## Challenge: area of a parallelogram

And now for something completely different!1

Suppose we have a parallelogram with one corner at the origin, and two adjacent corners at coordinates $(a,b)$ and $(c,d)$. What is the area of the parallelogram? 1. …or is it? Associate Professor of Computer Science at Hendrix College. Functional programmer, mathematician, teacher, pianist, follower of Jesus.
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### 18 Responses to Challenge: area of a parallelogram

1. Andrew Sansom says:

An easy way is simply to take the magnitude of the cross product of $(a,b,0)$ and $(c,d,0)$!

• Brent says:

Yes – but why?

2. Sylvain B. says:

Directly from the cross product: S = |ad-bc|
The parallelogram can also be seen as the sum of two triangles with the same area, or four right triangles with the same area.

• Brent says:

Wait, how do you cut it up into four right triangles with the same area?

• Sylvain B. says:

huuuu. Yes, you are right, there are not right triangles with the same area for a general parallelogram. Actually, it is always possible to build up two right triangles in the semi-parallelogram, but they have different areas. This may be visualized easily taking a rectangle, that is b = c = 0 or a = d = 0 (Geogebra is helpful).

3. Tom Rose says:

Area = (a + c – b)*d – b^2

Mark a rectangle with bottom left corner at (0,0) and top right at (a+c, b+d).
ie The top right corner of the parallelogram and rectangle match.

Obtain parallelogram area by:

Area of rectangle minus area of the triangles formed between the rectangle and parallelogram on the 4 sides.

• Master Blaster says:

There’s a typo in the above.

Area = a*(b + d) – b*(a + c)

4. blaisepascal2014 says:

Two parallelograms between the same parallels that share a base have the same area.

Let’s call the parallelogram $OABC$ with $O$ at the origin and the rest of the vertices counterclockwise. The line $AB$ passes through the points $(a,b), (a+c, b+d)$, has a slope of $d/c$, and an $x-$intercept of $A' = (a-bc/d, 0)$. The point $B' = (a+c-bc/d, d)$ lies on the same line.

The two parallelograms $OABC, OA'B'C$ are both between the parallels $OC, AB$, and share $OC$ as a base. Therefore they have the same area.

The area of a parallelogram is equal to the product of the length of a base and the length of its altitude. For parallelogram $OA'B'C$, the length of side $OA'$ is $a-bc/d$ and the altitude is of length $d$, giving an area of $(a-bc/d)(d) = ad-bc$

5. Jan Van lent says:

I like the shoelace formula, which works for any polygon. To see why it works is much the same as understanding the final formula for the parallellogram though. I like to think about it as the trapezium rule, but extended from a function to a curve.

points: $x = (0, a, a+c, c, 0)$, $y = (0, b, b+d, d, 0)$.

shoelace formula: $A = \sum_{i=1}^4 x_i y_{i+1} - x_{i+1} y_i$.

trapezium rule: $A = \sum_{i=0}^4 \frac{y_i + y_{i+1}}{2} (x_{i+1}-x_i)$.

• Brent says:

Yes, one of the places I’d eventually like to go is deriving a proof of the shoelace formula. Understanding the formula for the (signed) area of a parallelogram is one of the first building blocks along the way!

6. Dave says:

I enclosed the parallelogram in a rectangle that has a length of a+c and a width of b+d. I calculated the rectangle’s area and then split the negative space outside the original parallelogram into 2 congruent trapezoids and 2 congruent triangles. Subtracting the areas of those 4 figures from the area of the rectangle yields ad – bc.

7. Jan Van lent says:

The volume interpretation of the determinant gives some nice generalisations. The determinant is the factor involved in the change of variables formula. This gives $ad-bc$ for the area, because it is the determinant of the matrix $\begin{bmatrix} a & c \\ b & d \end{bmatrix}$ which maps the unit square (of area 1) to the parallelogram. And, for example, the area of the ellipse inscribed in the parallelogram is $\frac{\pi}{4}(ad-bc)$, since the circle inscribed in the unit square has area $\frac{\pi}{4}$. The determinant also generalizes to higher dimensions.

8. Jack Keynes says:

We want an operation on $v,w$ that gives us the area of the parallelogram they span. Let’s call it $A(v,w)$. We want to say that $A$ is linear—that is, $A(v+u,w) = A(v,w) + A(u,w)$, $A(v, w+u) = A(v,w) + A(v,u)$, and for any real number $r$, $A(rv,w) = rA(v,w) = A(v,rw)$ (you can see this for natural $r$ by tiling parallelograms). But for this to work with negative $r$, we’ll actually have to consider *signed area*, or area with specified orientation, so that $A(-v,w) = -A(v,w)$. Now we can see that we want $A(w,v) = -A(v,w)$ (which implies $A(v,v)=0$), and we’ll normalise by saying the unit square has area 1: $A(i,j) = 1$ where $i,j$ are standard unit vectors.

Now, we can calculate the (signed) area of the parallelogram spanned by (a,b) and (c,d): $A(ai + bj, ci+dj) = acA(i,i) + adA(i,j) + bcA(j,i) + bdA(j,j) = ad - bc$.

So the unsigned area is $|ad-bc|$.

• Brent says:

This is a neat analysis. The only part I don’t follow is — why should we believe that e.g. $A(v+u,w) = A(v,w) + A(u,w)$?

…hmm, I guess the right-hand side corresponds to two stacked parallelograms, and the left-hand side corresponds to taking the common side and sliding it (leaving the other ends fixed) until you get one giant parallelogram. Sliding one side of a parallelogram along the line containing it doesn’t change its area, of course.

• Jack Keynes says:

That’s exactly the right picture! If I could have drawn it in my comment I would have 😀

• Brent says:

I will draw it in my next post! =)

9. Thibault Vroonhove says:

I found that using polar coordinates (and a little trigonometry) gives a really nice proof. Let us rewrite the points $(a,b)$ and $(c,d)$ with polar coordinates: $a = r \cos\alpha$ $b = r \sin\alpha$ $c = s \cos\gamma$ $d = s \sin\gamma$
We thus have the lengths of the parallelogram’s vertices $r$ and $s$, and the angle between them is given by $\gamma -\alpha$. We can compute the area using the formula $A = rs\sin(\gamma-\alpha) = rs (\sin \gamma \cos \alpha - \cos \gamma \sin \alpha) = ad - bc$

Using polar coordinates was nice in the fact that we did not have to express the lengths and the angle in the formula explicity in terms of $a$, $b$, $c$ and $d$, and it simplified very nicely instead.