Mensuration (2D)

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

Overview Learn (5)Formulas (26)Examples (4)Practice (62)

Circles, sectors and rings

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

The idea in one minute

A circle of radius r has area pi times r squared and circumference two pi r, with pi as 22/7 on these exams. Sectors, arcs, rings and wheels are fractions of those two formulas. Quadrants and semicircles are just sectors of 90 and 180 degrees.

The two core formulas

K=\pi r^2,\qquad C=2\pi r

Use $\pi=\dfrac{22}{7}$ unless the paper says otherwise. Radius first, everything after is arithmetic. Halve any diameter before squaring; the formulas live on the radius.

Circumference $132$ cm: $r=\dfrac{132\times7}{2\times22}=21$ , so $K=\dfrac{22}{7}\times441=1386$ sq cm.

Rule: Every circle question starts by fixing $r$ . Reverse questions divide by $2\pi$ to get it.

Sectors and arcs

A sector is a pizza slice; the arc is its crust. Both take the fraction $\dfrac{\theta}{360}$ of the full shape:

\text{arc}=\frac{\theta}{360}\times2\pi r,\qquad \text{sector area}=\frac{\theta}{360}\times\pi r^2

Radius $35$ , angle $72^\circ$ : arc $=\dfrac{72}{360}\times220=44$ cm, sector $=\dfrac{72}{360}\times3850=770$ sq cm.

Perimeter of a sector adds the two radii: $44+70=114$ cm. A $90^\circ$ sector of the same circle has arc $\dfrac14\times220=55$ cm and perimeter $55+70=125$ cm.

A ring between radii $17.5$ and $10.5$ : $R^2-r^2=306.25-110.25=196$ , so $K=\dfrac{22}{7}\times196=616$ sq cm. Difference of squares keeps it clean.

Tip: $\dfrac{72}{360}=\dfrac15$ . Simplify the fraction before multiplying; big numbers shrink fast.

Rings and rolling wheels

A ring between radii $R$ and $r$ has area $\pi(R^2-r^2)$ . Outer $14$ , inner $7$ : $\dfrac{22}{7}\times147=462$ sq cm.

A wheel of diameter $70$ cm covers $C=\dfrac{22}{7}\times70=220$ cm per turn. Rolling $231$ m $=23100$ cm:

\text{revolutions}=\frac{23100}{220}=105

The reverse works too: $150$ turns cover $150\times220=33000$ cm, which is $330$ m.

Watch: Convert metres to centimetres before dividing. The wheel diameter is in centimetres.

Circles and squares together

Circle inscribed in a square: diameter $=$ square side. Side $14$ gives $r=7$ and $K=154$ .
Square inside a circle: the square's diagonal is the circle's diameter.

Remember: Inscribed means the circle touches all four sides from inside, so side and diameter match.

Quadrants and semicircles

A quadrant is a quarter circle ( $90^\circ$ ), a semicircle half ( $180^\circ$ ). Same fraction rules.

Quadrant with $r=14$ : area $=\dfrac{1}{4}\times\dfrac{22}{7}\times196=154$ sq cm. Arc $=\dfrac14\times88=22$ cm, so its perimeter is $14+14+22=50$ cm. A semicircle with $r=7$ has area $\dfrac12\times154=77$ sq cm and perimeter $14+22=36$ cm.

Tip: Perimeter of a quadrant is two radii plus the quarter arc. Semicircle: two radii make a diameter, plus the half arc.

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

Area and circumference, forward and reverse

How to spot it:

One of area or circumference given, the other asked.

Method

From circumference: $r=\dfrac{C}{2\pi}$ .
From area: $r=\sqrt{\dfrac{K}{\pi}}$ .
Substitute into the asked formula.

Why it works:

Both formulas share the single unknown $r$ , so one conversion bridges the two.

Try this

The circumference of a circle is 132 cm. Its area is:

Show solution

$r=\dfrac{132\times7}{44}=21$ cm.
$K=\dfrac{22}{7}\times21^2=1386$ sq cm.

Answer

1386 sq cm

Type 2very common2 practice Q

Sector area and arc length

How to spot it:

A sector angle with a radius; arc, sector area or perimeter asked.

Method

Simplify $\dfrac{\theta}{360}$ to lowest terms.
Arc: multiply by $2\pi r$ .
Sector area: multiply by $\pi r^2$ .
Perimeter adds $2r$ to the arc.

Why it works:

One fraction serves both quantities, and simplified fractions keep the arithmetic small.

Try this

Find the arc length of a sector of angle $72^\circ$ in a circle of radius 35 cm.

Show solution

$\dfrac{72}{360}=\dfrac15$ and $2\pi r=220$ cm.
Arc $=\dfrac{220}{5}=44$ cm.

Answer

44 cm

Type 3common2 practice Q

Ring (annulus) area and wheel revolutions

How to spot it:

Two concentric radii, or a wheel rolling a distance.

Method

Ring: $K=\pi(R^2-r^2)$ .
Wheel: one turn $=$ one circumference.
Divide total distance by the circumference, same units.

Why it works:

Both ideas reuse the circumference and area formulas with zero new theory.

