Geometry

high importance~4 Q in Tier 137 formulas⚡ 19 shortcuts6 subtopics

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Quadrilaterals and polygons

⏱ 3 min read🧩 5 question types🎯 12 practice Q

The idea in one minute

The four angles of any quadrilateral add to $360^\circ$ . Key areas: parallelogram $=b\times h$ , rhombus $=\dfrac{1}{2}d_1d_2$ , trapezium $=\dfrac{1}{2}(a+b)h$ . In a cyclic quadrilateral (all four corners on one circle) opposite angles add to $180^\circ$ . A regular polygon with $n$ sides has exterior angle $\dfrac{360^\circ}{n}$ , interior angle $180^\circ$ minus that, and $\dfrac{n(n-3)}{2}$ diagonals.

The parallelogram family

Parallelogram: opposite sides parallel and equal; opposite angles equal; adjacent angles add to $180^\circ$ .
Rectangle: parallelogram with all angles $90^\circ$ ; equal diagonals that bisect each other.
Rhombus: parallelogram with all sides equal; diagonals cross at $90^\circ$ and bisect each other.
Square: rectangle and rhombus together; diagonal $=a\sqrt2$ .
Trapezium: exactly one pair of parallel sides, $a$ and $b$ ; an isosceles trapezium also has equal slanted sides and equal base angles.

A rectangle $8\times6$ has diagonal $10$ and area $48$ . Every claim above turns into a one-line sum like that.

Rule: In a parallelogram, ratio questions on adjacent angles are just co-interior angles: the parts add to $180^\circ$ .

Rhombus: work with half-diagonals

The diagonals of a rhombus cut each other in half at right angles. So the halves and one side form a right triangle:

\text{side}=\sqrt{\left(\frac{d_1}{2}\right)^2+\left(\frac{d_2}{2}\right)^2}

Diagonals 10 and 24: halves 5 and 12, side 13, perimeter 52. Area $=\dfrac{1}{2}d_1d_2=\dfrac{1}{2}\times10\times24=120$ .

Tip: Halves of diagonals are almost always a triplet: (5,12,13), (8,6,10), (12,35,37).

Angles: parallelogram and cyclic quadrilateral

Adjacent angles of a parallelogram add to $180^\circ$ . Ratio $1:3$ gives $\dfrac{1}{4}\times180^\circ=45^\circ$ and $135^\circ$ .

A cyclic quadrilateral has all four corners on one circle. Its opposite angles add to $180^\circ$ . Angles in ratio $2:3:4:3$ : opposite parts must match, $2+4=3+3=6$ . So $k=30^\circ$ and the angles are $60^\circ, 90^\circ, 120^\circ, 90^\circ$ .

The exterior angle of a cyclic quadrilateral equals the interior angle at the opposite corner.

Regular polygons: everything from n

For $n$ equal sides:

each exterior angle $=\dfrac{360^\circ}{n}$
each interior angle $=180^\circ-$ exterior
interior angle sum $=(n-2)\times180^\circ$
diagonals $=\dfrac{n(n-3)}{2}$

Interior angle $150^\circ$ means exterior $30^\circ$ , so $n=\dfrac{360}{30}=12$ . A decagon has $\dfrac{10\times7}{2}=35$ diagonals. A pentagon: exterior $72^\circ$ , interior $108^\circ$ , diagonals $\dfrac{5\times2}{2}=5$ .

Tip: Travel through the exterior angle. It is the fastest bridge: interior $\to$ exterior $\to$ $n$ .

Trapezium

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

Parallel sides 13 and 17 with area 105: $h=\dfrac{2\times105}{30}=7$ .

The segment joining the midpoints of the two slanted sides (the midsegment) $=\dfrac{a+b}{2}=\dfrac{13+17}{2}=15$ . No height needed.

Question types you will see

Each type: how to recognise it, the method step by step, and one question to try.

Type 1very common2 practice Q

Regular polygon: angle to number of sides

How to spot it:

An interior or exterior angle, an angle sum, or a diagonal count is given; find $n$ or another angle.

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

Method

Turn the interior angle into the exterior one: $180^\circ-$ interior.
Divide $360^\circ$ by the exterior angle to get $n$ .
From a sum: $n=\dfrac{\text{sum}}{180^\circ}+2$ . From diagonals: solve $n(n-3)=2d$ .

Why it works:

All these quantities are fixed by $n$ , and the exterior angle is the quickest link between them.

Try this

Each interior angle of a regular polygon is $150^\circ$ . How many sides does it have?

Show solution

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

Answer

Type 2very common2 practice Q

Rhombus from its diagonals

How to spot it:

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

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

Method

Halve both diagonals.
Side $=$ hypotenuse of the halves (spot the triplet).
Perimeter $=4\times$ side; area $=\dfrac{1}{2}d_1d_2$ .
Given area and one diagonal: $d_2=\dfrac{2K}{d_1}$ first.

Why it works:

Rhombus diagonals cross at right angles and bisect each other, so the halves make a right triangle.

Try this

The diagonals of a rhombus are 10 cm and 24 cm. Find its perimeter.

Show solution

Halves: $5$ and $12$ .
Side $=\sqrt{25+144}=13$ .
Perimeter $=4\times13=52$ cm.

