Edexcel IGCSE Math A Higher 4.9 Mensuration of 2d Shapes Questions

Use this mensuration page to calculate perimeters, areas, circle circumferences and sectors, including shapes built from rectangles, triangles and circles.

Exam points

Calculate perimeters of rectangles, compound shapes and shapes drawn on grids.
Find areas of triangles, rectangles, trapezia and compound shapes from diagrams.
Use circle formulae for circumference, area, sectors or shaded regions.

Covered syllabus points

4.9.A Measurements within the metric system 4.9.B Perimeters of triangles and rectangles 4.9.C Areas of triangles and rectangles 4.9.D The area of parallelograms and trapezia 4.9.E Circumferences and areas of circles using relevant 4.9.HA Perimeters and areas of sectors of circles Radian

[Maximum number: 4]

The diagram shows rectangle ABCD

Figure for Question 1 — Edexcel IGCSE Math A Higher

Diagram NOT accurately drawn

Work out the perimeter of the rectangle.
Show your working clearly.
cm

5x-1=3x+7.4x=4.2

Perimeter:
$2\times24+2(5\times4.2-1)=88$

Final answer:
88

[Maximum number: 6]

The diagram shows a quadrilateral ABCD

Figure for Question 2 — Edexcel IGCSE Math A Higher

Diagram NOT accurately drawn

In the diagram, ABC and DAC are right-angled triangles.

B C=6 \mathrm{~cm} \quad A C=7.5 \mathrm{~cm}

The area of quadrilateral ABCD is $31.5 \mathrm{~cm}^{2}$
Work out the length of AD
cm

$AB=\sqrt{7.5^{2}-6^{2}}=4.5$ $\text{Area }ABC=\frac12\times6\times4.5=13.5$ $\text{Area }DAC=31.5-13.5=18$ $AD=\frac{18}{7.5}\div0.5=4.8$

Final answer:
4.8

Accept 4.78 to 4.81.

[Maximum number: 5]

ABCD is a trapezium with one line of symmetry.

Figure for Question 3 — Edexcel IGCSE Math A Higher

Diagram NOT accurately drawn

\text { angle } A D C=60^{\circ} \quad A D=12 \mathrm{~cm} \quad D C=47 \mathrm{~cm}

Work out the area of the trapezium.
Give your answer correct to 3 significant figures.
Show your working clearly.
$\mathrm{cm}^{2}$

Figure for Answer to Question 3 — Edexcel IGCSE Math A Higher

Height of the trapezium:
$12\sin60^\circ=6\sqrt3=10.392\ldots$

Horizontal base of each end triangle:
$12\cos60^\circ=6$

AB=47-6-6=35

Area of the trapezium:
$\frac12(47+35)(6\sqrt3)=426.084\ldots$

Final answer:
$426\text{ cm}^2$

M1 for a method to find the height of the trapezium or the area of triangle ADC. M1 independently for a method to find the base of the triangle. M1 dependent on the previous method mark for finding AB. M1 for a complete method. A1 dependent on M1; allow 420 to 427 from correct working. Working required.

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Covered syllabus points

Question 1

Question 2

Question 3