Mensuration of 2D Shapes · Study Notes · Edexcel IGCSE 4MA1 Higher

Areas and perimeters. Memorise the formulae for triangles, quadrilaterals, circles and sectors.

At a glance

Triangle: $A = \dfrac{1}{2}bh$ .
Rectangle: $A = lw$ .
Parallelogram: $A = bh$ .
Trapezium: $A = \dfrac{1}{2}(a + b)h$ .
Circle: $A = \pi r^{2}$ , $C = 2\pi r$ .
Sector: fraction of circle.
Always use PERPENDICULAR height.

What you’ll learn

Mapped to the Pearson Edexcel IGCSE 4MA1 syllabus (2026 onwards).

4.4 — Find areas and perimeters of standard 2D shapes.
4.4 — Use circle formulae (area, circumference).
4.4 — Calculate arc length and sector area.
4.4 — Solve compound shape problems.

Standard area formulae

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Memorise these. PERPENDICULAR height matters.

Standard formulae:

Shape	Area
Square	$s^{2}$
Rectangle	$lw$
Triangle	$\dfrac{1}{2}bh$
Parallelogram	$bh$
Trapezium	$\dfrac{1}{2}(a + b)h$
Rhombus	$\dfrac{1}{2}d_1 d_2$ (diagonals)

$h$ = PERPENDICULAR HEIGHT (90° to base). Not slant.

If diagram shows slant length, may need Pythagoras to find perpendicular height first.

Always use the height measured at 90° to the base, never the slant side.

Worked example. Triangle, base 10, slant side 13, perpendicular height to that base is unknown.

If you know base 10 and slant side 13 (and these don't form a right angle directly): use Pythagoras with half-base if isosceles, or use other methods.

Edexcel tip. Diagrams sometimes have $h$ marked clearly. If not, identify the perpendicular drop from the apex to the base.

Triangle: $\dfrac{1}{2}bh$ .
Trapezium: $\dfrac{1}{2}(a+b)h$ .
All use PERPENDICULAR height.
Compound: split into rectangles + triangles.

Circle formulae

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$A = \pi r^{2}$ , $C = 2\pi r$ .

Circle formulae:

Area: $A = \pi r^{2}$ .
Circumference: $C = 2\pi r$ or $\pi d$ .

$r$ = radius (centre to edge). $d$ = diameter (across, $= 2r$ ).

The diameter is twice the radius — check which one the question gives.

Worked example. Radius $7$ cm.

Area: $\pi \times 49 = 49\pi \approx 153.94$ cm².
Circumference: $14\pi \approx 43.98$ cm.

Exact vs decimal. Often Edexcel asks 'in terms of $\pi$ ': leave answer as $49\pi$ , not $153.94$ .

Use the right value of $\pi$ . Calculators have built-in $\pi$ . Otherwise, use $3.14$ or $3.142$ .

Worked qualitative. Why is circumference $\pi d$ ?

$\pi$ is defined as the ratio: circumference / diameter.
So circumference = $\pi \times$ diameter.
Equivalent: $C = 2\pi r$ (since $d = 2r$ ).

Edexcel tip. Always check: does the question give RADIUS or DIAMETER? Misreading is the #1 error.

$A = \pi r^{2}$ .
$C = 2\pi r = \pi d$ .
$r$ = half of $d$ .
Leave in $\pi$ if asked.

Sectors and arcs

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Sector = fraction of full circle. Multiply by $\theta / 360$ .

Sector = pie-slice of circle, bounded by two radii and an arc.

Arc length:

$L = \dfrac{\theta}{360} \times 2\pi r$

where $\theta$ = angle of the sector (degrees).

Sector area:

$A = \dfrac{\theta}{360} \times \pi r^{2}$

Reasoning. A full circle has angle $360°$ . A sector with angle $\theta$ is the fraction $\theta/360$ of the whole. So area and arc length are scaled by the same fraction.

A sector is the fraction θ⁄360 of the whole circle.

Worked example. Sector with $r = 10$ cm, $\theta = 72°$ .

