Quadrilaterals — square, rectangle, parallelogram, rhombus, trapezium, kite: Worked examples

This page is a training ground. We take every kind of question a quadrilateral can throw at you — angles, areas, missing sides, degenerate shapes, word problems, exam twists — and work each one from the ground up. Before you read a solution, guess the answer at the "Forecast" line. Guessing wrong is how the shape of the problem sticks in your memory.

If a symbol here feels unfamiliar, it was built in the parent note. We lean on three earlier tools: triangles (a quadrilateral is two triangles glued together), the Pythagorean theorem (for missing sides in right triangles), and parallel-line angle rules (for the angle chases).

Definition Two extra tools we build from scratch on this page

A couple of examples need helpers that the parent note did not use. Rather than assume them, here is each one in plain words, anchored to a picture you will meet below.

The "vertical slice" of a slanted side (Example 5). When a side of length $L$ leans away from the ground at an angle, its vertical height — how far up it reaches — is shorter than $L$ . The exact rule that turns a slanted length into its vertical height is the sine of the angle, written $sin (angle)$ : it is defined as $\frac{opposite side}{hypotenuse}$ inside a right triangle. For the special angle $30°$ it happens that $sin 30° = \frac{1}{2}$ (the vertical reach is exactly half the slant). We only use this one value, and Figure s03 shows the exact right triangle it comes from — so nothing is taken on faith.
Measuring a side from two corner points (Example 8). If two corners of a shape are given as grid positions like $(0, 0)$ and $(4, 0)$ , the distance between them is found by making a right triangle out of the horizontal gap and the vertical gap, then using the Pythagorean theorem: $distance = (across)^{2} + (up)^{2}$ . This is the distance formula — but notice it is just Pythagoras in disguise, not a new law.

The scenario matrix

Every quadrilateral problem lands in one of these cells. Our examples below cover every row.

Cell	What makes it tricky	Example
A. Angle chase	opposite-equal / consecutive-supplementary	Ex 1
B. Straightforward area	plug into a formula	Ex 2
C. Missing length from area	run a formula backwards	Ex 3
D. Two tools combined	area and Pythagoras in one shape	Ex 4
E. Slant-side trap	perpendicular height hidden inside a triangle	Ex 5
F. Degenerate / limiting input	a side or diagonal shrinks to $0$	Ex 6
G. Real-world word problem	translate a story into a shape	Ex 7
H. Coordinate-geometry twist	corners given as points, not lengths	Ex 8
I. Exam trap (classify)	"which quadrilateral is it, really?"	Ex 9

Example 1 — Cell A: chasing all four angles

Figure — Quadrilaterals — square, rectangle, parallelogram, rhombus, trapezium, kite

Step 1 — Write the "consecutive" rule as an equation. In a parallelogram, side $A B$ and side $C D$ are parallel. Look at the figure: $∠ A$ and $∠ B$ sit on the same side of the slanted line $A D$ (a transversal). Angles on the same side of a transversal between parallel lines add to $180°$ — that is a co-interior angle rule. $∠ A + ∠ B = 180°$ Why this step? It links the two unknowns so we can solve for one.

Step 2 — Translate the words into a second equation. " $∠ A$ is $40°$ more than twice $∠ B$ " becomes $∠ A = 2∠ B + 40°.$ Why this step? Two unknowns need two equations; the story gives the second.

Step 3 — Substitute and solve. Replace $∠ A$ in Step 1: $(2∠ B + 40°) + ∠ B = 180° \Rightarrow 3∠ B = 140° \Rightarrow ∠ B = \frac{140}{3} ° = 46. \overline{6} °.$ Then $∠ A = 180° - 46. \overline{6} ° = 133. \overline{3} °$ . Why this step? Pure algebra; the geometry is finished.

Step 4 — Fill in the opposite angles. Opposite angles of a parallelogram are equal, so $∠ C = ∠ A = 133. \overline{3} °$ and $∠ D = ∠ B = 46. \overline{6} °$ .

Verify: All four add up: $133. \overline{3} + 46. \overline{6} + 133. \overline{3} + 46. \overline{6} = 360°$ . ✓ And $∠ A$ is indeed the big one, matching a parallelogram tilted so one corner is sharp.

Example 2 — Cell B: straightforward trapezium area

Step 1 — Identify the parts. Parallel sides are the bases: $a = 6$ , $b = 10$ . The perpendicular gap is the height $h = 4$ . Why this step? The trapezium formula only accepts the perpendicular distance, never a slanted leg.

