Quadrilateral area and perimeter

A tiled floor, a kite in the sky, a farmer's field with one slanted edge — all of them are quadrilaterals, and no two use the same formula. A square and a rectangle you already trust from primary school, but a parallelogram, a rhombus, and a trapezium each hide a different geometric trick, and an irregular field with no name at all needs a trick of its own: cut it into triangles. This section collects every area and perimeter rule Class 9 ICSE mensuration expects for quadrilaterals, shows why the rhombus formula $\tfrac12 d_1 d_2$ works rather than just stating it, and drills the two skills examiners actually test — picking the correct "height" in a diagram, and splitting a general four-sided field along a diagonal using Heron's formula. By the end you should be able to look at any quadrilateral, name its type, and reach for the right formula without hesitation.

The core insight

Every quadrilateral's area formula is really a triangle-area formula in disguise — a rectangle is two triangles, a rhombus is four, and any irregular quadrilateral can always be split along one diagonal into two triangles and measured that way.

The core idea

Five named shapes cover almost every board question, plus one general method for anything else.

Square and rectangle need only their sides. For a square of side $a$ , and a rectangle of length $l$ and breadth $b$ :

P_{\text{square}} = 4a, \quad A_{\text{square}} = a^2, \qquad P_{\text{rect}} = 2(l+b), \quad A_{\text{rect}} = l \times b

where: $a$ is the side of the square, $l$ and $b$ are the length and breadth of the rectangle, $P$ is perimeter, $A$ is area.

Parallelogram. Opposite sides are equal, so the perimeter is twice the sum of the two adjacent sides. The area, however, needs a perpendicular height, not a slant side:

A = b \times h

where: $b$ is any side chosen as the base and $h$ is the perpendicular distance from that base to the opposite side — not the length of the slanting adjacent side.

Rhombus. All four sides are equal, and the two diagonals cross each other at right angles, cutting the shape into four congruent right triangles. That perpendicularity is what makes this formula work:

A = \frac{1}{2} \times d_1 \times d_2

where: $d_1$ and $d_2$ are the lengths of the two diagonals. Because a square is a special rhombus, this same formula also gives a square's area from its diagonals.

Trapezium. Only one pair of sides is parallel, and the height must be measured between those two parallel sides:

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

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

Any irregular quadrilateral — no special name, no parallel or equal sides guaranteed — is handled by drawing one diagonal, splitting the figure into two triangles, finding each triangle's area (using Heron's formula when only sides are known), and adding them. This is the one universal method that never fails, because every quadrilateral is exactly two triangles glued along a diagonal.

Figure

Four quadrilateral shapes side by side showing the base and height used in each area formula, with a fifth irregular quadrilateral cut by a dashed diagonal into two triangles

Derivation AWhy the rhombus area formula is half the product of the diagonals

The goal is to show, using only the properties of a rhombus already known from the Geometry chapter — that its diagonals bisect each other at right angles — why the area of a rhombus with diagonals $d_1$ and $d_2$ works out to exactly $\tfrac12 d_1 d_2$ , rather than simply memorising the result.

Step 1. Draw a rhombus $ABCD$ with diagonals $AC$ and $BD$ meeting at point $O$ . Since the diagonals of a rhombus bisect each other, $AO = OC = \tfrac{d_1}{2}$ and $BO = OD = \tfrac{d_2}{2}$ , where $d_1 = AC$ and $d_2 = BD$ .

Step 2. Since the diagonals also meet at right angles, the point $O$ splits the rhombus into four right triangles: $\triangle AOB$ , $\triangle BOC$ , $\triangle COD$ , $\triangle DOA$ , each with legs $\tfrac{d_1}{2}$ and $\tfrac{d_2}{2}$ .

Step 3. Each right triangle has area $\tfrac12 \times \tfrac{d_1}{2} \times \tfrac{d_2}{2} = \tfrac{d_1 d_2}{8}$ . All four triangles are congruent (equal legs, right angle), so the total area of the rhombus is four times one triangle's area.

Step 4. Add the four equal pieces: $A = 4 \times \dfrac{d_1 d_2}{8} = \dfrac{d_1 d_2}{2} = \dfrac12 d_1 d_2$ . The formula is therefore not a separate rule to memorise but a direct consequence of the diagonals being perpendicular bisectors of each other — which is also why the same $\tfrac12 d_1 d_2$ trick works for any quadrilateral whose diagonals happen to be perpendicular, not rhombuses alone.