Arc: $\dfrac{72}{360} \times 2\pi \times 10 = \dfrac{1}{5} \times 20\pi = 4\pi$ cm.
Area: $\dfrac{72}{360} \times \pi \times 100 = \dfrac{1}{5} \times 100\pi = 20\pi$ cm².

Common sector angles:

$\theta$	Fraction
$90°$	$1/4$
$60°$	$1/6$
$45°$	$1/8$
$30°$	$1/12$
$180°$	$1/2$ (semicircle)

Worked qualitative. What's the perimeter of a sector?

Two radii ( $2r$ ) PLUS the arc length.
$P = 2r + L = 2r + \dfrac{\theta}{360} \times 2\pi r$ .

Edexcel tip. Show the fraction $\theta/360$ explicitly. Mark scheme awards M1 for the formula, A1 for the answer.

Sector = fraction of circle.
Arc: $\dfrac{\theta}{360} \times 2\pi r$ .
Sector area: $\dfrac{\theta}{360} \times \pi r^{2}$ .
Perimeter = arc + 2r.

Compound shapes

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Split into simpler shapes. Sum or subtract areas.

Compound shape = made of multiple simpler shapes.

Method:

Identify simpler shapes (rectangles, triangles, semicircles).
Find each area.
Sum (or subtract).

Example: L-shape.

Whole rectangle $10 \times 8 = 80$ . Cut $4 \times 3 = 12$ . L-area: $80 - 12 = 68$ .

Example: Garden with circular pond.

Square $20 \times 20 = 400$ m². Circular pond, $r = 5$ , area $= 25\pi \approx 78.5$ m². Garden minus pond: $400 - 78.5 = 321.5$ m².

Worked qualitative. Why split rather than try one formula?

Compound shapes don't have direct formulae.
Adding/subtracting simpler areas is the standard technique.
Edexcel mark schemes credit each sub-area calculation.

Edexcel tip. Label each sub-shape clearly. Show all dimensions. Sum at the end.

Split into simpler shapes.
Sum or subtract areas.
Label each sub-area.
Watch for hidden semicircles or sectors.

Quick recap

Triangle: $\dfrac{1}{2}bh$ .
Trapezium: $\dfrac{1}{2}(a+b)h$ .
Circle: $A = \pi r^{2}$ , $C = 2\pi r$ .
Sector: fraction × full circle.
Compound: split, sum or subtract.

Memorise this

Verbatim phrases and definitions Edexcel mark schemes credit.

Triangle: $\dfrac{1}{2}bh$ .
Trapezium: $\dfrac{1}{2}(a + b)h$ .
Circle: $A = \pi r^{2}$ , $C = 2\pi r$ .
Sector arc: $\dfrac{\theta}{360} \times 2\pi r$ .
Sector area: $\dfrac{\theta}{360} \times \pi r^{2}$ .

How it’s examined

2D mensuration on every Higher Tier paper (4-7 marks). Often combined with circle properties or compound shapes. Examiner reports flag (1) using slant for perpendicular height, (2) circumference using radius (or area using diameter), (3) wrong fraction for sectors.

Sources: Pearson Edexcel International GCSE Mathematics A 4MA1 specification (2026 onwards); 4MA1 Examiner Reports — Jan and May/Jun 2022-2024 series; Past Papers — 4MA1/2H May/Jun 2024. Last reviewed 2026-05-07.

Step-by-step solutions to past-paper-style questions on mensuration of 2d shapes, written exactly the way a tutor would explain them at the board.