Step 2 — Apply the formula. $A = \frac{1}{2} (a + b) h = \frac{1}{2} (6 + 10) \times 4.$ Why this step? Two copies of a trapezium form a parallelogram of base $a + b$ ; half of it is one trapezium (see the parent derivation).

Step 3 — Compute. $A = \frac{1}{2} \times 16 \times 4 = 32 m^{2} .$

Verify: The average base is $\frac{6 + 10}{2} = 8 m$ , and $8 \times 4 = 32 m^{2}$ — same answer, so the "average side × height" intuition holds. Units: metre $\times$ metre $= m^{2}$ . ✓

Example 3 — Cell C: running a formula backwards

Step 1 — Start from the forward formula. $A = b \times h .$ Why this step? We know two of the three quantities; the formula pins the third.

Step 2 — Rearrange for the unknown. Divide both sides by $b$ : $h = \frac{A}{b} = \frac{84}{12} .$ Why this step? Isolating $h$ is the whole point of a "backwards" problem.

Step 3 — Compute. $h = 7 cm .$

Verify: Plug back forwards: $12 \times 7 = 84 cm^{2}$ . ✓ The height ( $7$ ) is less than the base ( $12$ ), which is sensible.

Example 4 — Cell D: rhombus needing area and Pythagoras

Step 1 — Area from the diagonals. $A = \frac{1}{2} d_{1} d_{2} = \frac{1}{2} \times 16 \times 30 = 240 cm^{2} .$ Why this step? In a rhombus the diagonals cross at right angles, so the shape splits into four right triangles whose total area is $\frac{1}{2} d_{1} d_{2}$ .

Step 2 — Find the half-diagonals. The diagonals bisect each other, so each right triangle has legs $\frac{16}{2} = 8 and \frac{30}{2} = 15.$ Why this step? The side of the rhombus is the hypotenuse of one such triangle — we need its two legs first (look at the red right-angle mark in the figure).

Step 3 — Apply the Pythagorean theorem. $s = 8^{2} + 1 5^{2} = 64 + 225 = 289 = 17 cm .$ Why this step? The two half-diagonals are perpendicular legs, so their squares add to the square of the side.

Verify: $8^{2} + 1 5^{2} = 64 + 225 = 289 = 1 7^{2}$ . ✓ And $17 > 15$ , matching the forecast. All four sides equal $17 cm$ , confirming it is a rhombus.

Example 5 — Cell E: the slant-side trap

Step 1 — Spot the trap. Area is $b \times h$ where $h$ is the perpendicular height, not the slant side $6 cm$ . The slant side leans over, so the vertical drop is shorter than $6$ . Why this step? Multiplying by the slant length always overcounts.

Step 2 — Drop a perpendicular to build a right triangle. The slant side, the base, and the vertical drop form a right triangle (green line in the figure). The $6 cm$ slant is the hypotenuse; the height $h$ is the side opposite the $30°$ angle. Using the vertical-slice tool from the top of this page — $sin (angle) = \frac{opposite}{hypotenuse}$ , and for $30°$ that ratio is $\frac{1}{2}$ — we get $h = 6 \times sin 30° = 6 \times \frac{1}{2} = 3 cm .$ Why this step? $sin 30°$ is exactly the fraction of the slant length that points straight up — precisely the perpendicular height the area formula demands.

Step 3 — Now use the area formula correctly. $A = b \times h = 10 \times 3 = 30 cm^{2} .$

Verify: $30 cm^{2}$ is exactly half of the naive $60$ , and $sin 30° = \frac{1}{2}$ — the two facts agree. Less than $60$ , as forecast. ✓

Example 6 — Cell F: the degenerate / limiting case

Step 1 — Write area as a function of $a$ . $A (a) = \frac{1}{2} (a + 7) \times 4 = 2 (a + 7) = 2 a + 14.$ Why this step? Turning the formula into a function of $a$ lets us watch a limit — how the shape behaves as one input approaches a boundary value.

Step 2 — Evaluate at the two ends.

At $a = 7$ (both bases equal): $A = 2 (7) + 14 = 28 cm^{2}$ . This is a parallelogram (or rectangle-ish shape) — two equal parallel sides.
At $a = 0$ : $A = 2 (0) + 14 = 14 cm^{2}$ . The top base has vanished to a single point.

Why this step? The endpoints of a range reveal degenerate shapes the general formula still handles correctly.

Step 3 — Interpret the $a = 0$ shape. When one base shrinks to a point, the trapezium becomes a triangle with base $7$ and height $4$ . Its triangle area is $\frac{1}{2} \times 7 \times 4 = 14 cm^{2}$ — identical. Why this step? It shows the trapezium formula smoothly contains the triangle formula as a limiting case.

