Geometry

high importance~4 Q in Tier 136 formulas⚡ 19 shortcuts6 subtopics

Overview Learn (6)Formulas (36)Examples (4)Practice (73)

Subtopic 5 of 6·← Pythagoras theorem and triplets·Circles: chords, tangents, secants and cyclic angles →

Quadrilaterals and polygons

Quadrilateral: angles add to $360^\circ$ . Key areas: parallelogram $=b\times h$ ; rhombus $=\frac12 d_1d_2$ (with side $=\sqrt{(d_1/2)^2+(d_2/2)^2}$ ); trapezium $=\frac12(a+b)h$ . Diagonals of a rectangle are equal and bisect each other; of a rhombus they bisect at right angles.

Cyclic quadrilateral: all four vertices on a circle; opposite angles are supplementary. Its area (Brahmagupta) is $\sqrt{(s-a)(s-b)(s-c)(s-d)}$ .

Regular polygon with $n$ sides: interior angle $=\frac{(n-2)180^\circ}{n}$ , each exterior $=\frac{360^\circ}{n}$ , diagonals $=\frac{n(n-3)}{2}$ .

📖 Notes 🧩 5 question types ⚡ Formulas 🎯 12 practice Q

Detailed notes

The polygon angle toolkit (all of it is two formulas)

For a regular polygon of $n$ sides: each exterior angle $= \frac{360^\circ}{n}$ , and each interior angle $= 180^\circ - \text{exterior} = \frac{(n-2) \times 180^\circ}{n}$ . Total interior sum $= (n-2) \times 180^\circ$ ; sum of exterior angles (any polygon) $= 360^\circ$ . Number of diagonals $= \frac{n(n-3)}{2}$ . Almost every polygon question is: given one quantity (interior angle, exterior angle, sum, diagonals), recover $n$ . Work through the exterior angle — it is the fastest hub: interior $150^\circ$ → ext $30^\circ$ → $n = 12$ .

Parallelogram family

Parallelogram: opposite sides parallel and equal; diagonals bisect each other; adjacent angles are supplementary, opposite angles equal. If $\angle A = k\angle B$ , then $k\angle B + \angle B = 180^\circ$ .
Rectangle: parallelogram + all $90^\circ$ ; diagonals equal and bisect each other.
Rhombus: parallelogram + all sides equal; diagonals bisect at $90^\circ$ (not equal). Side $= \sqrt{\left(\frac{d_1}{2}\right)^2 + \left(\frac{d_2}{2}\right)^2}$ ; area $= \frac12 d_1 d_2$ .
Square: all of the above; diagonal $= a\sqrt2$ .
Trapezium: one pair of parallel sides $a, b$ ; area $= \frac12 (a+b)h$ ; the segment joining the midpoints of the non-parallel sides (midsegment) $= \frac{a+b}{2}$ .

Cyclic quadrilaterals

All four vertices on one circle → opposite angles are supplementary: $\angle A + \angle C = \angle B + \angle D = 180^\circ$ . Ratio questions: the opposite pairs' ratio parts must each total the same count (e.g. $2:3:4:3$ works since $2+4 = 3+3 = 6$ ); solve $k$ from $180^\circ$ , then answer the asked angle. The exterior angle of a cyclic quad equals the interior opposite angle — the same fact re-stated.

The two standard computation shells

Rhombus from diagonals: halves of the diagonals form a right triangle with the side as hypotenuse — a triplet family in half-size. E.g. diagonals 10, 24 → halves 5, 12 → side 13.
Rhombus area ↔ diagonal: area $= \frac12 d_1 d_2$ gives the missing diagonal, then the side from the halves.

Quick revision

Regular $n$ : ext $= \frac{360}{n}^\circ$ , interior $= 180 -$ ext; sum $= (n-2) \times 180^\circ$ ; diagonals $= \frac{n(n-3)}{2}$ .
Parallelogram: adjacent angles sum to $180^\circ$ ; opposite angles equal.
Rhombus: side $= \sqrt{(\frac{d_1}{2})^2 + (\frac{d_2}{2})^2}$ ; area $= \frac12 d_1 d_2$ ; diagonals ⊥ bisect.
Cyclic quad: opposite angles sum to $180^\circ$ .
Trapezium area $= \frac12 (a+b)h$ ; midsegment $= \frac{a+b}{2}$ .

Types of questions asked

Every way this subtopic shows up in exams — how to recognise it, the formula or logic to use, and a solved example.

Type 1: Regular polygon: angle ↔ number of sidesvery common2 practice Q

How to spot it:

A regular polygon's interior/exterior angle, angle sum, or diagonal count is given — find $n$ (or another of these).

\text{ext} = \frac{360^\circ}{n},\quad \text{int} = 180^\circ - \text{ext},\quad d = \frac{n(n-3)}{2}

Convert the given interior angle to exterior: $180^\circ -$ interior.
$n = \frac{360^\circ}{\text{ext}}$ ; or from the sum, $n = \frac{\text{sum}}{180^\circ} + 2$ ; or solve $d = \frac{n(n-3)}{2}$ .
Answer what was asked — sometimes it is another angle, not $n$ .

Why: all polygon quantities are functions of $n$ , and the exterior angle is the quickest bridge between them.

Example: Each interior angle of a regular polygon is $150^\circ$ . How many sides has it?

Exterior $= 180 - 150 = 30^\circ$ ; $n = \frac{360}{30} = 12$ .