Worked example

Problem. A rhombus-shaped kite has diagonals of length 16 cm and 12 cm, and its side length is 10 cm. Find its area and perimeter.

Step 1. All four sides of a rhombus are equal, so the perimeter is

P = 4 \times 10 = 40 \text{ cm}

Step 2. Apply the diagonal formula just derived:

A = \frac{1}{2} \times d_1 \times d_2 = \frac{1}{2} \times 16 \times 12 = \frac{192}{2} = 96 \text{ cm}^2

Step 3 — check. The diagonals bisect each other at right angles, so each side is the hypotenuse of a right triangle with legs $8$ cm and $6$ cm: $\sqrt{8^2+6^2} = \sqrt{64+36} = \sqrt{100} = 10$ cm, matching the given side and confirming the figure is consistent. The rhombus has area 96 cm² and perimeter 40 cm.

Competitive stretch

Problem. A field $ABCD$ has $AB = 15$ m, $BC = 20$ m, $CD = 9$ m, $DA = 10$ m, and diagonal $AC = 17$ m, with $\angle ABC = 90^\circ$ . Find the area of the field.

Step 1. Split the quadrilateral along diagonal $AC$ into $\triangle ABC$ and $\triangle ACD$ .

Step 2. Since $\angle ABC = 90^\circ$ , sides $AB = 15$ m and $BC = 20$ m are the legs of a right triangle:

A(\triangle ABC) = \frac{1}{2} \times 15 \times 20 = 150 \text{ m}^2

Check: $15^2 + 20^2 = 225 + 400 = 625 = 17^2$ , confirming $AC = 17$ m is indeed the hypotenuse.

Step 3. For $\triangle ACD$ , only the three sides $AC = 17$ m, $CD = 9$ m, $DA = 10$ m are known, so use Heron's formula:

s = \frac{17+9+10}{2} = 18 \text{ m}, \qquad A = \sqrt{18(18-17)(18-9)(18-10)} = \sqrt{18 \times 1 \times 9 \times 8} = \sqrt{1296} = 36 \text{ m}^2

Step 4. Add the two triangle areas: $A(ABCD) = 150 + 36 = \textbf{186 m}^2$ .

Common traps

Using a slant side as the "height" of a parallelogram — the height in $A = b \times h$ must be the perpendicular distance to the opposite side, never the length of the slanting adjacent side.
Applying $\tfrac12 d_1 d_2$ to an ordinary parallelogram — this shortcut only works when the diagonals meet at right angles (rhombus or square), not for a general parallelogram.
Measuring a trapezium's height along the slanting leg instead of perpendicular between the two parallel sides.
Splitting an irregular quadrilateral without checking the two resulting triangles are valid — each "side" after the split must be less than the sum of the other two, as in the field example above.

Ask the tutor: Want a step-by-step walkthrough of picking the right height in a tricky parallelogram or trapezium diagram, or practice splitting an irregular field into triangles?

Why does half d1 d2 work for a rhombus but not for a general parallelogram?

The formula relies entirely on the diagonals meeting at right angles, because that is what splits the shape into four congruent right triangles whose combined area is $\tfrac12 d_1 d_2$ — shown step by step in the derivation above. A rhombus (and its special case, the square) always has perpendicular diagonals, so the shortcut applies. An ordinary parallelogram's diagonals usually cross at some other angle, so the same triangles are not right triangles and the shortcut breaks down; you must fall back on $A = b \times h$ instead.

How do I know which side to call the base when finding a parallelogram's height?

Any side can serve as the base — the formula $A = b \times h$ gives the same area whichever side you pick, as long as $h$ is measured perpendicular to that chosen base. In a diagram, look for the dashed line marked with a right-angle symbol; that segment, not any slanting side, is your $h$ . If a diagram gives only the four sides with no perpendicular marked, you cannot find the area from the parallelogram formula alone — you would need to split it into triangles instead.

When splitting a field into two triangles, how do I decide whether to use Heron's formula or base-height?

Use base-height whenever you can identify a right angle or a clearly marked perpendicular height for a triangle — it is faster and less error-prone, as in $\triangle ABC$ of the field example, where the right angle at $B$ made $\tfrac12 \times 15 \times 20$ immediate. Use Heron's formula when you only know the three side lengths of a triangle with no right angle or height given, as with $\triangle ACD$ there. Always compute the semi-perimeter $s$ first, then $\sqrt{s(s-a)(s-b)(s-c)}$ , and double-check by confirming the three sides can actually form a triangle.

Quadrilateral area and perimeter

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads