Area of trapezium

A trapezium is a quadrilateral with exactly one pair of parallel sides, and finding its area is one of the key mensuration skills in ICSE Class 8 Mathematics. You already know how to find the area of a triangle, rectangle, square and parallelogram — the trapezium formula builds directly on the parallelogram idea, by showing that two congruent trapeziums placed back to back always form a parallelogram. Once that single picture is fixed in your mind, the formula $A = \frac{1}{2}(a+b)h$ stops being something to memorise and becomes something you can rebuild from scratch in an exam. This topic also sets up the next two topics in the chapter: finding the area of a general quadrilateral using a diagonal and offsets, and finding the area of an irregular polygon by splitting it into triangles and trapeziums.

The core insight

The area of a trapezium is half the sum of its two parallel sides multiplied by the perpendicular height between them, $A = \tfrac12(a+b)h$ , because two congruent trapeziums always join to form a parallelogram of base $(a+b)$ and height $h$ .

The core idea

A trapezium has exactly one pair of parallel sides. Call their lengths $a$ and $b$ ; call the perpendicular distance between them the height, $h$ . Unlike a rectangle or parallelogram, the two non-parallel sides can be of any length and need not be equal — they play no role in the area formula at all. The area is given by:

A = \frac{1}{2}(a+b)\,h

where: $A$ is the area of the trapezium, $a$ and $b$ are the lengths of the two parallel sides, and $h$ is the perpendicular height between them, all measured in the same unit so that $A$ comes out in that unit squared (for example $\text{cm}^2$ or $\text{m}^2$ ). A few points worth fixing before you use the formula:

$a$ and $b$ must be the two parallel sides in the figure — not any two sides you happen to pick.
$h$ is the shortest (perpendicular) distance between the parallel sides, never the length of a slanted side.
The formula quietly contains the parallelogram formula as a special case: if $a=b$ , the trapezium becomes a parallelogram and $A=\tfrac12(a+a)h = ah$ , which matches base $\times$ height. Checking that a new formula collapses to an old, familiar one is a useful habit whenever you meet a new mensuration result.

Figure

A trapezium with parallel sides a and b and height h, next to a rotated copy of itself joined to it to form a parallelogram of base a plus b and the same height h

Deriving the area formula

Derivation ADeriving the Area of a Trapezium

The goal of this derivation is to show, using only the already-known area formula for a parallelogram, that a trapezium with parallel sides $a$ and $b$ and height $h$ has area exactly $\tfrac12(a+b)h$ , without assuming the result in advance.

Step 1. Take a trapezium with parallel sides $a$ and $b$ and perpendicular height $h$ . Make an exact congruent copy of it.

Step 2. Rotate the copy by $180^\circ$ and place it next to the original trapezium so that one slanted side of the copy exactly matches up with one slanted side of the original. Because the copy is congruent and rotated by a half-turn, the two shapes fit together with no gap and no overlap.

Step 3. Look at the combined figure. Its two long sides are made from the side of length $a$ of one trapezium joined end to end with the side of length $b$ of the other, so each long side of the combined figure has length $(a+b)$ . Both long sides are parallel to each other (since they came from the original parallel sides), and the perpendicular distance between them is still $h$ , because rotation by $180^\circ$ does not change perpendicular distances. This combined figure is therefore a parallelogram with base $(a+b)$ and height $h$ .

Step 4. The area of a parallelogram is base times height, so the combined figure has area $(a+b)\times h$ .

Step 5. Since the combined figure is made of two congruent (equal-area) trapeziums, one trapezium is exactly half of that total area. Dividing by $2$ gives $A = \frac{1}{2}(a+b)h$ , which is the formula for the area of a single trapezium.

Worked example

Problem: The parallel sides of a trapezium are $18\ \text{cm}$ and $12\ \text{cm}$ , and the perpendicular distance between them is $7\ \text{cm}$ . Find its area.

Step 1. Identify the known values: $a = 18\ \text{cm}$ , $b = 12\ \text{cm}$ , $h = 7\ \text{cm}$ .

Step 2. Write the formula: $A = \frac{1}{2}(a+b)h$ .

Step 3. Substitute the values: $A = \frac{1}{2}(18+12)\times 7 = \frac{1}{2}\times 30\times 7$ .

