๐Ÿ”ต Circles, Surface Area and Volume

GCSE Maths ยท Geometry and Measures (G17)

Ages 15โ€“16 ยท Foundation & Higher

โ† Back to topic overview
1 Circumference and Area of a Circle
The two circle formulae
Circumference:  $C = 2\pi r = \pi d$
Area:  $A = \pi r^2$
Do not mix them up. The one with the square is the area. A useful memory line: "Twinkle twinkle little star, circumference equals two pi r."
What is $\pi$? It is the number you get when you divide any circle's circumference by its diameter โ€” always about $3.14159$. That is exactly why $C = \pi d$.
Worked Example 1 โ€” Circumference and area

A circle has radius $9$ cm. Find its circumference and area, each to 1 d.p.

โ‘ $C = 2\pi r = 2 \times \pi \times 9 = 18\pi = 56.5$ cm
โ‘ก$A = \pi r^2 = \pi \times 81 = 81\pi = 254.5\text{ cm}^2$
Worked Example 2 โ€” Working from the diameter

A circular pond has diameter $7$ m. Find its area, to 2 d.p.

โ‘ Halve the diameter first: $r = 7 \div 2 = 3.5$ m.
โ‘ก$A = \pi \times 3.5^2 = \pi \times 12.25 = 38.48\text{ m}^2$
Never square the diameter. $\pi \times 7^2$ would give four times the correct area.
Worked Example 3 โ€” Working backwards

A circle has area $200\text{ cm}^2$. Find its radius, to 2 d.p.

โ‘ $\pi r^2 = 200$
โ‘ก$r^2 = \dfrac{200}{\pi} = 63.662$
โ‘ข$r = \sqrt{63.662} = 7.98$ cm
"Leave your answer in terms of $\pi$" means do not press the $\pi$ button at the end โ€” write $81\pi$, not $254.5$. This is exact and often easier.
2 Semicircles and Composite Shapes
ShapeAreaPerimeter
Semicircle$\tfrac{1}{2}\pi r^2$$\pi r + 2r$  (curve $+$ diameter)
Quarter circle$\tfrac{1}{4}\pi r^2$$\tfrac{1}{2}\pi r + 2r$
The perimeter of a semicircle is not half the circumference. You must add the straight diameter as well, or the shape would not be closed.
diameter = 2r curve = ฯ€r Perimeter = ฯ€r + 2r A running track: rectangle + two semicircles
Worked Example 4 โ€” Semicircle

A semicircle has radius $8$ cm. Find its area and its perimeter, each to 1 d.p.

โ‘ Area $= \tfrac{1}{2}\pi r^2 = \tfrac{1}{2} \times \pi \times 64 = 32\pi = 100.5\text{ cm}^2$
โ‘กCurved part $= \tfrac{1}{2} \times 2\pi r = \pi \times 8 = 25.13$ cm
โ‘ขStraight part (the diameter) $= 16$ cm
โ‘ฃPerimeter $= 25.13 + 16 = 41.1$ cm
Worked Example 5 โ€” A composite shape

A shape is a rectangle $20$ cm by $12$ cm with a semicircle of diameter $12$ cm attached to one short end. Find its area, to 1 d.p.

โ‘ Rectangle: $20 \times 12 = 240\text{ cm}^2$
โ‘กSemicircle radius $= 12 \div 2 = 6$ cm
โ‘ขSemicircle area $= \tfrac{1}{2}\pi \times 36 = 18\pi = 56.55\text{ cm}^2$
โ‘ฃTotal $= 240 + 56.55 = 296.5\text{ cm}^2$
3 Volume and Surface Area of Solids
SolidVolumeSurface area
Cuboid$lwh$$2(lw + lh + wh)$
Prismcross-section $\times$ length$2\times$cross-section $+$ perimeter $\times$ length
Cylinder$\pi r^2 h$$2\pi r^2 + 2\pi r h$
Cone$\tfrac{1}{3}\pi r^2 h$$\pi r^2 + \pi r l$  ($l$ = slant height)
Sphere$\tfrac{4}{3}\pi r^3$$4\pi r^2$
Hemisphere (solid)$\tfrac{2}{3}\pi r^3$$2\pi r^2 + \pi r^2 = 3\pi r^2$
Pyramid$\tfrac{1}{3} \times$ base area $\times h$base $+$ area of the triangular faces
The sphere and cone formulae are usually given on the exam formulae sheet. The cuboid, prism and cylinder formulae are not โ€” learn those.
Curved surface of a cone: $\pi r l$. Note the slant height $l$, not the vertical height $h$. If you are given $h$, find $l$ first using $l^2 = r^2 + h^2$.
Worked Example 6 โ€” Sphere

A sphere has radius $6$ cm. Find its volume and surface area, in terms of $\pi$ and as decimals to 1 d.p.