1Area of a triangle
Foundation• triangle, area
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Question
Triangle base $8$ cm, height $5$ cm. Area?
Step-by-step solution
1. Step 1
  $A = \dfrac{1}{2} \times \text{base} \times \text{height}$ .
2. Step 2
  $A = \dfrac{1}{2} \times 8 \times 5 = 20$ cm².
Answer
$20$ cm²
2Area of a trapezium
Foundation• trapezium
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Question
Parallel sides $7$ cm and $11$ cm, height $5$ cm. Area?
Step-by-step solution
1. Step 1
  $A = \dfrac{1}{2}(a + b)h$ .
2. Step 2
  $A = \dfrac{1}{2}(7 + 11)(5) = \dfrac{1}{2}(18)(5) = 45$ cm².
Answer
$45$ cm²
3Area and circumference of a circle
Foundation• Adapted from 4MA1/2H May/Jun 2024 Q9• circle
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Question
Circle with radius $7$ cm. Find (a) circumference, (b) area. Use $\pi = 3.14$ .
Step-by-step solution
1. Step 1
  (a) $C = 2\pi r = 2 \times 3.14 \times 7 = 43.96$ cm.
2. Step 2
  (b) $A = \pi r^{2} = 3.14 \times 49 = 153.86$ cm².
Answer
$C \approx 44.0$ cm, $A \approx 153.9$ cm²
Examiner tip
If asked for EXACT, leave in terms of $\pi$ : $C = 14\pi$ , $A = 49\pi$ . Otherwise compute.
4Sector area and arc length
Higher• sector
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Question
Sector with radius $10$ cm and angle $72°$ . Find arc length and area.
Step-by-step solution
1. Step 1
  Arc length: $\dfrac{\theta}{360} \times 2\pi r = \dfrac{72}{360} \times 2\pi \times 10 = \dfrac{1}{5} \times 20\pi = 4\pi$ cm.
2. Step 2
  Sector area: $\dfrac{\theta}{360} \times \pi r^{2} = \dfrac{72}{360} \times 100\pi = 20\pi$ cm².
Answer
Arc length: $4\pi \approx 12.57$ cm. Sector area: $20\pi \approx 62.83$ cm².
Examiner tip
Sector formulas: $\dfrac{\theta}{360}$ × full circumference (for arc) or × full area (for sector area).
5Compound shape area
Higher• compound shape
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Question
An L-shape. The whole rectangle is $10 \times 8$ . A $4 \times 3$ rectangle is cut from the corner. Find the area.
Step-by-step solution
1. Step 1
  Whole rectangle: $10 \times 8 = 80$ .
2. Step 2
  Cut rectangle: $4 \times 3 = 12$ .
3. Step 3
  L-shape area: $80 - 12 = 68$ cm².
Answer
$68$ cm²
6Diameter of a circle from an exact area [1 mark]
Getting started• Paper 1H/2H style — 1 mark• circles, area, exact answers
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Question
The area of a circle is $64\pi$ cm².

Write down the diameter of the circle. [1]
Step-by-step solution
1. Step 1
  The area is $\pi r^{2}$ , so compare $\pi r^{2}$ with $64\pi$ : the $\pi$ matches, which leaves $r^{2} = 64$ and $r = 8$ cm. No calculator is needed.
  $\pi r^{2} = 64\pi \;\Rightarrow\; r^{2} = 64 \;\Rightarrow\; r = 8$
2. Step 2
  B1 The question asks for the diameter, which is twice the radius.
  $d = 2 \times 8 = 16$
Answer
$16$ cm
Examiner tip
Examiner tip: Neither the area nor the circumference of a circle is on the 4MA1 formula sheet. Learn $\pi r^{2}$ and $2\pi r$ and know which is which. When the area is given in terms of $\pi$ , the number in front of $\pi$ is $r^{2}$ .

Common mistake: Stopping at the radius and writing $8$ cm, or treating $r^{2}$ as $2r$ and giving the diameter as $64$ cm.
7Area of an obtuse-angled triangle [2 marks]
Getting started• Paper 1H/2H style — 2 marks• triangles, area, perpendicular height
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Question
In triangle $PQR$ , $QR = 9$ cm, $PQ = 11$ cm and angle $PQR$ is obtuse.

The perpendicular distance from $P$ to the line $QR$ extended is $7.4$ cm.

Work out the area of triangle $PQR$ . [2]
Step-by-step solution
1. Step 1
  M1 Take $QR$ as the base. The perpendicular height is the distance from $P$ to the line through $Q$ and $R$ , even though it lands outside the triangle. $PQ$ is a slant side and is not needed.
  $\tfrac{1}{2} \times 9 \times 7.4$
2. Step 2
  A1 Work it out.
  $33.3$
Answer
$33.3$ cm²
Examiner tip
Examiner tip: For an obtuse-angled triangle the height is drawn to the base line extended. The base and the height must still be at right angles to each other.