Verify: Triangle check $\frac{1}{2} \times 7 \times 4 = 14$ matches $A (0) = 14$ . ✓ And $A (7) = 28$ is exactly twice $A (0)$ , since $2 (a + 7)$ is linear. No jump, no undefined step — the formula never breaks. ✓

Example 7 — Cell G: real-world word problem

Step 1 — Translate the story into diagonals. Spine $= d_{1} = 1.2 m$ , cross-beam $= d_{2} = 0.8 m$ , perpendicular — exactly the data a kite-area formula wants. Why this step? Word problems are solved the moment you spot which lettered quantity each phrase means.

Step 2 — Compute the area. $A = \frac{1}{2} d_{1} d_{2} = \frac{1}{2} \times 1.2 \times 0.8 = 0.48 m^{2} .$ Why this step? A kite's diagonals are perpendicular, so the same $\frac{1}{2} d_{1} d_{2}$ rule as the rhombus applies.

Step 3 — Multiply by the unit cost. $Cost = 0.48 \times 250 = ₹120.$ Why this step? Cost is area $\times$ price-per-area — the units $m^{2} \times \frac{₹}{m ^{2}} = ₹$ cancel cleanly.

Verify: $0.48 < 1 m^{2}$ as forecast. Sanity check on cost: a full $1 m^{2}$ would be $₹250$ , and we have just under half of that, so $₹120$ is right. ✓

Example 8 — Cell H: coordinates instead of lengths

Step 1 — Measure the sides with the distance-from-Pythagoras helper. Using the distance formula built at the top of this page — which is just the Pythagorean theorem applied to the horizontal and vertical gaps — $P Q = (4 - 0)^{2} + (0 - 0)^{2} = 4, QR = (4 - 4)^{2} + (3 - 0)^{2} = 3.$ By the same formula $S R = 4$ and $P S = 3$ . Why this step? With corners as points, lengths aren't given — this helper recovers them from grid positions.

Step 2 — Confirm right angles. Side $P Q$ runs horizontally (its $y$ stays $0$ ); side $QR$ runs vertically (its $x$ stays $4$ ). Horizontal meets vertical at $90°$ . All four corners work the same way. Why this step? Rectangle needs opposite sides equal and all angles $90°$ — we now have both.

Step 3 — Area and diagonal. $A = l \times w = 4 \times 3 = 12 square units .$ $P R = (4 - 0)^{2} + (3 - 0)^{2} = 16 + 9 = 25 = 5.$ Why this step? Area uses the two side lengths; the diagonal is a distance between opposite corners — a $3$ – $4$ – $5$ right triangle.

Verify: $3^{2} + 4^{2} = 25 = 5^{2}$ , so the diagonal $5$ obeys Pythagoras. ✓ Area $12$ matches the visible $4 \times 3$ grid. ✓

Example 9 — Cell I: the classify-it exam trap

Step 1 — Check the rhombus condition. All four sides equal $\Rightarrow$ it is at least a rhombus. True so far. Why this step? Establish the baseline family before hunting for a stronger one.

Step 2 — Use the right angle to upgrade. A rhombus with one $90°$ angle forces all angles to $90°$ (opposite angles equal, consecutive supplementary: if one is $90°$ , its neighbour is $180° - 90° = 90°$ too). Equal sides and all right angles is the definition of a square. Why this step? "Square" is the most specific — hence most informative — correct label.

Step 3 — Confirm it satisfies every stronger definition. Sides all $6$ : ✓ equal. Angles all $90°$ : ✓. So it is simultaneously a rhombus, a rectangle, and a square. The best name is the tightest one: square.

Verify: Check all four angles are forced to $90°$ and hence sum correctly: each consecutive pair gives $90° + 90° = 180°$ (the parallelogram supplementary rule), and the total is $90° \times 4 = 360°$ (the interior-angle-sum rule). Both angle laws hold with all angles equal to $90°$ , so "square" is fully justified — the exam's "rhombus" is true but not the sharpest name, since every square is a rhombus but not the reverse. ✓

Recall Quick self-test

One angle of a parallelogram is $100°$ . What are the other three? ::: $80°, 100°, 80°$ A trapezium's bases are $5$ and $9$ , height $6$ . Area? ::: $\frac{1}{2} (5 + 9) \times 6 = 42$ A rhombus has diagonals $6$ and $8$ . Side length? ::: $3^{2} + 4^{2} = 5$ As a trapezium's shorter base shrinks to $0$ , what shape appears? ::: a triangle All sides equal and one angle $90°$ — best name? ::: a square

See also: Mensuration - Area and Perimeter for how these area formulas feed into larger composite-shape problems.