The Average of the Two Parallel Sides Does the Work

Because a trapezium is a rectangle that has been tilted at one end — and the tilt lends from one side exactly what it gives to the other.

If the parallel sides are $10\text{cm}$ and $6\text{cm}$, the trapezium covers the same area as a rectangle of width $8\text{cm}$ — the average — on the same height.

That is the key formula of the second part of the CBSE Class 8 Mathematics chapter on area. This page derives it, then uses the same splitting idea on any quadrilateral, any polygon, and shapes with a hole cut out.

Draw a diagonal, drop a perpendicular from each of the other two vertices onto it, and add the two triangles.

Both triangles sit on the same base — the diagonal — so the working is short:

$$\text{Area}=\frac{1}{2}\times d\times (h_1+h_2)$$

Worked example. A quadrilateral has diagonal $d=15\text{cm}$, with perpendiculars of $6\text{cm}$ and $8\text{cm}$ dropped onto it from the two opposite vertices:

$$\text{Area}=\frac{1}{2}\times 15\times (6+8)=\frac{1}{2}\times 15\times 14=105{\text{cm}}^2$$

Checking it as two separate triangles:

$$\frac{1}{2}\times 15\times 6=45{\text{cm}}^2,\frac{1}{2}\times 15\times 8=60{\text{cm}}^2$$

and $45+60=105{\text{cm}}^2$. Correct.

When the diagonals are perpendicular. If a quadrilateral's diagonals cross at right angles — as in a rhombus, a square or a kite — each diagonal acts as the other's height, and the formula shortens to

$$\text{Area}=\frac{1}{2}\times d_1\times d_2$$

With diagonals $12\text{cm}$ and $16\text{cm}$:

$$\frac{1}{2}\times 12\times 16=96{\text{cm}}^2$$

That is the same $\frac{1}{2}\times d\times (h_1+h_2)$ rule in disguise, with $h_1$ and $h_2$ being the two halves of the other diagonal.

Split the shape into triangles, rectangles and trapeziums and add; or subtract the missing piece from a bigger shape. Choose whichever needs fewer measurements.

A polygon by splitting. A six-sided figure can often be cut by two lines into a rectangle with a triangle on each side. Say the rectangle measures $10\text{cm}$ by $6\text{cm}$, and each triangle stands on the rectangle's $6\text{cm}$ side with height $4\text{cm}$:

$$\text{Rectangle}=10\times 6=60{\text{cm}}^2$$

$$\text{Two triangles}=2\times \frac{1}{2}\times 6\times 4=24{\text{cm}}^2$$

$$\text{Total}=60+24=84{\text{cm}}^2$$

A path around a garden, by subtracting. A rectangular garden is $20\text{m}$ by $15\text{m}$, with a path $2\text{m}$ wide running around the outside.

The outer rectangle gains $2\text{m}$ at each end, so it measures $20+4=24\text{m}$ by $15+4=19\text{m}$:

$$\text{Outer}=24\times 19=456{\text{m}}^2$$

$$\text{Garden}=20\times 15=300{\text{m}}^2$$

$$\text{Path}=456-300=156{\text{m}}^2$$

The trap is adding $2\text{m}$ once instead of twice, giving $22\times 17=374$ and a path of only $74{\text{m}}^2$. The path wraps both sides of every dimension.

Checking it another way. Treat the path as four strips plus four corner squares:

$$2\times (20\times 2)+2\times (15\times 2)+4\times (2\times 2)=80+60+16=156{\text{m}}^2$$

The two methods agree, which is the strongest check available on this kind of question.

Substitute into the formula and solve the equation. The trapezium formula has three inputs, so a question can hide any one of them.

Example 1 — a missing parallel side. A trapezium has area $84{\text{cm}}^2$, height $6\text{cm}$, and one parallel side $10\text{cm}$. Find the other.

$$84=\frac{1}{2}\times (10+b)\times 6=3(10+b)$$

$$10+b=28,b=18\text{cm}$$

Checking: $\frac{1}{2}\times (10+18)\times 6=\frac{1}{2}\times 28\times 6=84{\text{cm}}^2$. Correct.

Example 2 — a missing height. A trapezium has area $180{\text{cm}}^2$ with parallel sides $16\text{cm}$ and $14\text{cm}$:

$$180=\frac{1}{2}\times (16+14)\times h=15h,h=12\text{cm}$$

Example 3 — a missing diagonal. A rhombus has area $120{\text{cm}}^2$ and one diagonal $20\text{cm}$:

$$120=\frac{1}{2}\times 20\times d_2=10d_2,d_2=12\text{cm}$$

Example 4 — a cost question. Flooring the $84{\text{cm}}^2$ trapezium above costs, at a rate of $\text{₹}45$ per square unit,

$$84\times 45=\text{₹}3780$$

The area is always the middle step in a cost question: find it first, then multiply by the rate.