Common mistake: Using the side $PQ$ as the height, $\tfrac{1}{2} \times 9 \times 11 = 49.5$ cm². A slant side is always longer than the perpendicular height.
8Perimeter of a quarter circle [3 marks]
Building confidence• Paper 1H/2H style — 3 marks• circles, perimeter, arc length
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Question
A quarter circle has a radius of $7.5$ cm.

Work out the perimeter of the quarter circle. Give your answer correct to $3$ significant figures. [3]
Step-by-step solution
1. Step 1
  M1 The curved edge is a quarter of the circumference of a circle of radius $7.5$ cm.
  $\tfrac{1}{4} \times 2 \times \pi \times 7.5 = 11.780\ldots$
2. Step 2
  M1 A quarter circle also has two straight edges, both radii.
  $11.780\ldots + 7.5 + 7.5 = 26.780\ldots$
3. Step 3
  A1 Round to $3$ significant figures.
  $26.8 \text{ cm}$
Answer
$26.8$ cm
Examiner tip
Examiner tip: A quarter circle is a sector with angle $90°$ , so its perimeter is arc $+ \, 2r$ , just like any other sector.

Common mistake: Adding only one radius ( $19.3$ cm), or none ( $11.8$ cm). Sketch the shape and count the edges: one curve and two straight lines.
9Area swept by a wiper blade, in square metres [4 marks]
Building confidence• Paper 1H/2H style — 4 marks• sectors, shaded regions, area units
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Question
A windscreen wiper arm is $50$ cm long and turns about one end, $O$ , through an angle of $110°$ .

The rubber blade covers the part of the arm from $15$ cm to $50$ cm from $O$ . The blade wipes the region between two arcs, both with centre $O$ .

Work out the area wiped by the blade. Give your answer in m², correct to $3$ significant figures. [4]
Step-by-step solution
1. Step 1
  M1 Area of the large sector, radius $50$ cm.
  $\dfrac{110}{360} \times \pi \times 50^{2} = 2399.82\ldots$
2. Step 2
  M1 Subtract the small sector, radius $15$ cm, that the blade does not reach.
  $2399.82\ldots - \dfrac{110}{360} \times \pi \times 15^{2} = 2399.82\ldots - 215.98\ldots$
3. Step 3
  A1 The wiped area in cm².
  $2183.84\ldots \text{ cm}^{2}$
4. Step 4
  B1 Change to m²: $1$ m² $= 100 \times 100 = 10\,000$ cm².
  $2183.84\ldots \div 10\,000 = 0.218 \text{ m}^{2}$
Answer
$0.218$ m²
Examiner tip
Examiner tip: Factorising first, $\dfrac{110}{360}\pi(50^{2} - 15^{2})$ , gives the same result in one calculation. Note $50^{2} - 15^{2} = 2275$ , not $35^{2}$ .

Common mistake: Using a single sector of radius $35$ cm (the blade length), which gives $1176$ cm². The region is the difference of two sectors. Dividing by $100$ instead of $10\,000$ at the end is the other common loss.
10Region between three semicircles [5 marks]
Stretch• Paper 1H/2H style — 5 marks• semicircles, shaded regions, exact answers
▼
Question
$A$ , $C$ and $B$ lie in that order on a straight line, with $AC = 10$ cm and $CB = 6$ cm.

A semicircle is drawn on $AB$ as diameter. Two smaller semicircles are drawn on $AC$ and on $CB$ as diameters, on the same side of $AB$ and inside the large semicircle.

The region $R$ is inside the large semicircle and outside both small semicircles.

