Area and perimeter

Perimeter and area

Two different measurements describe a flat shape.

• The perimeter is the total distance around the outside edge. It is a length, so it is measured in mm, cm, m or km.
• The area is the amount of flat surface inside the shape. It is measured in square units: ${\text{cm}}^2$, ${\text{m}}^2$ and so on. An area of $12{\text{cm}}^2$ means the shape could be covered by 12 squares of side 1 cm (cut up and rearranged if necessary).

To find a perimeter, add the lengths of all the sides. Make sure you have every side: some may be marked equal, or be equal because the shape is a rectangle or parallelogram, and are not labelled.

On the exam paper, only the formula for the area of a triangle is printed in the List of formulas. You must know the formulas for the rectangle, the parallelogram and the trapezium. Understanding where each one comes from makes them easy to remember, and they all come from the rectangle.

Rectangles

A rectangle of length $l$ and width $w$ can be filled with $w$ rows of $l$ unit squares, so

$$\text{Area}=l\times w$$

Its perimeter is two lengths and two widths:

$$\text{Perimeter}=2l+2w=2(l+w)$$

For a rectangle 7 cm by 4.5 cm: area $=7\times 4.5=31.5{\text{cm}}^2$ and perimeter $=2\times (7+4.5)=23$ cm. A square is a rectangle with all sides equal, so a square of side $s$ has area $s^2$ and perimeter $4s$.

Triangles

For a triangle with base $b$ and perpendicular height $h$:

$$\text{Area}=\frac{1}{2}bh$$

The perpendicular height is measured at right angles to the base, from the base (or the base extended) to the opposite corner. It is usually not one of the sides.

Why it works. Draw the rectangle with the same base and height around the triangle. The height line splits the triangle into two right-angled triangles, and each of those is exactly half of one part of the rectangle. So the triangle is half the rectangle, $\frac{1}{2}\times b\times h$.

Any side can be the base, as long as you use the height that is perpendicular to it. In a right-angled triangle the two sides that meet at the right angle are a base and its height, so the area is half their product.

The height $h$ meets the base at a right angle. The sloping sides are longer than $h$ and are not used for the area.

Parallelograms

For a parallelogram with base $b$ and perpendicular height $h$:

$$\text{Area}=bh$$

Why it works. Cut the right-angled triangle off one end of the parallelogram and slide it to the other end. The pieces fit exactly, because opposite sides of a parallelogram are parallel and equal, and the result is a rectangle $b$ long and $h$ high. Nothing was added or removed, so the area is $bh$.

The sloping side of a parallelogram is not its height. A parallelogram with base 10 cm, sloping sides 6 cm and height 5 cm has area $10\times 5=50{\text{cm}}^2$, and perimeter $2\times (10+6)=32$ cm. The perimeter uses the sloping sides; the area uses the height.

The shaded triangle moves to the dashed position on the right, turning the parallelogram into a rectangle $b$ by $h$.

Trapeziums

A trapezium has one pair of parallel sides. Call their lengths $a$ and $b$, and the perpendicular distance between them $h$. Then

$$\text{Area}=\frac{1}{2}(a+b)h$$

In words: add the parallel sides, multiply by the distance between them and halve.

Why it works. Take two copies of the trapezium and turn one upside down beside the other, as in the diagram. Together they make a parallelogram with base $a+b$ and height $h$, of area $(a+b)h$. One trapezium is half of that.

Another way to see it: $\frac{1}{2}(a+b)$ is the mean of the parallel sides, so a trapezium has the same area as a rectangle whose length is the average of its two parallel sides.

Work out the bracket first. For parallel sides 6 cm and 10 cm, 5 cm apart: $\frac{1}{2}\times (6+10)\times 5=\frac{1}{2}\times 16\times 5=40{\text{cm}}^2$.

Two identical trapeziums, one turned round, make a parallelogram with base $a+b$ and height $h$.

Working backwards from an area or perimeter

If you know the area, or the perimeter, you can find a missing length. Write the formula with what you know substituted, then solve the equation.

• A rectangle has area $54{\text{cm}}^2$ and length 9 cm. $9w=54$, so $w=6$ cm.
• A triangle has area $36{\text{cm}}^2$ and base 8 cm. $\frac{1}{2}\times 8\times h=36$, so $4h=36$ and $h=9$ cm. Do not forget the half: dividing 36 by 8 gives 4.5, which is wrong.
• A rectangle has perimeter 30 cm and length 9 cm. The two lengths use 18 cm, leaving 12 cm for the two widths, so each width is 6 cm.

With a trapezium, the bracket stays together. If the area is $75{\text{cm}}^2$ and the parallel sides are 7 cm and 13 cm, then $\frac{1}{2}\times 20\times h=75$, so $10h=75$ and $h=7.5$ cm.

Choosing the right height, and the right units

Most lost marks in this topic come from two slips.

Using a sloping side as the height. For the area of a triangle, parallelogram or trapezium, the height must be perpendicular to the base (or to the parallel sides). On a diagram it is usually drawn dashed, with a right-angle mark. Sloping sides are only used for the perimeter. If a diagram shows both, check which is marked with a right angle.

Mixing units. If one length is in metres and another in centimetres, change one of them before multiplying. A rectangle 2 m by 45 cm is $200\times 45=9000{\text{cm}}^2$, or $2\times 0.45=0.9{\text{m}}^2$, never $2\times 45=90$. Areas are in square units and perimeters in ordinary units: write ${\text{cm}}^2$ for an area and cm for a perimeter.

On a grid, count squares to find a base and a height. For a triangle whose top corner is beyond the end of its base, the perpendicular height falls outside the triangle, but the formula $\frac{1}{2}bh$ still works: measure the height straight up from the line of the base.