1
2-D shapes (polygons + circles).
Rectangle
5 cm
Area = width × height
Parallelogram
2 cm
2 cm  5 cm  10 cm2
5 cm
Area = base width × vertical height
2
2 cm 2 cm  5 cm  10 cm
Trapezium
b
 ab
Area  
h
 2 
h
a
2m
3m
6
 42
2
Area  
  3   3  3 3  9 m
2
 2 
4m
Triangle
Area = base width × vertical height ÷ 2
h6 m
A
wh 10  6 60


 30 m2
2
2
2
w  10 m
Collections of shapes: “add the areas”:
5m
Area  5  5  6  2  25  12  37 m2
5m
6m
2m
2
3D shapes.
Surface area: add up the area of each surface.
Cube with 1 m sides:
1m
1m
1m
Each face has area 1m × 1m = 1m2.
The cube has 6 faces (like a dice) so the total surface area = 6m2.
Cuboid:
2m
1m
Add up all the areas:
Front area = 5×2 = 10 m2,
Top area = 5×1 = 5 m2,
5m
front + back = 10+10 = 20 m2.
top + bottom = 5+5 = 10 m2
End area = 1×2 = 2 m2, left end + right end = 2+2 = 4 m2.
Total 20 + 10 + 4 = 34 m2
or, unwrap it to make a net:
Area = 5  6+2 1 2  34 m2
1 2 1 2  6 m
5m
3
Volume of a cuboid.
2m
1m
5m
Volume = width × depth × height
= 5 × 1 × 2 = 10 m3.
This is just a special case of the general rule:
Volume of a prism = (base area) × height
(A prism is anything with a constant cross-section, like a stick of Blackpool rock).
Extension (for completeness, not needed now, will see much later):
Pyramids


1
 base area   height
3
1 2
o e.g. volume of a cone =  r h where h is the vertical height.
3
Surface area of a cone (the curved surface only) =  rl where l is the slant
Volume of any pyramid =
height.
Spheres
4 3
r
3

Volume of a sphere =

Surface area of a sphere = 4 r 2
4
Examples
Find the surface area and volume of this triangular prism.
3m
5m
4m
wh 3  4

 6 m2
Cross-section area =
2
2
6m
Volume = A×L= 6 × 6 = 36 m3.
Surface area = 2×6 + (3+4+5) ×6 = 12 + 12×6 = 84 m2.
Circles
Circumference is the distance around the outside of the circle (for any other shape we
would call this the perimeter).
Circumference   D
or
Circumference  2 r
(same thing!)
Area   r 2
Perimeter and area of sectors.

A sector may be half a circle, a quarter of a circle etc.
For perimeter:
 Find the diameter D
 Write a formula for a fraction of
the circumference  D + 1D for
the straight edges.
 Put numbers in the formula
Half a circle:
5m
For area:
 Find the radius
 Write a formula for a fraction of
the circle area  r
Put numbers in the formula.
2

5
D=5m
Perimeter =
D
2
D
 5
2
r = 2.5 m
Area of half a circle =
 5  12.85 m
 r2
2

  2.52
2
 19.6 m2
Cylinder
r 3m
(i) Diameter D=6m, height h=5 m
Volume  r h    3  5  141.4 m
2
2

3



2
2
2
Surface area 2  r  2 rh  2 r  rh  150.8 m
(ii) Half of this cylinder, sliced vertically
Volume
1 2
 r h    32  5  2  70.7 m3 (half the whole-cylinder value)
2
1
2




Surface area 2   r 2    rh  2rh   r 2  rh  2rh  105.4 m2 (half the whole cylinder
plus the cut surface (rectangle 2r×h)
6
Problems where a length must be found

Decide which of the above formulae you need

Write an equation in the form “formula = value”

Solve the equation
Examples:
(a) The area of a circle = 100 cm2. Find the radius.
 r 2  100 (equation)
r2 
100

on calculator, then r 
ANS  5.642 cm
(b) A cylinder has radius 5 m and volume 200 m3. Find the length.
 r 2 L  200 (equation)
L
200
 2.546 m
  52
(c) A cylinder has volume 20 m3 and length 3 m. Find the radius.
 r 2 L  20 (equation)
r2 
20 20

on calculator, then r  ANS  1.457 m
 L 3
(d) The surface area of this cuboid = 112 cm2. Find the volume.
First we must find the height x.
x
4m
5m
Surface area = 2  4  5   4  5  4  5 x  40  18 x
Hence we need 40  18x  112 (equation)
18x  112  40  72
x
72
4m
18
nb. Each of these lines takes an expression and simplifies it.
7
Please do not tell me that 40+18x = 112 – 40 = 72÷18 = 4, this is horribly wrong!
If you do this you are abusing the equals sign. You cannot use = unless the values
are the same on both sides.
Now we know x, the volume is 4×4×5 = 80 m3.
(e) The volume of this trapezoidal prism = 3780 cm3. Find the surface area.
10 cm
13 cm
12 cm
13 cm
L
20 cm
 ab
 10  20 
2
h  
 12  15 12  180 cm
2
2




Area of the cross-section (the trapezium) = 
We know the volume, so can find the length L:
AL  180L  3780 (equation)
L
3780
 21 cm
180
Now we find the surface area:
Surface area = 2 180  10 13  20 13   21  360 1176 1536 cm 2
(= two trapeziums plus a ribbon of the four rectangular sides).