Try this

A wheel of diameter 70 cm rolls a distance of 231 m. The number of revolutions it makes is:

Show solution

$C=\dfrac{22}{7}\times70=220$ cm; $231$ m $=23100$ cm.
$\dfrac{23100}{220}=105$ .

Answer

105 revolutions

Type 4common2 practice Q

Circles inscribed in / circumscribed about squares

How to spot it:

A circle touching a square from inside, or a square inside a circle.

Method

Circle in square: diameter $=$ side, so $r=\dfrac{a}{2}$ .
Square in circle: square diagonal $=$ circle diameter.
Substitute and compute the asked area.

Why it works:

The touch conditions translate straight into one length equality, so the shape pair collapses to one line.

Try this

A circle is inscribed in a square of side 14 cm. The area of the circle is:

Show solution

$r=\dfrac{14}{2}=7$ cm.
$K=\dfrac{22}{7}\times49=154$ sq cm.

Answer

154 sq cm

Type 5common

Quadrant and semicircle pieces

How to spot it:

A quarter or half circle, area or perimeter asked.

Method

Quadrant: take $\dfrac14$ of circle area; semicircle $\dfrac12$ .
Arc is the same fraction of the circumference.
Perimeter adds the straight edges.

Why it works:

These are sectors of $90^\circ$ and $180^\circ$ , so the fraction rules already known do all the work.

Try this

A quadrant has radius 14 cm. Find its area and its perimeter.

Show solution

Area $=\dfrac14\times\dfrac{22}{7}\times196=154$ sq cm.
Arc $=\dfrac14\times88=22$ cm.
Perimeter $=14+14+22=50$ cm.

Answer

Area 154 sq cm, perimeter 50 cm

Formula sheet

Circle

K=\pi r^2,\quad C=2\pi r

Take pi as 22/7 unless stated.

Arc and sector

\text{arc}=\frac{\theta}{360}2\pi r,\quad \text{sector}=\frac{\theta}{360}\pi r^2

Same fraction of circumference and area.

Ring

K=\pi(R^2-r^2)

R = outer radius, r = inner radius.

Wheel revolutions

N=\frac{D}{C}=\frac{\text{distance}}{2\pi r}

Count turns by dividing distance by one circumference.

Quadrant / semicircle

\text{quad}=\frac{\pi r^2}{4},\quad \text{semi}=\frac{\pi r^2}{2}

Quarter and half of the circle area.

Shortcuts that save time

⚡ Radius first, always

Circumference, diameter or area given: convert to the radius before anything else. Every formula lives on r.

Example

The circumference of a circle is 132 cm. Its area is:

Show solution

$r=\dfrac{132\times7}{2\times22}=21$ cm.
$K=\dfrac{22}{7}\times21\times21$ .
$K=1386$ sq cm.

Answer

1386 sq cm

⚡ Simplify the sector fraction

Reduce theta over 360 first: 72/360 is 1/5, 90/360 is 1/4. Then one multiplication finishes.

Example

Find the arc length of a sector of angle $72^\circ$ in a circle of radius 35 cm.

Show solution

$2\pi r=2\times\dfrac{22}{7}\times35=220$ cm.
$\dfrac{72}{360}=\dfrac15$ .
Arc $=\dfrac{220}{5}=44$ cm.

Answer

44 cm

⚡ Wheels: count the turns

One turn covers one circumference. Divide total distance by it, in the same units.

Example

A wheel of diameter 70 cm rolls a distance of 231 m. The number of revolutions it makes is:

Show solution

$C=\dfrac{22}{7}\times70=220$ cm.
$231\text{ m}=23100$ cm.
$\dfrac{23100}{220}=105$ turns.

Answer

105

Mistakes to avoid

Where most students lose marks on this subtopic.

Mistake 01

Using the diameter as $r$ .

Diameter 70 means $r=35$ . Halve before squaring.

Mistake 02

Taking $\pi$ as 3.14 when the numbers are multiples of 7.

Radius 7, 14, 21, 35 with $\pi=\dfrac{22}{7}$ cancels the 7s cleanly.

Mistake 03

Forgetting the two radii in a sector perimeter.

Sector perimeter $=$ arc $+2r$ . Radius 35, arc 44 gives 114.

Mistake 04

Computing ring area as $\pi(R-r)^2$ .

It is $\pi(R^2-r^2)$ ; for 14 and 7 that is $\dfrac{22}{7}\times147=462$ .

Mistake 05

Mixing metres and centimetres in wheel counts.

231 m is 23100 cm; match the wheel's units before dividing.

Quick revision

Read this the night before the exam.

$K=\pi r^2$ , $C=2\pi r$ ; fix the radius before anything else.
Arc and sector are the same fraction $\dfrac{\theta}{360}$ of circumference and area.
Sector perimeter adds the two radii to the arc.
Ring area is $\pi(R^2-r^2)$ , the difference of squares.
Wheel revolutions $=$ distance divided by one circumference, same units.
Quadrant $=\dfrac14$ of the circle; semicircle $=\dfrac12$ ; perimeter includes the straight edges.

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.

Continue learning