Answer

52 cm

Type 3very common2 practice Q

Parallelogram angles

How to spot it:

Adjacent angles of a parallelogram given as a ratio or as $\angle A=k\angle B$ ; one angle asked.

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

Method

Add the ratio parts to $180^\circ$ .
Scale each part to get the two angles.
Opposite angles repeat; check which letter is asked.

Why it works:

The parallel sides make each adjacent pair co-interior, so they supplement.

Try this

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

Show solution

Parts $1+3=4$ share $180^\circ$ .
Larger $=\dfrac{3}{4}\times180^\circ$ .
$=135^\circ$ .

Answer

135°

Type 4common2 practice Q

Cyclic quadrilateral opposite angles

How to spot it:

Four corners on a circle; one angle or an angle ratio given, a missing angle asked.

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

Method

Find the corner opposite the given one.
Subtract the given angle from $180^\circ$ .
In ratio questions, check which parts sit opposite each other; each pair totals $180^\circ$ .

Why it works:

Opposite corners of a cyclic quadrilateral split the circle's arc between them, so their angles supplement.

Try this

In a cyclic quadrilateral $ABCD$ , $\angle A=70^\circ$ . Find $\angle C$ .

Show solution

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

Answer

110°

Type 5common2 practice Q

Trapezium area and midsegment

How to spot it:

Parallel sides with height or area given; area, height or midsegment asked.

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

Method

Average the two parallel sides.
Multiply by the height for the area.
Height from area: $h=\dfrac{2K}{a+b}$ .
Midsegment $=$ the average, no height needed.

Why it works:

A trapezium is its average width times its height.

Try this

A trapezium has parallel sides 13 cm and 17 cm and area 105 sq cm. Find its height.

Show solution

$h=\dfrac{2\times105}{13+17}$ .
$=\dfrac{210}{30}$ .
$=7$ cm.

Answer

7 cm

Formula sheet

Quadrilateral angle sum

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

Any four-sided figure.

Parallelogram angles

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

Adjacent angles supplement; opposite angles equal.

Cyclic quadrilateral

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

Opposite corners on one circle.

Rhombus

K=\frac{1}{2}d_1d_2,\quad a=\sqrt{\left(\frac{d_1}{2}\right)^2+\left(\frac{d_2}{2}\right)^2}

d1, d2 = diagonals; they cross at right angles.

Trapezium

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

a and b are the two parallel sides; h is the gap between them.

Parallelogram

K=bh

Height is measured perpendicular to the base.

Regular polygon

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

n = number of sides.

Shortcuts that save time

⚡ Exterior angle to number of sides

Exterior angle $=180^\circ-$ interior, and $n=\dfrac{360^\circ}{\text{exterior}}$ .

Example

The interior angle of a regular polygon is $156^\circ$ . How many sides does it have?

Show solution

Exterior $=180-156=24^\circ$ .
$n=\dfrac{360}{24}$ .
$n=15$ .

Answer

15 sides

⚡ Rhombus side from half-diagonals

Halve both diagonals. They are the legs of a right triangle whose hypotenuse is the side.

Example

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

Show solution

Halves: $8$ and $6$ .
Side $=\sqrt{64+36}=10$ cm.
Area $=\dfrac{1}{2}\times16\times12=96$ sq cm.

Answer

Side 10 cm, area 96 sq cm

⚡ Count diagonals in one line

Each of the $n$ corners joins $n-3$ others, and every diagonal is counted twice, giving $\dfrac{n(n-3)}{2}$ .

Example

How many diagonals does a decagon have?

Show solution

$\dfrac{10\times(10-3)}{2}$ .
$=\dfrac{70}{2}$ .
$=35$ .

Answer

Mistakes to avoid

Where most students lose marks on this subtopic.

Mistake 01

Using $\dfrac{1}{2}d_1d_2$ for a rectangle or parallelogram.

That area formula is for a rhombus (or a kite with perpendicular diagonals).

Mistake 02

Adding two neighbouring angles of a cyclic quadrilateral to $180^\circ$ .

Opposite corners add to $180^\circ$ . Neighbours add to whatever is left of $360^\circ$ .

Mistake 03

Using $(n-2)\times180^\circ$ as one interior angle.

That is the whole sum. Divide by $n$ for one angle of a regular polygon.

Mistake 04

Forgetting the half in $\dfrac{1}{2}(a+b)h$ for a trapezium.

Average the parallel sides first, then multiply by the height.

Mistake 05

Using full diagonals as legs of the rhombus right triangle.

Halve both diagonals first. The side is the hypotenuse of the halves.

Quick revision

Read this the night before the exam.

Quadrilateral angles total $360^\circ$ ; parallelogram adjacent angles total $180^\circ$ .
Cyclic quadrilateral: opposite angles total $180^\circ$ .
Rhombus: side $=\sqrt{(d_1/2)^2+(d_2/2)^2}$ , area $=\dfrac{1}{2}d_1d_2$ .
Trapezium: area $\dfrac{1}{2}(a+b)h$ ; midsegment $\dfrac{a+b}{2}$ .
Regular polygon: ext $=\dfrac{360^\circ}{n}$ , int $=180^\circ-$ ext, sum $(n-2)\times180^\circ$ .
Diagonals $=\dfrac{n(n-3)}{2}$ .

Practice: 12 questions

Sets of 10, mixed across the question types above. Every answer has a step-by-step explanation.

Topic test · 12 questions

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

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