2009, 05:40 AM
Hello thereddevils
Quote:

Originally Posted by thereddevils
Prove that parallelograms on the same base and betweem the same parallels are equal in area .

Divide the parallelogram into two equal triangles, using a diagonal. Then the area of the parallelogram = twice the area of one of the triangles. Area of triangle = half base times height. So?

• Nov 18th 2009, 05:51 AM
masters
Quote:

Originally Posted by thereddevils
Prove that parallelograms on the same base and betweem the same parallels are equal in area .

Hi thereddevils,

Draw parallelogram ABCD with AB being the top side. Extend AB to E such that BE = DC.

Connect D to B and C to E.

You have just constructed another parallelogram BECD with same base as ABCD and between the same parallels, as required.

In parallelogram ABCD with diagonal BD, we can prove that the two triangles ABC and CDB are congruent.

We also have sufficient evidence to prove that triangles CDB and BEC are congruent.

Now all three triangles are congruent to each other. They have equal areas.
• Nov 18th 2009, 06:03 AM
thereddevils
(Bow)oh thansk a lot , Grandad and masters . i was just a little confused with the meaning of between the parallels .
• Nov 18th 2009, 07:15 AM
Hello everyone
Quote:

Originally Posted by masters
Hi thereddevils,

Draw parallelogram ABCD with AB being the top side. Extend AB to E such that BE = DC.

Connect D to B and C to E.

You have just constructed another parallelogram BECD with same base as ABCD and between the same parallels, as required.

In parallelogram ABCD with diagonal BD, we can prove that the two triangles ABC and CDB are congruent.

We also have sufficient evidence to prove that triangles CDB and BEC are congruent.

Now all three triangles are congruent to each other. They have equal areas.

While this is true, it's a very special case. You should really consider two general parallelograms on the same base and between the same parallels, as I have indicated in the attached drawing. I outlined the proof in my first posting. Here are a few more details.

In the diagram, $\displaystyle ABCD$ and $\displaystyle CDE F$ are any two parallelograms on the base CD, with the points $\displaystyle A,B, E,F$ in a straight line.

Construct the diagonals $\displaystyle BD$ and $\displaystyle CE$. Triangles $\displaystyle ABD$ and $\displaystyle CDB$ can be easily proven congruent; as can triangles $\displaystyle DEC$ and $\displaystyle FCE$.

Drop perpendiculars from $\displaystyle B, E$ to meet $\displaystyle CD$ at points $\displaystyle X, Y$. $\displaystyle BEYX$ is then a rectangle. Therefore $\displaystyle BX = EY$.

Area $\displaystyle ABCD = 2\times$ area $\displaystyle \triangle CDB = CD \times BX = CD \times EY =$ area $\displaystyle CDE F$.