Type 2: Rhombus from its diagonalsvery common2 practice Q

How to spot it:

Diagonals of a rhombus given (or area + one diagonal) — side, perimeter, or the other diagonal asked.

s = \sqrt{\left(\frac{d_1}{2}\right)^2 + \left(\frac{d_2}{2}\right)^2},\quad K = \frac{d_1 d_2}{2}

Halve both diagonals; they form a right triangle whose hypotenuse is the side.
Spot the half-size triplet: diagonals 10, 24 → 5-12-13 → side 13.
Area given instead? $d_2 = \frac{2K}{d_1}$ first, then the side.

Why: diagonals of a rhombus bisect at right angles — the geometry is literally Pythagoras with half-diagonals.

Example: The diagonals of a rhombus are $10$ cm and $24$ cm. Its perimeter is:

Half-diagonals $5, 12$ → side $= 13$ ; perimeter $= 4 \times 13 = 52$ cm.

Type 3: Parallelogram angles (adjacent supplementary)very common2 practice Q

How to spot it:

Adjacent angles of a parallelogram are given as a ratio or as $\angle A = k\angle B$ — find one angle, or an opposite/related one.

\angle A + \angle B = 180^\circ

Set the adjacent relation: ratio $a:b$ → angles $\frac{a}{a+b} \times 180^\circ$ and $\frac{b}{a+b} \times 180^\circ$ .
Opposite angles equal the ones found; adjacent differ by $180^\circ -$ .
Read the letter asked — opposite vs adjacent flips the answer.

Why: a pair of parallel sides makes each pair of adjacent angles co-interior, hence supplementary.

Example: Two adjacent angles of a parallelogram are in the ratio $1:3$ . The larger angle is:

$\frac{3}{4} \times 180^\circ = 135^\circ$ (the smaller is $45^\circ$ ).

Type 4: Cyclic quadrilateral opposite anglescommon2 practice Q

How to spot it:

Four vertices on a circle; one angle or the ratio of angles given — a missing angle asked.

\angle A + \angle C = \angle B + \angle D = 180^\circ

Identify the opposite pair of the given angle; subtract from $180^\circ$ .
For ratios, check which ratio parts form opposite pairs; each pair sums to $180^\circ$ .
The exterior angle at one vertex equals the interior opposite angle — a shortcut when a side is produced.

Why: opposite angles of a cyclic quad subtend complementary arcs totalling the full circle, so the inscribed angles sum to $180^\circ$ .

Example: In a cyclic quadrilateral $ABCD$ , $\angle A = 70^\circ$ . The angle $\angle C$ is:

$\angle C = 180^\circ - 70^\circ = 110^\circ$ ( $A$ and $C$ are opposite).

Type 5: Trapezium area and midsegmentcommon2 practice Q

How to spot it:

Parallel sides and height given (or area + parallel sides) — area, height, or the midsegment asked.

K = \frac{1}{2}(a+b)h,\quad m = \frac{a+b}{2}

Add the parallel sides; halve; multiply by height for the area.
Height from area: $h = \frac{2K}{a+b}$ .
Midsegment (midpoints of the slanted sides) is just $\frac{a+b}{2}$ — no height needed.

Why: a trapezium is the average of its parallel sides times the height — the triangle-and-rectangle split proves it.

Example: A trapezium has parallel sides $13$ cm and $17$ cm and area $105$ sq cm. Its height is:

$h = \frac{2 \times 105}{13+17} = \frac{210}{30} = 7$ cm.

Formulas

Quadrilateral angle sum

A+B+C+D=360^\circ

Cyclic quadrilateral

A+C=180^\circ,\quad B+D=180^\circ

Rhombus

K=\frac12 d_1d_2,\quad a=\sqrt{\left(\frac{d_1}{2}\right)^2+\left(\frac{d_2}{2}\right)^2}

Trapezium

K=\frac12(a+b)h

Parallelogram

K=bh

Regular polygon

\text{int}=\frac{(n-2)180^\circ}{n},\quad \text{ext}=\frac{360^\circ}{n},\quad \text{diagonals}=\frac{n(n-3)}{2}

Brahmagupta

K=\sqrt{(s-a)(s-b)(s-c)(s-d)}

Shortcut tricks

⚡ Exterior angle → number of sides

For regular polygons the exterior angle is the fastest route: $n=\frac{360^\circ}{\text{ext}}$ , and $\text{ext}=180^\circ-\text{int}$ .

Example: The interior angle of a regular polygon is $156^\circ$ . How many sides has it?

ext $=24^\circ$ , so $n=360/24=15$ .

⚡ Rhombus side from diagonals

Half-diagonals form a right triangle with the side: side $=\sqrt{(d_1/2)^2+(d_2/2)^2}$ . If the halves are 8 and 6, the side is a 6-8-10 triplet.

Example: The diagonals of a rhombus are 16 cm and 12 cm. Find its side.

Halves 8 and 6 → side $=10$ cm (area $=96$ sq cm).

⚡ Diagonal count one-liner

$n$ vertices, each joins $n-3$ others (not itself, not two neighbours), each diagonal counted twice: $\frac{n(n-3)}{2}$ .

Example: How many diagonals does a decagon have?

$\frac{10\times7}{2}=35$ .

Where students lose marks

Rhombus area $\frac12 d_1d_2$ applied to a rectangle or parallelogram.
Cyclic quadrilateral: adding non-opposite angles to $180^\circ$ .
Interior/exterior angle confused; also $(n-2)\times180^\circ$ misapplied as $n\times180^\circ$ .

Practice sets — 12 questions

Sets of 10, mixed across the question types above. Each answer comes with a step-by-step explanation.

Topic test · 12 questions

Suggested time 8 min · wrong answers go to your mistake notebook automatically.