Areas of Parallelograms

(a) $6$ units squared ($=3\times 2-0\times 5$)

(b) $12$ units squared ($=3\times 4-2\times 0$)

(c) $9$ units squared ($=4\times 3-1\times 3$)

(d)$10$ units squared ($=2\times 3-4\times -1$)

Here's one way to work that out.

Remember that $\mathbf{p}=\left(\begin{array}{c}a \\ b\end{array}\right)$ and $\mathbf{q}=\left(\begin{array}{c}c\\d\end{array}\right)$.

 

The diagram shows the parallelogram inside a rectangle with base $a+c$ and height $b+d$. This rectangle has area $(a+c)(b+d)$. If we could find the areas of the triangles outside the parallelogram, then we could subtract them from the area of the rectangle to find the area of the parallelogram.

 

The bottom triangle has height $b$ and base $a+c$, so has area $\frac{1}{2}b(a+c)$.

 

The triangle on the left has (if you imagine it on its side so that its base is at the bottom) base $b+d$ and height $c$, so has area $\frac{1}{2}c(b+d)$.

 

The triangle at the top is congruent to the triangle at the bottom (it has been rotated by $180^{\circ}$), so has area $\frac{1}{2}b(a+c)$.

 

The triangle on the right is congruent to the triangle on the left, so has area $\frac{1}{2}c(b+d)$.

 

So the area of the parallelogram is $$(a+c)(b+d)-2\times\frac{1}{2}b(a+c) -2\times\frac{1}{2}c(b+d)=$$
$$a b+b c+a d+c d-a b-b c-b c-c d=a d-b c$$

 

just as James, Jamie and George said.