(a) Show that the area of $R$ is $15\pi$ cm². [3]

(b) Find the perimeter of $R$ . Give your answer in terms of $\pi$ . [2]
Step-by-step solution
1. Step 1
  M1 (a) $AB = 16$ cm, so the large semicircle has radius $8$ cm.
  $\tfrac{1}{2} \times \pi \times 8^{2} = 32\pi$
2. Step 2
  M1 The small semicircles have radii $5$ cm and $3$ cm.
  $\tfrac{1}{2} \times \pi \times 5^{2} + \tfrac{1}{2} \times \pi \times 3^{2} = 12.5\pi + 4.5\pi = 17\pi$
3. Step 3
  A1 Subtract, keeping every term in terms of $\pi$ so the result is exact.
  $32\pi - 17\pi = 15\pi$
4. Step 4
  M1 (b) The boundary of $R$ is the three curved edges. There are no straight edges, because $AC$ and $CB$ are covered by the small semicircles. Each arc is half of $\pi d$ .
  $\tfrac{1}{2}\pi(16) + \tfrac{1}{2}\pi(10) + \tfrac{1}{2}\pi(6)$
5. Step 5
  A1 Collect the terms.
  $8\pi + 5\pi + 3\pi = 16\pi$
Answer
(a) $15\pi$ cm² (shown) (b) $16\pi$ cm
Examiner tip
Examiner tip: In a "Show that" with an exact answer, never switch to decimals. $47.12\ldots$ does not show that the area is $15\pi$ .

Common mistake: Using the diameters as radii ( $\tfrac{1}{2}\pi \times 16^{2}$ and so on), which makes every area four times too big. In (b), adding the straight line $AB$ to the perimeter: it lies along the edge of the two small semicircles, not the edge of $R$ .

Question 1
Paper 1H/2H style — 1 mark1 mark
The area of a square is $0.36$ m².

Write down the length of one side of the square in cm. [1]
Model answer
$\sqrt{0.36} = 0.6 \text{ m}$

$0.6$ m $= 60$ cm

$60$ cm
Why this scores
Mark scheme:
- B1 for $60$ cao
Why this scores full marks: The candidate square-roots while the area is still in m², then converts a length with the length factor $100$ . Converting the area with the length factor first ( $0.36 \times 100 = 36$ cm²) and then square-rooting gives $6$ cm, the usual wrong answer.
Question 2
Paper 1H/2H style — 2 marks2 marks
A triangle has an area of $27$ cm². The base of the triangle is $7.5$ cm.

Work out the perpendicular height of the triangle. [2]
Model answer
$\tfrac{1}{2} \times 7.5 \times h = 27$

$h = \dfrac{2 \times 27}{7.5} = 7.2$

$7.2$ cm
Why this scores
Mark scheme:
- M1 for $\tfrac{1}{2} \times 7.5 \times h = 27$ or $2 \times 27 \div 7.5$ oe
- A1 for $7.2$ cao
Why this scores full marks: The candidate writes the area formula with the unknown height in it before rearranging, so the $\tfrac{1}{2}$ is not lost. Dividing $27$ by $7.5$ alone gives $3.6$ cm, which is the height of a parallelogram with that area, not a triangle.
Question 3
Paper 1H/2H style — 2 marks2 marks
A rectangle has an area of $58.5$ cm². The length of the rectangle is $9$ cm.

Work out the perimeter of the rectangle. [2]
Model answer
Width $= 58.5 \div 9 = 6.5$ cm

$2 \times (9 + 6.5) = 31$

$31$ cm
Why this scores
Mark scheme:
- M1 for $58.5 \div 9$ ( $= 6.5$ )
- A1 for $31$ cao
Why this scores full marks: The candidate finds the missing side before attempting the perimeter and adds all four sides. Stopping at $6.5$ , or giving $9 + 6.5 = 15.5$ , is the usual loss.
Question 4
Paper 1H/2H style — 3 marks3 marks
The parallelogram $PQRS$ has $PQ = 12$ cm and $QR = 8.5$ cm. The area of $PQRS$ is $84$ cm².

(a) Work out the perpendicular distance between $PQ$ and $SR$ . [1]

(b) Work out the perpendicular distance between $QR$ and $PS$ . Give your answer correct to $3$ significant figures. [2]
Model answer
(a) $84 \div 12 = 7$

$7$ cm

(b) Using $QR$ as the base:

$84 \div 8.5 = 9.882\ldots$

$9.88$ cm
Why this scores
Mark scheme:
- B1 (a) for $7$
- M1 (b) for $84 \div 8.5$
- A1 for awrt $9.88$
Why this scores full marks: The candidate sees that either pair of parallel sides can be the base, and that the area is the same both ways. Each height is the area divided by the base it is perpendicular to.
Question 5
Paper 1H/2H style — 3 marks3 marks
The area of a circle is $200$ cm².

