Area of parallelograms, rhombuses and trapeziums

A rectangle's area is easy: length times breadth. But a park is rarely a perfect rectangle — it might lean over into a parallelogram, taper into a trapezium, or turn into the diamond shape of a rhombus. Class 8 mathematics answers a natural question: can the same "base times height" idea, already mastered for rectangles and triangles, be stretched to cover these three shapes too? It can, and the reasoning is delightfully physical rather than purely algebraic — you cut a piece off a shape and slide it somewhere else, and the area does not change one bit, only the outline does. This article builds the area formulas for a parallelogram, a trapezium, and a rhombus from that single cutting-and-rearranging idea, shows why the three formulas are secretly cousins of each other rather than three separate things to memorise, and walks through fully worked numerical examples of the kind that show up in board exams and everyday problems — tiling a floor, costing a plot of land, or pricing a garden bed.

The core insight

Every one of these three area formulas comes from cutting a shape apart and rearranging the pieces into a rectangle without gaining or losing any area, so $A = b\times h$ for a parallelogram, $A = \tfrac12(a+b)h$ for a trapezium, and $A = \tfrac12 d_1 d_2$ for a rhombus.

The core idea

Parallelogram — slide a triangle to make a rectangle. Take a parallelogram and drop a perpendicular from one of the top vertices down to the base (extending the base if needed). This slices off a right triangle. Slide that triangle around to the other end of the parallelogram: it fits exactly, and the whole shape becomes a rectangle whose length equals the parallelogram's base $b$ and whose breadth equals the parallelogram's height $h$ — the perpendicular distance between the two parallel sides, not the slanting side. Since cutting and sliding never changes area,

A_{\text{parallelogram}} = b \times h

where: $b$ is the base of the parallelogram and $h$ is the perpendicular height measured at right angles to that base.

Trapezium — two copies make a parallelogram. A trapezium has exactly one pair of parallel sides, usually called $a$ and $b$ , separated by a perpendicular height $h$ . Take a second, identical copy of the trapezium, flip it $180°$ , and glue it onto the first along one of the non-parallel sides. The two copies together form a parallelogram whose base is the sum $a+b$ and whose height is still $h$ . That combined parallelogram has area $(a+b)h$ by the formula above, and since it is made of two equal trapeziums, one trapezium is exactly half of it:

A_{\text{trapezium}} = \frac12 (a+b)\, h

where: $a$ and $b$ are the lengths of the two parallel sides and $h$ is the perpendicular distance between them.

Rhombus — diagonals build a rectangle. A rhombus is a parallelogram with all four sides equal, and a special property: its two diagonals always cross each other at right angles and bisect each other. Draw a rectangle around the rhombus so that the rectangle's sides pass through the four vertices of the rhombus, parallel to the diagonals. That bounding rectangle has dimensions $d_1$ by $d_2$ , the two diagonal lengths, and the rhombus turns out to occupy exactly half of that rectangle's area — the four corner triangles left outside the rhombus are, together, equal in area to the four triangles that make up the rhombus itself. So

A_{\text{rhombus}} = \frac12\, d_1 d_2

where: $d_1$ and $d_2$ are the lengths of the rhombus's two diagonals.

Figure

A parallelogram with a right triangle sliced from its left end sliding to the right end to form a rectangle of the same base and height

It helps to see how the three formulas connect rather than treating them as unrelated rules. A rhombus is a special parallelogram, so its area could equally be found as $b\times h$ if its base and height were known — the diagonal formula $\tfrac12 d_1 d_2$ is just a more convenient route when the diagonals are given instead. A parallelogram, in turn, is a trapezium whose two "parallel sides" happen to be equal, $a=b$ ; substituting into the trapezium formula gives $\tfrac12(b+b)h = bh$ , exactly the parallelogram formula. Keeping this chain in mind — trapezium generalises to parallelogram generalises to rhombus's alternative form — makes all three formulas far easier to recall than memorising them as three isolated facts.

Why it works

Derivation AWhy cutting and rearranging never changes area

The goal is to justify, with nothing beyond the idea that area is additive over non-overlapping pieces, why a parallelogram's area equals base times height even though its slanted sides make it "look" different from a rectangle of the same footprint, and why this same reasoning is what ultimately licenses the trapezium and rhombus formulas above.

Step 1. Start from the one fact everyone already accepts: the area of a rectangle is length times breadth, and if a region is cut into pieces that are rearranged without overlapping and without gaps, the total area of the pieces is unchanged, because area is just "how much surface is covered," and covering the same amount of surface in a different arrangement covers the same amount.

Step 2. Apply this to the parallelogram. Dropping a perpendicular from a top vertex to the base line splits the parallelogram into a right triangle and a right trapezium-shaped remainder (a quadrilateral with two right angles). Sliding the triangle to the opposite end and reattaching it along the parallel side it was cut from produces a four-sided figure with all four angles equal to $90°$ — by construction, since the triangle's cut edge was perpendicular to the base and it now lines up with the opposite perpendicular edge. A quadrilateral with four right angles and opposite sides equal is a rectangle.