โ‘ $V = \tfrac{4}{3}\pi r^3 = \tfrac{4}{3} \times \pi \times 216 = 288\pi = 904.8\text{ cm}^3$
โ‘ก$SA = 4\pi r^2 = 4 \times \pi \times 36 = 144\pi = 452.4\text{ cm}^2$
Worked Example 7 โ€” Cone with vertical height given

A cone has radius $5$ cm and vertical height $12$ cm. Find its total surface area, to 1 d.p.

โ‘ Slant height: $l^2 = 5^2 + 12^2 = 169$, so $l = 13$ cm.
โ‘กBase: $\pi r^2 = 25\pi$
โ‘ขCurved surface: $\pi r l = \pi \times 5 \times 13 = 65\pi$
โ‘ฃTotal $= 25\pi + 65\pi = 90\pi = 282.7\text{ cm}^2$
Worked Example 8 โ€” Pyramid

A pyramid has a square base of side $8$ cm and vertical height $9$ cm. Find its volume.

โ‘ Base area $= 8 \times 8 = 64\text{ cm}^2$
โ‘ก$V = \tfrac{1}{3} \times 64 \times 9 = \tfrac{1}{3} \times 576 = 192\text{ cm}^3$
4 Composite Solids
Surface area of a composite solid is not the sum of the two surface areas. The faces where the pieces meet are inside the solid and must not be counted.
Worked Example 9 โ€” Cylinder with a hemisphere on top

A solid consists of a cylinder of radius $4$ cm and height $10$ cm with a hemisphere of the same radius on top. Find its volume and total surface area, both in terms of $\pi$.

โ‘ Cylinder volume $= \pi r^2 h = \pi \times 16 \times 10 = 160\pi$
โ‘กHemisphere volume $= \tfrac{2}{3}\pi r^3 = \tfrac{2}{3} \times \pi \times 64 = \dfrac{128\pi}{3}$
โ‘ขTotal volume $= 160\pi + \dfrac{128\pi}{3} = \dfrac{480\pi + 128\pi}{3} = \dfrac{608\pi}{3}\text{ cm}^3$
โ‘ฃSurfaces: the flat circular base, the cylinder's curved surface, and the hemisphere's curved surface. The cylinder's top is hidden.
โ‘คBase $= \pi r^2 = 16\pi$; cylinder curved $= 2\pi r h = 80\pi$; hemisphere curved $= 2\pi r^2 = 32\pi$
โ‘ฅTotal surface area $= 16\pi + 80\pi + 32\pi = 128\pi\text{ cm}^2$
Worked Example 10 โ€” A hole drilled through

A cuboid $10$ cm by $10$ cm by $6$ cm has a cylindrical hole of radius $2$ cm drilled all the way through its $6$ cm depth. Find the remaining volume, to 1 d.p.

โ‘ Cuboid volume $= 10 \times 10 \times 6 = 600\text{ cm}^3$
โ‘กCylinder removed $= \pi r^2 h = \pi \times 4 \times 6 = 24\pi = 75.40\text{ cm}^3$
โ‘ขRemaining $= 600 - 75.40 = 524.6\text{ cm}^3$
5 Quick Reference

Circle

$C = 2\pi r = \pi d$; $A = \pi r^2$. The squared one is the area.

Diameter first

Halve it to get $r$ before using any formula.

Semicircle

Area $\tfrac{1}{2}\pi r^2$; perimeter $\pi r + 2r$.

Cylinder

$V = \pi r^2 h$; $SA = 2\pi r^2 + 2\pi r h$.

Cone

$V = \tfrac{1}{3}\pi r^2 h$; curved surface $= \pi r l$.

Sphere

$V = \tfrac{4}{3}\pi r^3$; $SA = 4\pi r^2$.

Pyramid

$V = \tfrac{1}{3} \times$ base $\times$ height.

Composite

Add volumes; count only the visible faces for surface area.

Exact answers

Leave in terms of $\pi$ unless told otherwise.

6 Practice Questions
Question 1

A circle has radius $12$ cm. Find (a) its circumference, (b) its area, each in terms of $\pi$.

โ–ถ Show solution

(a) $C = 2\pi r = 24\pi$ cm

(b) $A = \pi r^2 = 144\pi\text{ cm}^2$

Question 2

A circular table has diameter $1.4$ m. Find its area, to 3 significant figures.

โ–ถ Show solution

$r = 1.4 \div 2 = 0.7$ m

$A = \pi \times 0.7^2 = 0.49\pi = 1.539\ldots$

$1.54\text{ m}^2$ (3 s.f.)