Work out the circumference of the circle. Give your answer correct to $3$ significant figures. [3]
Model answer
$\pi r^{2} = 200 \Rightarrow r = \sqrt{\dfrac{200}{\pi}} = 7.9788\ldots$

$2 \times \pi \times 7.9788\ldots = 50.132\ldots$

$50.1$ cm
Why this scores
Mark scheme:
- M1 for $r^{2} = \dfrac{200}{\pi}$ or $r = 7.97\ldots$
- M1 for $2 \times \pi \times$ their $r$
- A1 for awrt $50.1$
Why this scores full marks: The candidate goes back to the radius first and keeps it unrounded. Forgetting the square root (using $r = 63.66$ ) is the most common error.
Question 6
Paper 1H/2H style — 3 marks3 marks
A trapezium has parallel sides of length $1.2$ m and $85$ cm. The perpendicular distance between the parallel sides is $60$ cm.

Work out the area of the trapezium in m². [3]
Model answer
$85$ cm $= 0.85$ m, $60$ cm $= 0.6$ m

$\tfrac{1}{2}(1.2 + 0.85) \times 0.6 = \tfrac{1}{2} \times 2.05 \times 0.6$

$0.615$ m²
Why this scores
Mark scheme:
- M1 for all three lengths in the same unit
- M1 for $\tfrac{1}{2}(1.2 + 0.85) \times 0.6$ oe (or $\tfrac{1}{2}(120 + 85) \times 60 = 6150$ cm²)
- A1 for $0.615$ cao
Why this scores full marks: The candidate converts every length to metres before substituting, so no area conversion is needed. Working in cm is fine, but $6150$ cm² must then be divided by $10\,000$ , not $100$ .
Question 7
Paper 1H/2H style — 4 marks4 marks
A circular pond has a radius of $3.5$ m. A path of width $1.2$ m goes all the way round the pond.

Work out the area of the path. Give your answer correct to $3$ significant figures. [4]
Model answer
Outer radius $= 3.5 + 1.2 = 4.7$ m

$\pi \times 4.7^{2} = 69.397\ldots, \quad \pi \times 3.5^{2} = 38.484\ldots$

$69.397\ldots - 38.484\ldots = 30.913\ldots$

$30.9$ m²
Why this scores
Mark scheme:
- M1 for the outer radius $4.7$
- M1 for $\pi \times 4.7^{2}$ or $\pi \times 3.5^{2}$
- M1 for the difference of the two circle areas
- A1 for awrt $30.9$
Why this scores full marks: The path is a ring, so the candidate subtracts the pond from the larger circle. The path adds $1.2$ m to the radius once, so the outer radius is $4.7$ m; adding $2 \times 1.2$ to the radius (giving $5.9$ m) is the usual error.
Question 8
Paper 1H/2H style — 4 marks4 marks
A sector of a circle has an angle of $48°$ . The arc length of the sector is $10$ cm.

Work out the area of the sector. Give your answer correct to $3$ significant figures. [4]
Model answer
$\dfrac{48}{360} \times 2\pi r = 10$

$r = \dfrac{10 \times 360}{96\pi} = 11.936\ldots$

$\dfrac{48}{360} \times \pi \times (11.936\ldots)^{2} = 59.683\ldots$

$59.7$ cm²
Why this scores
Mark scheme:
- M1 for $\dfrac{48}{360} \times 2\pi r = 10$ oe
- A1 for $r = 11.9\ldots$
- M1 for $\dfrac{48}{360} \times \pi \times$ their $r^{2}$
- A1 for awrt $59.7$
Why this scores full marks: The candidate uses the arc-length formula to find $r$ and then switches to the sector-area formula, keeping $r$ unrounded. Check: sector area $= \tfrac{1}{2} \times r \times \text{arc} = \tfrac{1}{2} \times 11.936\ldots \times 10$ .
Question 9
Paper 1H/2H style — 5 marks5 marks
A circle has centre $O$ and radius $7.5$ cm. $AB$ is a chord of the circle of length $12$ cm.

