Mensuration (2D)

high importance~2 Q in Tier 126 formulas⚡ 15 shortcuts5 subtopics

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Regular polygons and inscribed figures

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

The idea in one minute

A regular polygon has equal sides and equal angles, so one side length fixes everything. The regular hexagon is the star: it is six equilateral triangles glued together. Exterior angles, interior angles and diagonal counts all come from the number of sides alone.

Hexagon: six equilateral triangles

Join the centre to all six corners. Six equilateral triangles of side $a$ appear, so

K=\frac{3\sqrt3}{2}a^2 = 6\times\frac{\sqrt3}{4}a^2

Side $6$ cm: each small triangle has area $9\sqrt3$ , total $54\sqrt3$ sq cm.

Given a hexagon perimeter instead, divide by $6$ first: perimeter $36$ cm means side $6$ cm and area $54\sqrt3$ sq cm.

A regular octagon has area $2(1+\sqrt2)a^2$ . Side $4$ cm gives $32+32\sqrt2$ sq cm, but the six-triangle picture is the one that pays off weekly.

Rule: Hexagon side, radius (centre to corner) and side are equal: $R=a$ . The apothem is $\dfrac{\sqrt3}{2}a$ .

Any regular polygon from perimeter and apothem

K=\frac{1}{2}\times\text{perimeter}\times\text{apothem}

The apothem is the perpendicular from the centre to a side. Perimeter $72$ cm, apothem $6$ cm: $K=\dfrac{1}{2}\times72\times6=216$ sq cm.

For the hexagon of side $6$ : apothem $=\dfrac{\sqrt3}{2}\times6=3\sqrt3$ , and $\dfrac12\times36\times3\sqrt3=54\sqrt3$ , matching the six triangles. Two routes, one answer.

Slices from the centre

Join the centre of any regular polygon to its corners. You get $n$ identical slices, each an isosceles triangle with apex angle $\dfrac{360^\circ}{n}$ . That single picture generates the exterior angle, the apothem (slice height) and the area formula at once.

Tip: This one formula fits every regular polygon, hexagon included. It is the triangle formula applied to all the slices.

Angles from the side count

Each exterior angle $=\dfrac{360^\circ}{n}$ (the turn at each corner while walking the boundary).
Each interior angle $=180^\circ-$ exterior.
Interior angle sum $=(n-2)\times180^\circ$ .

Octagon: exterior $=\dfrac{360}{8}=45^\circ$ , interior $=135^\circ$ , angle sum $=6\times180^\circ=1080^\circ$ .

Pentagon: exterior $72^\circ$ , interior $108^\circ$ , sum $540^\circ$ . Decagon: exterior $36^\circ$ , interior $144^\circ$ . Reverse works too: interior $144^\circ$ means exterior $36^\circ$ , so $n=\dfrac{360}{36}=10$ .

Watch: The regular polygon interior angles are equal, so divide the sum by $n$ to check one angle.

Counting diagonals

From each corner, $n-3$ diagonals leave (skip itself and two neighbours). Each diagonal has two ends:

d=\frac{n(n-3)}{2}

A polygon with $90$ diagonals: $n^2-3n-180=0$ , which factors as $(n-15)(n+12)=0$ , so $n=15$ .

Small counts are worth memorising: hexagon $9$ , pentagon $5$ , octagon $20$ , decagon $35$ .

Hexagon special lengths

Longest diagonal (through the centre) $=2a$ . Side $8$ gives $16$ .
Short diagonal (skipping one corner) $=a\sqrt3$ .

Remember: In a hexagon, side $=$ radius and the longest diagonal $=$ diameter $=2a$ . Both fall out of the six triangles picture.

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

Hexagon as six equilateral triangles

How to spot it:

A regular hexagon with its side given.

Method

One equilateral triangle: $\dfrac{\sqrt3}{4}a^2$ .
Multiply by $6$ .
Or apply $K=\dfrac{3\sqrt3}{2}a^2$ directly.

Why it works:

The hexagon is the only regular polygon that splits into equilateral triangles, a free shortcut.

Try this

Find the area of a regular hexagon of side 6 cm.

Show solution

One triangle: $9\sqrt3$ .
$K=6\times9\sqrt3=54\sqrt3$ sq cm.

Answer

54*sqrt(3) sq cm

Type 2common2 practice Q

Area from perimeter and apothem

How to spot it:

A regular polygon with perimeter and apothem (inradius) given.

Method

Check the polygon is regular.
$K=\dfrac{1}{2}\times P\times a$ .
Multiply once.

Why it works:

The formula needs no side count and no angles, just the two given lengths.

Try this

A regular polygon has perimeter 72 cm and apothem 6 cm. Its area is:

Show solution

$K=\dfrac12\times72\times6$ .
$K=216$ sq cm.

Answer

216 sq cm

Type 3very common2 practice Q

Interior and exterior angles and the angle sum

How to spot it:

A regular polygon with an angle or a side count given.

