6. Perimeter, area and volume
Mathematics Standard Year 11
Area of a circle
The area of a circle is calculated using the
formula:
A = pr2
where r is the radius of the circle.
Example
Find the area of a circle with a radius of 6 metres. Give your
answer correct to 1 decimal place.
Solution
A = pr2
= p ´ 62
= 113.0973355
= 113.1 m 2
The area of the circle is 113.1 m2
Area of an annulus
A annulus is the area between a large and a
small circle with the same centre. The area
of an annulus is calculated using the
formula:
A = p (R2 - r 2 )
where R is the radius of the large circle and r is the radius
of the small circle.
Example
Alison draws two concentric circles whose radii are 5 cm
and 7 cm. What is the area of the annulus formed to the
nearest square centimetre?
Solution
R = 7 and r = 5
A =  (R2 − r 2 )
=   ( −  )
= 
=  cm 2
Area of the annulus is 75 cm2
Area of a sector
A sector is part of a circle. The area of a
sector is calculated using the formula:
A=
q
360
pr2
where q is the angle formed at the centre and r is the
radius of the circle
Example
Calculate the area of a sector whose angle at the centre is
60° and radius 9 m. (1 decimal place)
Solution
q = 60 and r = 9
A=
q
pr2
360
60
=
´ p ´ 92
360
= 42.41150082
= 42.4 m 2
Area of the sector is 42.4 m2
Area of composite shapes
❑ A composite
shape is made up of more than one simple
shape.
❑ Area of composite shapes can be found by adding or
subtracting the areas of the simple shapes.
Example
A decking for a house consists of a
square and a triangle. The square
has a side length of 6 metres and the
triangle is isosceles.
a. Use Pythagoras theorem to find the value of x.
b. Calculate the area of the shaded region.
Solution
a. x 2 + x 2 = 6
Pythagoras theorem
2x 2 = 6
x2 = 3
x= 3
b.
A = s 2 (Square)
= 62
= 36 m
2
bh
A=
(Triangle)
2
3´ 3
=
2
= 1.5 m 2
Shaded area = 36 + 1.5
= 37.5 cm 2
Example
The diagram shows a semi-circle cut out of a semi-circle.
What is the shaded area to the nearest square centimetre?
Solution
A = 12 p r 2 (Semi-circle)
= 12 ´ p ´ 52
25
= p cm 2
2
Example
The diagram shows a block of land that has been surveyed.
All measurements are in metres. Find the area of
quadrilateral ABCD.
Solution
Triangle ACD
1
A = bh
2
1
=  76  32
2
Triangle ABD
= 1216 m 2
= 988 m 2
1
A = bh
2
1
=  76  26
2
Total area = 1216 + 988 = 2204 m 2
Trapezoidal rule
h
A = (d f + dl )
2
A – Area of the shape.
h – Height or distance between the parallel sides.
df – Distance of the first parallel side.
dl – Distance of the second parallel side.
Example
1.
2.
Estimate the area by applying
trapezoidal rule once for the
this irregular field.
A block of land has a road as one
boundary. Apply the Trapezoidal
rule twice to approximate the area
of the block. Answer to the nearest
square metre.
Solution
1.
2.
h
A = (d f + dl )
2
48
= (25 + 12)
2
= 888 m 2
h
h
A = (d f + dl ) + (d f + dl )
2
2
21
21
= (23+ 25) + (25 + 24)
2
2
= 1018.5 m 2
Area of the block of a land is 1018.5 m2
Surface area of right prisms
1. Visualise the surfaces of the solid.
2. Write the formula for the surface area.
3. Substitute the values into the formula for the surface area
4. Use the calculator to find the surface area.
5. Write the answer to the specified level of accuracy.
Example
Find the surface area of this
rectangular prism
Solution
SA = (2 ´ l ´ b) + (2 ´ b ´ h) + (2 ´ l ´ h)
= (2 ´ 10 ´ 5) + (2 ´ 5 ´ 2) + (2 ´ 10 ´ 2)
= 160 cm 2
Surface area of the rectangular prism is 160 cm2
Surface area of cylinders
Surface area is the area of all the surfaces of a solid.
Open cylinder
SA = 2p rh
Curved surface only
SA = 2p r 2 + 2p rh
Closed cylinder
SA = p r 2
Area of the top
SA = p r 2
Area of the bottom
SA = 2p rh
Curved surface
Example
1. Find the surface area of a closed cylinder with a radius of
14 mm and a height of 24 mm. Answer correct to two
decimal places.
2. A car hose is an open cylinder with a radius of 3 cm and a
length of 14 cm. Calculate the area of the outer surface
correct to two decimal places.
