Year 8 Maths Area and Perimeter

Area of a Trapezium

Q: What is the formula for the area of a trapezium?

A = 1/2(a + b)h, where a and b are the parallel sides and h is the height.

Q: How do I find the height of a trapezium from its area?

Divide the area by half the sum of the parallel sides: h = A ÷ 1/2(a + b).

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Practice questions

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Theory

A trapezium has at least one pair of parallel sides, $a$ and $b$, a perpendicular distance $h$ apart. Its area is $A = \dfrac{1}{2}(a + b)h$: add the parallel sides, halve, then multiply by the height. With $a = 4$ cm, $b = 8$ cm and $h = 6$ cm, $A = \dfrac{1}{2} \times 12 \times 6 = 36~\text{cm}^2$. This Year 8 maths page shows how.

$A = \dfrac{1}{2}(a + b)h$, where $a$ and $b$ are the parallel sides and $h$ is the perpendicular height between them. Do the bracket first.

Why it works. Two copies fit together into a parallelogram with base $a + b$ and height $h$, so one trapezium is half of it. Splitting with two heights gives the same result: $\dfrac{1}{2}bh + ah + \dfrac{1}{2}ch = \dfrac{1}{2}h(a + (a + b + c))$, where $a + b + c$ is the bottom side.

Find the parallel sides first. Matching arrows, or two right angles on the same side, show which pair. A trapezium can sit on its side; the sloping sides are never used. If it leans, the height is drawn outside it.

Reverse problems. For $h$, divide the area by $\dfrac{1}{2}(a + b)$. For a missing parallel side, divide by $\dfrac{1}{2}h$ to get $a + b$, then subtract the side you know. If one side is twice the other, call them $x$ and $2x$.

Units, then the rest. One unit before you substitute. The answer may then be changed to hectares, costed per square metre, or divided by a coverage and rounded up.

The parallel sides $a$ and $b$ and the perpendicular height $h$ between them.

A second copy (orange), turned upside down, makes a parallelogram of base $a + b$ and height $h$. One trapezium is half of it.

Two heights split the trapezium into triangle $A_1$, rectangle $A_2$ and triangle $A_3$. Their total is $\dfrac{1}{2}h(a + d)$, where $d = a + b + c$ is the bottom side.

The trapezium formula:

\[ A = \dfrac{1}{2}(a + b)h \]

A = 1 / 2 (a + b) h

          Halve the sum, not one side. Parallel sides \(8\) m and \(14\) m with height \(5\) m: \(\dfrac{1}{2} \times 22 \times 5 = 55~\text{m}^2\).
        

Only the parallel sides go in the bracket. A sloping side is never a, b or h.

How to find the area of a trapezium

Find the parallel sides $a$, $b$ and the perpendicular height $h$, in one unit.
Add the parallel sides and halve the total.
Multiply by the height and give square units.

Example 1 — Straight into the formula

Find the area of a trapezium with a height of $6$ cm and parallel sides of $4$ cm and $8$ cm.

Solution

$A$	$=$	$\dfrac{1}{2} \times (4 + 8) \times 6$
	$=$	$6 \times 6 = 36$

36 cm

$36~\text{cm}^2$

Example 2 — Finding the height

A trapezium has an area of $60~\text{cm}^2$ and parallel sides of $10$ cm and $5$ cm. Find its height.

Solution

$60$	$=$	$\dfrac{1}{2} \times (10 + 5) \times h$
$120$	$=$	$15h$
$h$	$=$	$120 \div 15 = 8$

8 cm

The height is $8$ cm.

Example 3 — Larger numbers

Find the area of a trapezium with parallel sides of $14$ cm and $22$ cm, $9$ cm apart.

Solution

$A$	$=$	$\dfrac{1}{2} \times (14 + 22) \times 9$
	$=$	$18 \times 9 = 162$

162 cm

$162~\text{cm}^2$

Example 4 — Parallel sides upright

The side of a concrete ramp is a trapezium. Its upright ends are $0.4$ m and $1.1$ m, and they are $3$ m apart along the flat ground. Find its area.

Solution

The two upright ends are the parallel sides, so the $3$ m along the ground is the height.

