Perimeter, area, circles, arcs & sectors – IGCSE Maths exam questions with mark scheme

1Grade 1–3FoundationNon-calculator4 marks

(a) Change $3.6\ \text{m}^2$ into $\text{cm}^2$. [1]

Priya is tiling a rectangular floor that measures $4.8\ \text{m}$ by $3.2\ \text{m}$.
Each tile is a square of side $40\ \text{cm}$. The tiles do not overlap and none are cut.

(b) Work out the number of tiles Priya needs to cover the floor exactly. [3]

(a) $36\,000$B1
(b) consistent units for one dimension or area, e.g. $480$ and $320$ (cm) or $0.4$ (m) or tile area $1600\ \text{cm}^2$ / $0.16\ \text{m}^2$M1
(b) complete method, e.g. $\dfrac{480}{40} \times \dfrac{320}{40}$ ($= 12 \times 8$) or $\dfrac{153\,600}{1600}$ or $\dfrac{15.36}{0.16}$M1
(b) $96$A1
Answer: (a) $36\,000\ \text{cm}^2$ (b) $96$ tiles
Examiner tip: $1\ \text{m}^2 = 10\,000\ \text{cm}^2$, not $100$: square the length conversion $100 \times 100$.

2Grade 1–3FoundationNon-calculator5 marks

The diagram shows a shape made from a rectangle and an isosceles triangle.

The rectangle measures $8\ \text{cm}$ by $5\ \text{cm}$.
The two equal sides of the triangle are each $5\ \text{cm}$ long and the perpendicular height of the triangle is $3\ \text{cm}$.

(a) Work out the perimeter of the shape. [2]

(b) Work out the area of the shape. [3]

(a) adding the five outside edges, e.g. $8 + 5 + 5 + 5 + 5$ (do not award if the dashed $8\ \text{cm}$ line or the $3\ \text{cm}$ height is included)M1
(a) $28$ (cm)A1
(b) $8 \times 5$ ($=40$) or $\tfrac{1}{2} \times 8 \times 3$ ($=12$)M1
(b) $8 \times 5 + \tfrac{1}{2} \times 8 \times 3$M1
(b) $52$ ($\text{cm}^2$)A1
Answer: (a) $28\ \text{cm}$ (b) $52\ \text{cm}^2$
Examiner tip: Perimeter only uses the edges round the outside: the dashed lines inside the shape are not part of it.

3Grade 4–5Calculator4 marks

The diagram shows a trapezium.
The parallel sides are $7.4\ \text{cm}$ and $11.6\ \text{cm}$. The perpendicular distance between them is $5\ \text{cm}$.

(a) Work out the area of the trapezium. [2]

A parallelogram has the same area as the trapezium.
The base of the parallelogram is $7.6\ \text{cm}$.

(b) Work out the perpendicular height of the parallelogram. [2]

(a) $\tfrac{1}{2}(7.4 + 11.6) \times 5$ oe (e.g. rectangle + two triangles)M1
(a) $47.5$ ($\text{cm}^2$)A1
(b) their (a) $\div 7.6$M1
(b) $6.25$ (cm) (ft their (a) to 3 sf or better if M1 awarded in (a))A1ft
Answer: (a) $47.5\ \text{cm}^2$ (b) $6.25\ \text{cm}$
Examiner tip: Use the perpendicular height in both formulae; the slanted side of a parallelogram is never used for its area.

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4Grade 4–5Calculator3 marks

The wheel of a bicycle has a diameter of $64\ \text{cm}$.

Carlos rides the bicycle a distance of $500\ \text{m}$ along a straight road.

Work out the number of complete turns the wheel makes. [3]

$\pi \times 64$ ($= 201.06\ldots$ cm) or $\pi \times 0.64$ ($= 2.0106\ldots$ m)M1
$50\,000 \div$ their circumference in cm or $500 \div$ their circumference in m (units must be consistent) ($= 248.6\ldots$)M1
$248$ cao (do not accept $249$)A1
Answer: $248$
Examiner tip: The question asks for complete turns, so $248.68\ldots$ is truncated to $248$, not rounded up to $249$.

5Grade 4–5Calculator5 marks

The diagram shows the shape of a window.
The window is a rectangle $80\ \text{cm}$ wide and $120\ \text{cm}$ high with a semicircle on top. The diameter of the semicircle is $80\ \text{cm}$.

(a) Work out the area of the window.
Give your answer in $\text{m}^2$, correct to 2 decimal places. [3]

A strip of draught seal is fitted all the way round the outside edge of the window.

(b) Work out the length of the draught seal.
Give your answer correct to the nearest centimetre. [2]

(a) $\tfrac{1}{2} \times \pi \times 40^2$ ($= 2513.27\ldots$) (or in metres $\tfrac{1}{2}\pi \times 0.4^2 = 0.2513\ldots$)M1
(a) $80 \times 120 + \tfrac{1}{2}\pi \times 40^2$ ($= 12\,113.27\ldots\ \text{cm}^2$) oeM1
(a) $1.21$ ($\text{m}^2$)A1
(b) $120 + 80 + 120 + \tfrac{1}{2} \times \pi \times 80$ oe (do not include the dashed line)M1
(b) $446$ (cm) (M1 A0 for an unrounded $445.66\ldots$)A1
Answer: (a) $1.21\ \text{m}^2$ (b) $446\ \text{cm}$
Examiner tip: Divide by $10\,000$ (not $100$) to change $\text{cm}^2$ to $\text{m}^2$, and use half the circumference, $\tfrac{1}{2}\pi d$, for the curved edge.

