# Area of a parallelogram given one side and a perpendicular

This came up while I was doing some tutoring yesterday. Maybe it's simple, but it has been about 7 years since I last took geometry and I can't figure it out.

Given parallelogram $ABCD$ along with the length of $\overline{AB}$ and the length of the perpendicular from $D$ to $\overline{AB}$, how does one determine the area of $ABCD$?

If the perpendicular met $\overline{AB}$ at a particular point, such as its midpoint, then we could use the Pythagorean Theorem to find the hypotenuse of the right triangle that is formed, which is the other side of the parallelogram, and apply the usual area formula. However, I can find no result in the book that states any such thing, and drawing out some examples does not convince me either.

• Hint: Perpendiculars from $D$ to $AB$ and $B$ to $CD$ are the same. Now note some identical triangles there, and viola. – Lazar Ljubenović Feb 16 '12 at 19:12
• Ha. So basically if I tilted the paper I would have seen the answer. Thanks everyone. – Alex Petzke Feb 16 '12 at 21:23

It's the common formula, area is base times perpendicular. Let's prove it, starting from only formula for area of rectangle.

Let the foot of perpendicular from $D$ to $AB$ be $D'$ and from $B$ to $CD$ be $B'$. The given lengths are $AB=CD=a$ and $DD'=BB'=h$. We note that $AD'=CB'=x$, right angles at $D'$ and $B'$ and $DD'=BB'=h$, so triangles $ADD'$ and $CBB'$ are identical.

The area of rectangle is $x\times y$ if $x$ and $y$ are lengths of sides. If we draw diagonal, we get two triangles with obviously same areas, so the area of right triangle is $\frac{x\times y}2$, were $x$ and $y$ are sides.

We have two identical triangles here, so their area (summed) is $x\times h$. We also have a rectangle $D'BB'D$, which has sides $h$ and $a-x$, so the total area of parallelogram is $$A=h\times x+h\times (a-x)=h\times(x+a-x)=h\times a.$$

This is the most basic formula for parallelogram. Alternatively, but a bit less formally, you could "cut" $ADD'$ and snap it to $CB$, so you'd get a rectangle with sides $a$ and $h$.

The area is base times height.

Here is an only slightly dishonest proof. You will need scissors and paper. Produce a parallelogram $ABCD$. So that this parallelogram will look like the one I have in my head, the vertices $A$, $B$, $C$, $D$ are in counterclockwise order, the base $AB$ is at the bottom, and the point $A$ is on the left. Also, the parallelogram "leans" to the right, but does not lean by a ridiculous amount.

Let $P$ be the point where the perpendicular $D$ meets the line $AB$. If you are drawing the picture I have in mind, $P$ is between $A$ and $B$. Draw th perpendicular from $C$ to the extension of line segment $AB$. Suppose that this perpendicular meets $AB$ extended at $Q$.

OK, we are all set up. Note that triangles $APD$ and $BQC$ are congruent. This is easy, the angles match and $AD=BC$.

Now take a pair a scissors, cut out $\triangle APD$, and place the sliced off triangle so that it covers $\triangle BQC$. So we have sliced our parallelogram into two pieces, and reassembled the pieces to make rectangle $PQCD$. Thus the parallelogram and the rectangle have the same area. But the area of the rectangle is clearly $PQ$ times $DP$, which is $AB$ times $PD$.

Remark: My assertion that I cheated may be puzzling. But draw the parallelogram so that it leans ridiculously to the right, so that the perpendicular $DP$ meets the line $AB$ to the right of $B$. Then the argument just given breaks down. It can be fixed.

There is a lot of interesting mathematics connected with dissecting geometric figures, and rearranging the pieces to make another geometric figure. For example, there are very nice cut and paste proofs of the Pythagorean Theorem. You may also be interested in the Bolyai-Gerwien Theorem.

Look at the following figure.

$\hspace{2 in}$

$$\text{Area of the parallelogram}=\text{Area of}~~ \Delta ADB + \text{Area of}~~\Delta BCD$$

Now use the fact that area of the triangle is $\dfrac{1}{2}bh$ where $b$ is the length of the base on which the height of length $h$ stands.

Also, note that $AB=CD$ in a parallelogram and the distance between two parallels is the same. (i.e.) the length of the thicker height is the same ass that of the thinner one.

• Another way to look at it is that you can cut one ear of the parallelogram and attach it at the other side (so AD and BC align). You will get a rectangle with known sides. – Artium Feb 16 '12 at 19:30
• Sure. But, I am left with MS Paint to draw good pics. So, I can't draw this precisely to show what you want me to. But, nonetheless an answer in itself! :) (+1) – user21436 Feb 16 '12 at 19:34