CAIE IGCSE Mathematics Extended Compound Shape Mensuration

Practise decomposing compound two- and three-dimensional figures, including sectors, frustums and combined curved solids, before calculating measures.

Exam points

split a compound diagram into standard polygons, sectors or solids and identify shared dimensions
subtract a smaller similar pyramid or cone from the original solid to calculate a frustum volume
add only exposed surface areas or remaining volumes when solids or circular parts are joined or removed

Covered syllabus points

E5.5.1 Carry out calculations and solve problems involving perimeters and E5.5.2 Carry out calculations and solve problems involving surface areas and

[Maximum number: 4]

Figure for Question E5.5 Compound shapes and parts of shapes question 1 — CAIE IGCSE Mathematics Extended

NOT TO SCALE

The diagram shows a shape ABQP made from three straight lines and an arc of a sector of a circle. The sector has centre O and angle $90^{\circ}$ .
POQ is a straight line and $A P=P O=O Q=Q B=5 \mathrm{~cm}$ .
Find the area of ABQP.
Give your answer in the form $a+k \pi$ .
$\mathrm{cm}^{2}$

$25+\frac{25\pi}{2}$
M2 for $\frac{90}{360}\times\pi\times(\sqrt{5^2+5^2})^2$
or M1 for $\text{radius}^2=5^2+5^2$
M1 for triangle area $[2\times]\frac{1}{2}\times5\times5$ oe

[Maximum number: 6]

Figure for Question E5.5 Compound shapes and parts of shapes question 2 — CAIE IGCSE Mathematics Extended

The diagram shows a horizontal container for water with a uniform cross-section. The cross-section is a semicircle.
The radius of the semicircle is 0.45 m and the length of the container is 4 m .

NOT TO SCALE

The greatest depth of the water in the container is $0.3\,\mathrm{m}$. The diagram shows the cross-section.
Calculate the number of litres of water in the container.
Give your answer correct to the nearest integer.

Figure for Answer to Question E5.5 Compound shapes and parts of shapes question 2 — CAIE IGCSE Mathematics Extended

742 or 743
M5 for a method leading to the volume of water, e.g.
$4\left\{2\times\frac{\operatorname{invcos}(0.15/0.45)}{360}\times\pi\times0.45^2-\frac12\times0.45^2\times\sin\left(2\operatorname{invcos}(0.15/0.45)\right)\right\}$
OR M2 for a correct sector area expression and M2 for a correct triangle area expression
If 0 scored, SC1 for $\operatorname{invcos}(0.15/0.45)$ , $\operatorname{invsin}(0.15/0.45)$ , or $\sqrt{0.45^2-0.15^2}$ soi

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Covered syllabus points

E5.5 Compound shapes and parts of shapes question 1

E5.5 Compound shapes and parts of shapes question 2