You may also like

problem icon

Areas and Ratios

What is the area of the quadrilateral APOQ? Working on the building blocks will give you some insights that may help you to work it out.

problem icon

Six Discs

Six circular discs are packed in different-shaped boxes so that the discs touch their neighbours and the sides of the box. Can you put the boxes in order according to the areas of their bases?

problem icon


Given a square ABCD of sides 10 cm, and using the corners as centres, construct four quadrants with radius 10 cm each inside the square. The four arcs intersect at P, Q, R and S. Find the area enclosed by PQRS.

Areas of Parallelograms

Stage: 4 Challenge Level: Challenge Level:2 Challenge Level:2

Tony sent us his answers to the first parts.

(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 ($=2\times 3-4\times -1$)

James, Jamie and George from Gillingham School sent us their formula for the area of the general parallelogram.

$$a d-b c\ \text{ (or } bc-ad\text{ if } bc>ad)$$

Here's one way to work that out.

Parallelogram with rectangle and extra lines
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.