Mensuration

Units of measure

To convert m to cm, we multiply by 100. But because we have a power of 2 with the units, we multiply by $100^2 = 10\,000$:

$20 \times 10\,000 = 200\,000\ \text{cm}^2$

The same principle is used for other questions. For example, to convert $\text{cm}^3$ to $\text{m}^3$, we divide by $100^3 = 1\,000\,000$.

Area and perimeter

Perimeter is the sum of the lengths of all the sides of the shape.

$P = 3 + 5 + 5 + 6 = 19$ cm. Always remember to write the units.

Perimeter $=$ sum of all sides, so $14 = 4 + 3 + 2 + x$. Then $x = 14 - 9 = 5$ cm.

Area

• Area of A $= 4 \times 4 = 16\ \text{cm}^2$
• Area of B $= 8 \times 4 = 32\ \text{cm}^2$
• Area of C $= \frac{1}{2} \times 4 \times 5 = 10\ \text{cm}^2$
• Total area $= 16 + 32 + 10 = 58\ \text{cm}^2$

Circles, sectors and arcs

All answers are given to 3 significant figures.

Surface area and volume

You were introduced to various 3D shapes in the previous chapter. Now we will learn how to calculate the surface area and volume of those shapes.

The formulae in the table below will be given in the exam. However, it is preferable to memorise and understand them.

In the cone, $l$ is the slant height.

Volume $=$ area of face $\times$ length:

$V = \left(\tfrac{1}{2} \times 2.6 \times 3\right) \times 10 = 39\ \text{cm}^3$

Surface area $=$ areas of all faces $=$ 2 triangles $+$ 3 rectangles:

$SA = 2\left(\tfrac{1}{2} \times 2.6 \times 3\right) + 3(10 \times 3) = 7.8 + 90 = 97.8\ \text{cm}^2$

A compound solid

Volume $=$ volume of cone $+$ volume of hemisphere. The hemisphere is half a sphere, so its volume is $\frac{1}{2} \times \frac{4}{3}\pi r^3$.

$V = \tfrac{1}{3}\pi(4.5)^2(7.6) + \tfrac{1}{2} \times \tfrac{4}{3}\pi(4.5)^3 = 161.2 + 190.9 = 352\ \text{cm}^3$

Surface area $=$ curved surface of the cone $+$ curved surface of the hemisphere. Find the slant height $l$ with Pythagoras' theorem:

$l = \sqrt{4.5^2 + 7.6^2} = \sqrt{78.01} = 8.83\ \text{cm}$

The curved surface of the hemisphere is half of $4\pi r^2$, which is $2\pi r^2$:

$SA = \pi(4.5)(8.83) + 2\pi(4.5)^2 = 124.9 + 127.2 = 252\ \text{cm}^2$

Similar solids

Volumes scale with the cube of the lengths:

$\frac{V_1}{V_2} = \left(\frac{L_1}{L_2}\right)^3 \quad\Rightarrow\quad \frac{1000}{512} = \left(\frac{9}{x}\right)^3$

Take the cube root of both sides: $\dfrac{10}{8} = \dfrac{9}{x}$, so $x = 7.2$ cm.

A harder surface area

For questions with lots of faces, label the faces and calculate the area of each separately. The top and bottom faces are squares. Each of the four sides is a trapezium with parallel sides 110 and 80 and sloping edge 70.

Make a trapezium separately and find its height. The base is wider than the top by $110 - 80 = 30$, so each end sticks out by $15$:

$h = \sqrt{70^2 - 15^2} = \sqrt{4675} = 68.4$

$\text{Area of one trapezium} = \tfrac{1}{2}(110 + 80)(68.4) = 6496$

$SA = 110^2 + 80^2 + 4 \times 6496 = 12\,100 + 6400 + 25\,984 \approx 44\,500\ \text{square units}$

A frustum

By using similarity, we can find the radius of the smaller (missing) cone. It is half the height, so it is half as wide:

$\frac{8}{4} = \frac{4}{x} \quad\Rightarrow\quad x = 2\ \text{cm}$

Now we find the volume and surface area of both cones and subtract them to find the frustum.

$V_1 = \tfrac{1}{3}\pi(4)^2(8) = 134\ \text{cm}^3 \qquad V_2 = \tfrac{1}{3}\pi(2)^2(4) = 16.8\ \text{cm}^3$

$\text{Volume of frustum} = 134 - 16.8 = 117\ \text{cm}^3$

For the curved surface, $\pi r l$ with $l = \sqrt{r^2 + h^2}$:

$SA_1 = \pi \times 4 \times \sqrt{4^2 + 8^2} = 112\ \text{cm}^2 \qquad SA_2 = \pi \times 2 \times \sqrt{2^2 + 4^2} = 28\ \text{cm}^2$

The surface of the frustum is the curved part (the difference) plus the top circle and the bottom circle:

$SA = (112 - 28) + \pi(4)^2 + \pi(2)^2 = 84 + 50.3 + 12.6 = 147\ \text{cm}^2$