Step 4. Simplify: $A = 15\times 7 = 105\ \text{cm}^2$ .

Answer: the area of the trapezium is $105\ \text{cm}^2$ .

Competitive stretch

Problem: A trapezium-shaped field has an area of $480\ \text{m}^2$ . One of its parallel sides is $34\ \text{m}$ and the height of the field is $16\ \text{m}$ . Find the length of the other parallel side.

Step 1. Write the formula with the known values: $A = \frac{1}{2}(a+b)h$ , with $A = 480\ \text{m}^2$ , $a = 34\ \text{m}$ , $h = 16\ \text{m}$ , and $b$ unknown.

Step 2. Substitute: $480 = \frac{1}{2}(34+b)\times 16$ .

Step 3. Simplify the right-hand side: $480 = 8(34+b)$ .

Step 4. Divide both sides by $8$ : $60 = 34+b$ .

Step 5. Solve for $b$ : $b = 60-34 = 26\ \text{m}$ .

Answer: the other parallel side is $26\ \text{m}$ long.

Common traps

Using a slant side as the height. The height $h$ is always the perpendicular distance between the two parallel sides — never the length of one of the slanted (non-parallel) sides, even when it looks like the "obvious" side in a figure.
Adding all four sides instead of only the parallel ones. Only $a$ and $b$ , the pair of parallel sides, go inside the bracket $(a+b)$ ; the two non-parallel sides do not appear anywhere in the formula.
Forgetting the factor of one half. A common slip is computing $(a+b)\times h$ and stopping there, forgetting that a single trapezium is only half of the parallelogram formed by two congruent copies of it.
Mixing units. If one parallel side is given in metres and the height in centimetres, convert both to the same unit before substituting, otherwise the answer will be wrong by a large, easy-to-miss factor.
Confusing the trapezium with other quadrilaterals in a mixed problem set. A trapezium has exactly one pair of parallel sides, a parallelogram has two, and a general quadrilateral may have none — always check the figure or the given data before deciding which area formula applies, since applying the parallelogram or triangle formula to a trapezium by mistake gives a wrong answer even if the arithmetic is done correctly.

Because this topic sits inside a wider mensuration chapter, it is also worth connecting the trapezium formula back to ideas you already know well, and forward to ideas you are about to meet. Looking back, the trapezium formula is built entirely from the parallelogram formula, which itself was built from the rectangle formula — each new mensuration result in this course reduces to an older, simpler one under the right special case, and spotting that chain is often the fastest way to recall a formula you have half-forgotten during an exam. Looking forward, the next topic in this chapter uses a very similar "split the shape into simpler shapes" idea to find the area of a general quadrilateral from one of its diagonals and the perpendicular offsets to the other two vertices, and the topic after that extends the same idea again to irregular polygons, by cutting them into a fan of triangles and trapeziums whose areas you already know how to compute. Recognising that the trapezium formula is not an isolated fact but one link in this chain will make the rest of the chapter far easier to follow.

Ask the tutor: Not sure whether a particular side in your figure counts as the height or as a parallel side? Ask your AI tutor to walk through labelling the sides of your own trapezium diagram and check your area calculation step by step.

Why does the trapezium formula use only the parallel sides and not all four sides

Because the area depends only on how far apart the two parallel sides are and how long they are on average, not on how the non-parallel sides slant to connect them. Two trapeziums can have very different non-parallel side lengths yet the same $a$ , $b$ and $h$ , and they will always have the same area — the slanted sides only affect the shape's outline, not the enclosed area.

What if I am not told which sides are parallel in a diagram

Look for tick marks or arrow marks on the sides, or check the problem statement, since ICSE questions usually label the parallel sides directly or state their lengths as "the parallel sides are...". If neither is given, parallel sides are the ones drawn as the top and bottom of the trapezium shape, with the slanted left and right sides being the non-parallel ones.

Can I use this formula if the trapezium is drawn tilted or upside down on the page

Yes. The formula only cares about which two sides are parallel and the perpendicular distance between them — it does not matter how the trapezium is oriented on the page. Just make sure you correctly identify the pair of parallel sides as $a$ and $b$ , and measure $h$ perpendicular to them, regardless of how the figure is rotated.