Method

Exterior $=\dfrac{360}{n}$ .
Interior $=180-$ exterior.
Sum $=(n-2)\times180^\circ$ ; divide by $n$ for one angle.

Why it works:

One small table answers every angle question for any regular polygon.

Try this

Find the measure of each interior angle of a regular octagon.

Show solution

Exterior $=\dfrac{360}{8}=45^\circ$ .
Interior $=180-45=135^\circ$ .

Answer

135 degrees

Type 4common2 practice Q

Counting diagonals

How to spot it:

A diagonal count with the side count asked, or the reverse.

Method

Write $\dfrac{n(n-3)}{2}=\text{count}$ .
Expand into a quadratic.
Factor; keep the positive root.

Why it works:

The count formula is quadratic in $n$ , and exam values factor cleanly.

Try this

A polygon has 90 diagonals. The number of its sides is:

Show solution

$n(n-3)=180$ .
$(n-15)(n+12)=0$ .
$n=15$ .

Answer

Type 5occasional2 practice Q

Hexagon diagonals and side relations

How to spot it:

Hexagon questions asking a diagonal or linking radius to side.

Method

Side $=$ radius: $R=a$ .
Longest diagonal $=2a$ (through the centre).
Short diagonal $=a\sqrt3$ (skipping a corner).

Why it works:

All three lengths come from the six-equilateral-triangles picture, with no new formulas.

Try this

The length of the longest diagonal of a regular hexagon of side 8 cm is:

Show solution

Longest diagonal $=2a$ .
$=2\times8=16$ cm.

Answer

16 cm

Formula sheet

Regular hexagon

K=\frac{3\sqrt3}{2}a^2

Six equilateral triangles of side a.

Polygon from apothem

K=\frac{1}{2}\times P\times a

P = perimeter, a = apothem (centre to a side).

Exterior angle

\text{ext}=\frac{360^\circ}{n}

Equal turns around the boundary.

Interior angle

\text{int}=180^\circ-\text{ext},\quad \text{sum}=(n-2)180^\circ

One angle plus the total for all n.

Diagonals

d=\frac{n(n-3)}{2}

n sides give this many diagonals.

Shortcuts that save time

⚡ Hexagon = 6 equilateral triangles

Six triangles of side a. Use the equilateral area times six; the root-three factor is already familiar.

Example

Find the area of a regular hexagon of side 6 cm.

Show solution

One triangle: $\dfrac{\sqrt3}{4}\times36=9\sqrt3$ .
$K=6\times9\sqrt3$ .
$K=54\sqrt3$ sq cm.

Answer

54*sqrt(3) sq cm

⚡ Exterior angle finds n

Divide 360 by the exterior angle to get the side count; the interior angle is its partner to 180.

Example

Find the measure of each interior angle of a regular octagon.

Show solution

Exterior $=\dfrac{360}{8}=45^\circ$ .
Interior $=180-45=135^\circ$ .

Answer

135 degrees

⚡ Diagonal equation factors

Set n(n-3)/2 equal to the given count, then factor the quadratic. Exam answers are whole numbers.

Example

A polygon has 90 diagonals. The number of its sides is:

Show solution

$n(n-3)=180$ .
$n^2-3n-180=(n-15)(n+12)=0$ .
$n=15$ .

Answer

Mistakes to avoid

Where most students lose marks on this subtopic.

Mistake 01

Using $\dfrac{\sqrt3}{4}a^2$ once for the hexagon.

That is one triangle. Six of them give $\dfrac{3\sqrt3}{2}a^2$ .

Mistake 02

Interior angle $=\dfrac{360}{n}$ .

That is the exterior. Interior $=180-\dfrac{360}{n}$ ; for an octagon, $135^\circ$ .

Mistake 03

Forgetting to halve $n(n-3)$ .

Each diagonal was counted from both ends: $d=\dfrac{n(n-3)}{2}$ .

Mistake 04

Longest hexagon diagonal taken as $a\sqrt3$ .

$a\sqrt3$ is the short one; the longest through the centre is $2a$ .

Mistake 05

Using the side as the apothem.

The apothem is centre to the middle of a side; hexagon apothem $=\dfrac{\sqrt3}{2}a$ .

Quick revision

Read this the night before the exam.

Hexagon $=$ six equilateral triangles; area $\dfrac{3\sqrt3}{2}a^2$ .
Any regular polygon: $\dfrac12\times$ perimeter $\times$ apothem.
Exterior $=\dfrac{360}{n}$ ; interior $=180-$ exterior; sum $=(n-2)\times180^\circ$ .
Diagonals $=\dfrac{n(n-3)}{2}$ ; given a count, factor the quadratic.
Hexagon: side $=$ radius, longest diagonal $=2a$ , short diagonal $=a\sqrt3$ .

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 7 min · wrong answers go to your mistake notebook automatically.

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