Step 3. Identify the rectangle's dimensions. The rectangle's length is still the parallelogram's original base $b$ , because the base line was never cut, only the piece above it was rearranged; its breadth is the perpendicular height $h$ of the parallelogram, because the sliding motion was purely horizontal along the base direction and never changed how far the top edge sits above the bottom edge. So the rectangle has area $b\times h$ , and since it was built from exactly the same two pieces as the original parallelogram with nothing added or removed, the parallelogram's area is also $b \times h$ .

Step 4. The trapezium and rhombus arguments in the previous section lean on the identical principle — doubling a trapezium into a parallelogram, or inscribing a rhombus in a bounding rectangle — both rely on this same additive property of area under cutting, sliding and reflecting, which is why all three formulas can be trusted even though none of them is "obvious" the way a rectangle's is.

Worked example

A parallelogram-shaped park has a base of $25\ \text{m}$ and a height of $12\ \text{m}$ . Within it sits a rhombus-shaped flowerbed with diagonals $10\ \text{m}$ and $8\ \text{m}$ , and next to the park is a trapezium-shaped strip with parallel sides $22\ \text{m}$ and $14\ \text{m}$ and perpendicular width $9\ \text{m}$ . Find the area of the park, the flowerbed, and the strip, and the total cost of the strip at ₹1,500 per square metre.

Step 1. Find the park's area using $A = b\times h$ : $A_{\text{park}} = 25 \times 12 = 300\ \text{m}^2$ .

Step 2. Find the flowerbed's area using $A = \tfrac12 d_1 d_2$ : $A_{\text{bed}} = \tfrac12 \times 10 \times 8 = \tfrac12 \times 80 = 40\ \text{m}^2$ .

Step 3. Find the strip's area using $A = \tfrac12(a+b)h$ : $A_{\text{strip}} = \tfrac12(22+14)(9) = \tfrac12 \times 36 \times 9 = 18\times 9 = 162\ \text{m}^2$ .

Step 4. Find the strip's cost: $\text{Cost} = 162 \times 1500 = ₹2{,}43{,}000$ .

The park covers $300\ \text{m}^2$ , the flowerbed covers $40\ \text{m}^2$ , the strip covers $162\ \text{m}^2$ , and the strip costs ₹2,43,000 in total.

Competitive stretch

If the same rhombus flowerbed were instead described by its side length ( $\approx 6.4\ \text{m}$ , from the right triangles the diagonals form) and its height rather than its diagonals, you would use $A = b\times h$ instead of $\tfrac12 d_1 d_2$ — same shape, same $40\ \text{m}^2$ answer, different route, because a rhombus is simultaneously a parallelogram and a diagonal-defined quadrilateral, and either formula must agree.

Common traps

Using a slant side of the parallelogram or trapezium as the "height" instead of the true perpendicular distance between the base and the opposite side.
In the trapezium formula, adding the two non-parallel sides instead of the two parallel sides — only the parallel sides go into $(a+b)$ .
Forgetting the $\tfrac12$ in the rhombus formula and computing $d_1 \times d_2$ directly, which doubles the correct area.
Assuming any quadrilateral with equal sides has perpendicular diagonals — the $\tfrac12 d_1 d_2$ formula relies specifically on the diagonals crossing at right angles, a property of rhombuses and squares, not of general equal-sided quadrilaterals.

Ask the tutor: Want to see the parallelogram-to-rectangle slide or the trapezium-doubling trick drawn out step by step for your own numbers?

Why does the rhombus formula use diagonals instead of base and height like a parallelogram?

A rhombus is a parallelogram, so $A = b\times h$ works for it too, and both formulas must give the same answer for the same rhombus. The diagonal formula $\tfrac12 d_1 d_2$ exists because it is often easier to measure or is given directly in a problem, and it exploits the special fact — true only for rhombuses and squares — that the diagonals cross at right angles, letting the shape be rebuilt into a simple bounding rectangle of dimensions $d_1$ by $d_2$ , half of which is the rhombus.

Do all four of the trapezium's sides matter in the area formula?

No. Only the two parallel sides, usually labelled $a$ and $b$ , and the perpendicular height $h$ between them enter the formula $\tfrac12(a+b)h$ . The two non-parallel (slanting) sides can have any length and do not appear in the area calculation at all — they matter for finding the perimeter, not the area.

Can I use these formulas if I am only given the slanting side and not the perpendicular height?

Not directly — the perpendicular height must be found first, often using the Pythagorean relationship if you know the slant length and how far the top edge is offset horizontally from the base, or it may be given separately in the figure. Plugging a slant length in as if it were the height is one of the most common marks-losing mistakes in this chapter, so always check that the length you are using meets the base at a right angle before substituting it into $A=b\times h$ or $A=\tfrac12(a+b)h$ .