$A$	$=$	$\dfrac{1}{2} \times (0.4 + 1.1) \times 3$
	$=$	$0.75 \times 3 = 2.25$

2.25 m

$2.25~\text{m}^2$

Example 5 — Two units, then cost

A shade sail is a trapezium with parallel edges of $2.5$ m and $170$ cm, $2$ m apart. Fabric costs $\$28$ per square metre. Find the cost.

Solution

$A$	$=$	$\dfrac{1}{2} \times (2.5 + 1.7) \times 2$
	$=$	$4.2$
$4.2 \times 28$	$=$	$117.6$

$ 117.60

$170$ cm $= 1.7$ m. The area is $4.2~\text{m}^2$ and the fabric costs $\$117.60$.

Example 6 — Working backwards

A trapezium has an area of $96~\text{m}^2$ and parallel sides of $15$ m and $25$ m.
(i) Find its height.
(ii) A second trapezium has the same area and height, and one parallel side of $18$ m. Find its other parallel side.

Solution

$96$	$=$	$20 \times h$
$h$	$=$	$4.8$
$96$	$=$	$2.4 \times (18 + b)$
$18 + b$	$=$	$40$
$b$	$=$	$22$

4.8 m 22 m

(i) $4.8$ m
(ii) $22$ m. Check: $\dfrac{1}{2} \times 40 \times 4.8 = 96$.

Example 7 — One side twice the other

A trapezium has an area of $54~\text{cm}^2$ and a height of $4$ cm. One parallel side is twice as long as the other. Find both parallel sides.

Solution

$\dfrac{1}{2} \times (x + 2x) \times 4$	$=$	$54$
$6x$	$=$	$54$
$x$	$=$	$9$

9 cm and 18 cm

The parallel sides are $9$ cm and $18$ cm.

Common pitfalls

Using a sloping side. Only the parallel sides and the perpendicular height go into the formula.

Forgetting the brackets. $\dfrac{1}{2}(a + b)h$ adds $a$ and $b$ first; $\dfrac{1}{2}a + bh$ is wrong.

Mixing units. Change $170$ cm to $1.7$ m before substituting.

Assuming the top and bottom are parallel. Look for the arrows or the right angles; the parallel sides may be upright.

Frequently asked questions

What is the formula for the area of a trapezium?

$A = \dfrac{1}{2}(a + b)h$, where $a$ and $b$ are the parallel sides and $h$ is the height.

Why is there a half in the formula?

Two identical trapeziums make a parallelogram with base a + b, so one trapezium is half of it.

How do I find the height of a trapezium from its area?

Divide the area by half the sum of the parallel sides: $h = A \div \dfrac{1}{2}(a + b)$.

Is a parallelogram a trapezium?

Yes. It has at least one pair of parallel sides, and the formula still works.

Do I use the sloping sides?

No. Only the two parallel sides and the perpendicular height between them.

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Area of Rectangles, Triangles and Parallelograms

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Area of Kites and Rhombuses

\(A\)	\(=\)	\(\dfrac{1}{2} \times (4 + 8) \times 6\)
\(\)	\(=\)	\(6 \times 6 = 36\)

\(60\)	\(=\)	\(\dfrac{1}{2} \times (10 + 5) \times h\)
\(120\)	\(=\)	\(15h\)
\(h\)	\(=\)	\(120 \div 15 = 8\)

\(A\)	\(=\)	\(\dfrac{1}{2} \times (14 + 22) \times 9\)
\(\)	\(=\)	\(18 \times 9 = 162\)

\(A\)	\(=\)	\(\dfrac{1}{2} \times (0.4 + 1.1) \times 3\)
\(\)	\(=\)	\(0.75 \times 3 = 2.25\)

\(A\)	\(=\)	\(\dfrac{1}{2} \times (2.5 + 1.7) \times 2\)
\(\)	\(=\)	\(4.2\)
\(4.2 \times 28\)	\(=\)	\(117.6\)

\(96\)	\(=\)	\(20 \times h\)
\(h\)	\(=\)	\(4.8\)
\(96\)	\(=\)	\(2.4 \times (18 + b)\)
\(18 + b\)	\(=\)	\(40\)
\(b\)	\(=\)	\(22\)

\(\dfrac{1}{2} \times (x + 2x) \times 4\)	\(=\)	\(54\)
\(6x\)	\(=\)	\(54\)
\(x\)	\(=\)	\(9\)

Area of a Trapezium

📖 Theory

How to find the area of a trapezium

Common pitfalls

Frequently asked questions

Theory