6Grade 4–5Calculator4 marks

A circular rug has an area of $200\ \text{cm}^2$.

(a) Work out the circumference of the rug.
Give your answer correct to 3 significant figures. [3]

Sam says, “If you double the radius of a circle, you double its area.”

(b) Is Sam correct? Give a reason for your answer. [1]

(a) $\pi r^2 = 200$ leading to $r = \sqrt{\dfrac{200}{\pi}}$ ($= 7.978\ldots$)M1
(a) $2 \times \pi \times$ their $r$ or $\pi \times 2 \times$ their $r$M1
(a) $50.1$ (cm) (accept $50.13$)A1
(b) No, with a correct reason, e.g. the area is multiplied by $4$ because $\pi(2r)^2 = 4\pi r^2$, or a correct numerical example such as $r = 1$ gives $\pi$ but $r = 2$ gives $4\pi$C1
Answer: (a) $50.1\ \text{cm}$ (b) No: the area is multiplied by $4$
Examiner tip: Square-root after dividing by $\pi$ to find $r$; a common slip is to stop at $r^2 = 63.66$ and use that as the radius.

7Grade 6–7HigherCalculator5 marks

The diagram shows a sector $OAB$ of a circle with centre $O$.
The radius of the circle is $9\ \text{cm}$ and angle $AOB = 140^\circ$.

(a) Work out the area of the sector.
Give your answer correct to 3 significant figures. [2]

(b) Work out the perimeter of the sector.
Give your answer correct to 3 significant figures. [3]

(a) $\dfrac{140}{360} \times \pi \times 9^2$M1
(a) $99.0$ ($\text{cm}^2$) (accept $98.96$ to $98.97$)A1
(b) $\dfrac{140}{360} \times 2 \times \pi \times 9$ ($= 21.99\ldots$)M1
(b) $9 + 9 +$ their arc lengthM1
(b) $40.0$ (cm) (accept $39.99$)A1
Answer: (a) $99.0\ \text{cm}^2$ (b) $40.0\ \text{cm}$
Examiner tip: The perimeter of a sector includes both radii as well as the arc; many students give only the arc length.

8Grade 6–7HigherCalculator4 marks

A sector of a circle has an angle of $75^\circ$ at the centre.
The area of the sector is $60\ \text{cm}^2$.

Work out the perimeter of the sector.
Give your answer correct to 1 decimal place. [4]

$\dfrac{75}{360} \times \pi r^2 = 60$M1
$r = \sqrt{\dfrac{60 \times 360}{75\pi}}$ ($= 9.57\ldots$)M1
$2 \times$ their $r + \dfrac{75}{360} \times 2\pi \times$ their $r$M1
$31.7$ (cm) (accept answers that round to $31.7$)A1
Answer: $31.7\ \text{cm}$
Examiner tip: Keep the full calculator value of $r$; rounding $r$ to $9.6$ early gives $31.8$ and loses the accuracy mark.

9Grade 8–9HigherCalculator4 marks

The diagram shows a sector $OAB$ of a circle, centre $O$, radius $10\ \text{cm}$.
Angle $AOB = 110^\circ$.

The shaded region is the segment between the chord $AB$ and the arc $AB$.

Work out the area of the shaded region.
Give your answer correct to 3 significant figures. [4]

$\dfrac{110}{360} \times \pi \times 10^2$ ($= 95.99\ldots$)M1
$\tfrac{1}{2} \times 10 \times 10 \times \sin 110^\circ$ ($= 46.98\ldots$)M1
their sector $-$ their triangle (dep on both previous M marks)M1
$49.0$ ($\text{cm}^2$) (accept $49.01$)A1
Answer: $49.0\ \text{cm}^2$
Examiner tip: Use $\tfrac{1}{2}ab\sin C$ for triangle $OAB$; treating it as right-angled ($\tfrac{1}{2} \times 10 \times 10$) is a common error.

10Grade 6–7HigherNon-calculator5 marks

The diagram shows a shaded shape $ABCD$.
$OBC$ is a sector of a circle, centre $O$, radius $12\ \text{cm}$.
$OAD$ is a sector of a circle, centre $O$, radius $6\ \text{cm}$.
$OAB$ and $ODC$ are straight lines and angle $BOC = 60^\circ$.

(a) Show that the area of the shaded shape is $18\pi\ \text{cm}^2$. [3]

(b) Find the perimeter of the shaded shape.
Give your answer in the form $a + b\pi$, where $a$ and $b$ are integers. [2]

(a) $\dfrac{60}{360} \times \pi \times 12^2$ ($= 24\pi$) or $\dfrac{60}{360} \times \pi \times 6^2$ ($= 6\pi$)M1
(a) $\dfrac{60}{360}\pi \times 12^2 - \dfrac{60}{360}\pi \times 6^2$ oe, e.g. $\dfrac{1}{6}\pi(144 - 36)$M1
(a) $24\pi - 6\pi = 18\pi$ shown with no errors (both sector areas or $\tfrac{1}{6} \times 108\pi$ seen)A1
(b) arc lengths $\dfrac{60}{360} \times 2\pi \times 12$ ($= 4\pi$) and $\dfrac{60}{360} \times 2\pi \times 6$ ($= 2\pi$)M1
(b) $12 + 6\pi$ (cm) (accept $6\pi + 12$; do not accept a decimal answer)A1
Answer: (a) shown: $24\pi - 6\pi = 18\pi$ (b) $12 + 6\pi\ \text{cm}$
Examiner tip: The perimeter needs both straight edges $AB$ and $DC$ (each $6\ \text{cm}$) as well as both arcs; leave $\pi$ in exact answers on a non-calculator paper.

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