Question 3

A circle has circumference $50$ cm. Find its radius, to 2 d.p.

โ–ถ Show solution

$2\pi r = 50$

$r = \dfrac{50}{2\pi} = \dfrac{25}{\pi} = 7.96$ cm

Question 4

Find the perimeter of a semicircle of radius $10$ cm, to 1 d.p.

โ–ถ Show solution

Curved part $= \pi r = 10\pi = 31.42$ cm

Straight part $= 2r = 20$ cm

Perimeter $= 31.42 + 20 = 51.4$ cm

Question 5

A sphere has radius $3$ cm. Find its volume and surface area in terms of $\pi$.

โ–ถ Show solution

$V = \tfrac{4}{3}\pi \times 27 = 36\pi\text{ cm}^3$

$SA = 4\pi \times 9 = 36\pi\text{ cm}^2$

(A curious coincidence: for $r = 3$ the two numbers match, but the units differ.)

Question 6

A cone has radius $6$ cm and slant height $10$ cm. Find (a) its curved surface area, (b) its vertical height, (c) its volume. Give (a) and (c) in terms of $\pi$.

โ–ถ Show solution

(a) $\pi r l = \pi \times 6 \times 10 = 60\pi\text{ cm}^2$

(b) $h^2 = l^2 - r^2 = 100 - 36 = 64$, so $h = 8$ cm.

(c) $V = \tfrac{1}{3}\pi \times 36 \times 8 = 96\pi\text{ cm}^3$

Question 7

A cylinder has volume $1000\text{ cm}^3$ and radius $5$ cm. Find its height, to 2 d.p.

โ–ถ Show solution

$\pi r^2 h = 1000$

$25\pi h = 1000$

$h = \dfrac{1000}{25\pi} = \dfrac{40}{\pi} = 12.73$ cm

Question 8

A shape is a rectangle $14$ cm by $8$ cm with a semicircle of diameter $8$ cm attached to each short end. Find (a) the total area, (b) the total perimeter, each to 1 d.p.

โ–ถ Show solution

Semicircle radius $= 4$ cm. Two semicircles make one full circle.

(a) Rectangle $= 14 \times 8 = 112\text{ cm}^2$. Full circle $= \pi \times 16 = 50.27\text{ cm}^2$.

Total $= 112 + 50.27 = 162.3\text{ cm}^2$

(b) The perimeter is the two long rectangle sides plus the two curves (which together make a full circumference).

$= 2 \times 14 + 2\pi \times 4 = 28 + 25.13 = 53.1$ cm

Question 9

A solid hemisphere has radius $9$ cm. Find (a) its volume, (b) its total surface area (including the flat circular face), both in terms of $\pi$.

โ–ถ Show solution

(a) $V = \tfrac{2}{3}\pi r^3 = \tfrac{2}{3} \times \pi \times 729 = 486\pi\text{ cm}^3$

(b) Curved part $= 2\pi r^2 = 162\pi$; flat circle $= \pi r^2 = 81\pi$.

Total $= 162\pi + 81\pi = 243\pi\text{ cm}^2$

Question 10

A solid is made from a cylinder of radius $5$ cm and height $12$ cm with a cone of the same radius and slant height $13$ cm on top.

(a) Find the cone's vertical height.   (b) Find the total volume in terms of $\pi$.   (c) Find the total surface area in terms of $\pi$.   (d) The solid is made of a metal of density $7.8\text{ g/cm}^3$. Find its mass, to the nearest gram.

โ–ถ Show solution

(a) $h^2 = 13^2 - 5^2 = 169 - 25 = 144$, so $h = 12$ cm.

(b) Cylinder $= \pi \times 25 \times 12 = 300\pi$

Cone $= \tfrac{1}{3}\pi \times 25 \times 12 = 100\pi$

Total $= \mathbf{400\pi\text{ cm}^3}$

(c) Visible surfaces: the flat base, the cylinder's curved surface, the cone's curved surface. (The cylinder's top is covered by the cone.)

Base $= \pi \times 25 = 25\pi$

Cylinder curved $= 2\pi \times 5 \times 12 = 120\pi$

Cone curved $= \pi r l = \pi \times 5 \times 13 = 65\pi$

Total $= \mathbf{210\pi\text{ cm}^2}$

(d) Volume $= 400\pi = 1256.637\text{ cm}^3$

Mass $= 1256.637 \times 7.8 = 9801.8$, so about $\mathbf{9802}$ g (roughly $9.8$ kg).

Circles, Surface Area & Volume (G17) ยท GCSE Maths Revision ยท Created with MathJax