Work out the area of the minor segment cut off by the chord $AB$ . Give your answer correct to $3$ significant figures. [5]
Model answer
Let $M$ be the midpoint of $AB$ . $OM$ is perpendicular to $AB$ and $AM = 6$ .

$\sin AOM = \dfrac{6}{7.5} = 0.8 \Rightarrow AOM = 53.130\ldots°$

$AOB = 2 \times 53.130\ldots = 106.260\ldots°$

Sector: $\dfrac{106.260\ldots}{360} \times \pi \times 7.5^{2} = 52.160\ldots$

$OM = \sqrt{7.5^{2} - 6^{2}} = 4.5$ , so triangle $OAB = \tfrac{1}{2} \times 12 \times 4.5 = 27$

$52.160\ldots - 27 = 25.160\ldots$

$25.2$ cm²
Why this scores
Mark scheme:
- M1 for a correct method for angle $AOB$ (or half of it), e.g. $\sin^{-1}\left(\dfrac{6}{7.5}\right)$ or the cosine rule
- A1 for angle $AOB =$ awrt $106.3°$
- M1 for the sector area with their angle
- M1 for the triangle area, $27$ or $\tfrac{1}{2} \times 7.5^{2} \times \sin$ their angle
- A1 for awrt $25.2$
Why this scores full marks: The candidate uses the fact that the perpendicular from the centre bisects the chord, which gives a right-angled triangle for the angle and the height. The angle is kept unrounded in the sector area; the segment is then sector minus triangle.
Question 10
Paper 1H/2H style — 5 marks5 marks
A sector of a circle has radius $12$ cm and angle $150°$ .

A circle has the same area as the sector.

(a) Show that the radius of the circle is $\sqrt{60}$ cm. [3]

(b) Which has the greater perimeter, the sector or the circle? You must show your working. [2]
Model answer
(a) $\dfrac{150}{360} \times \pi \times 12^{2} = \dfrac{5}{12} \times 144\pi = 60\pi$

$\pi r^{2} = 60\pi \Rightarrow r^{2} = 60 \Rightarrow r = \sqrt{60}$

(b) Sector: $12 + 12 + \dfrac{150}{360} \times 2\pi \times 12 = 24 + 10\pi = 55.4\ldots \text{ cm}$

Circle: $2\pi\sqrt{60} = 48.66\ldots \text{ cm}$

The sector has the greater perimeter.
Why this scores
Mark scheme:
- M1 (a) for the sector area $\dfrac{150}{360} \times \pi \times 12^{2}$ ( $= 60\pi$ )
- M1 for $\pi r^{2} = 60\pi$ oe
- A1 for $r = \sqrt{60}$ from fully correct working
- M1 (b) for both perimeters, $24 + 10\pi$ and $2\pi\sqrt{60}$ oe
- A1 for "sector" with $55.4$ and $48.7$ (or better) seen
Why this scores full marks: In (a) the candidate keeps $\pi$ as a symbol so it cancels exactly; a decimal area such as $188.5$ makes the result approximate and loses the A mark. In (b) the sector's perimeter includes both radii, and the decision is backed by two values.

4MA1 Mensuration of 2D Shapes

At a glance

What you’ll learn

Standard area formulae

Circle formulae

Sectors and arcs

Compound shapes

Quick recap

Memorise this

How it’s examined

Worked ExamplesStep-by-step solutions to past-paper-style questions

1Area of a triangle

2Area of a trapezium

3Area and circumference of a circle

4Sector area and arc length

5Compound shape area

6Diameter of a circle from an exact area [1 mark]

7Area of an obtuse-angled triangle [2 marks]

8Perimeter of a quarter circle [3 marks]

9Area swept by a wiper blade, in square metres [4 marks]

10Region between three semicircles [5 marks]

Model AnswersFull mark-scheme-style answers

Question 1

Question 2

Question 3

Question 4

Question 5

Question 6

Question 7

Question 8

Question 9

Question 10