Solution
1.
SA= 2p r 2 + 2p rh
= 2 ´ p ´ 142 + 2 ´ p ´ 14 ´ 24
= 3342.654583
= 3342.65 mm 2
2. SA= 2p rh
= 2 ´ p ´ 3 ´ 14
= 263.8937829
= 263.89 cm 2
Surface area of pyramids
❑
Square pyramid has a square base and 4 triangular faces
with equal area.
s – side length of the base
l – slant height of triangular face
æ1 ö
SA = s + 4 ´ ç bh ÷
è2 ø
æ1 ö
= s 2 + 4 ´ ç sl ÷
è2 ø
2
Example
Find the surface area of the following square pyramid.
Answer correct to two decimal places.
Solution
Use Pythagoras theorem to find the slant height.
l 2 = h2 + r 2
= 1.52 + 12
l = 1.52 + 12
= 1.8027... » 1.80 m
Surface area of the pyramid
æ1 ö
SA = s + 4 ´ ç sl ÷
è2 ø
2
æ1
ö
= (2 ) + 4 ´ ç ´ 2 ´ 1.8027...÷
è2
ø
2
= 11.2111...
» 11.21 m 2
Surface area of cones
❑
A cone has a flat circular base and a curved surface.
r – radius of the circular base
l – slant height of curved surface
h – perpendicular height of cone
SA = p r 2 + p rl
Example
Find the surface area of the following cone.
Answer correct to two decimal places.
Solution
Use Pythagoras theorem to find the slant height.
l 2 = 6.42 + 12.12
l = 6.42 + 12.12
= 13.6883... » 13.69 cm
Surface area of the cone
SA =  r 2 +  rl
=   6.42 +   6.4 13.6883...
= 403.8988...
 403.90 cm 2
Volume of prisms and cylinders
Volume is the amount of space occupied by a threedimensional object.
Example
1. What is the volume of a rectangular prism with a base
area of 23 cm2 and a height of 15 cm?
2.
What is the volume of these prisms?
a.
b.
Solution
1.
V = Ah
= 23 15
= 345 cm3
2.
a.
1
A = bh
2
1
= 8 3
2
= 12 m 2
V = Ah
= 12 12
= 144 m3
b.
V =  r 2h
=   62 12
= 1357.168026 m3
 1357 m3
Volume of pyramids and cones
A pyramid fits exactly inside a prism and occupies one
third the volume of the prism.
Square pyramid
1
1 2
V = Ah = ´ s ´ h
3
3
Rectangular pyramid
1
1
V = Ah = ´ lb ´ h
3
3
Triangular pyramid
1
1 1
V = Ah = ´ bh ´ h
3
3 2
Cone
1
1
V = Ah = ´ p r 2 ´ h
3
3
Example
1. Find the volume of this square
pyramid of height 8 m and base
6 m. Give your answer correct to
the nearest cubic metre.
2. Find the volume of this cone.
Answer correct to one decimal
place.
Solution
1. V = 1 Ah
3
1 2
= s h
3
1 2
= ´6 ´8
3
= 96 m 3
Volume of the square
pyramid is 96 m3
1 2
2. V = p r h
3
1
= ´ p ´ 52 ´ 9
3
= 235.619449¼ m 3
» 235.6 m 3
Volume of the cone is
235.6 m3
Surface area of composite solids
1. Visualise the surfaces of the solid.
2. Calculate the area of each face using the appropriate area
formula.
3. Add the area of all the faces.
4. Write the answer to the specified level of accuracy.
Example
A storage shed consists of a rectangular
prism and half a cylinder as shown.
What is the surface area of the shed to
the nearest square centimetre?
Solution
A= (5 ´ 6) + (2 ´ 6 ´ 2) + (2 ´ 5 ´ 2) (Rectangular prism)
= 74 m 2
A = 12 ´ 2p rh (Half a cylinder)
1
= ´ 2 ´ p ´ 2.5 ´ 6
2
= 47.123889¼ m 2
SA = Rectangular prism + Half a cylinder
= 74 + 47.123889¼
= 121.123889¼
= 121 m 2
Surface area of the storage shed is 121 m2 = 1210000 𝑐𝑚2
Volume of composite solids
1. Divide the composite solid into two or more common
solids.
2. Calculate the volume of each solid.
3. Add or subtract the volumes of the common solids.
4. Write the answer to the specified level of accuracy with
correct units.
Example
A storage shed consists of a rectangular
prism and half a cylinder as shown.
What is the volume of the shed to the
nearest cubic centimetre?
Solution
V = Ah (Rectangular prism)
= 5´ 6 ´ 2
= 60 m 3
1 2
V =  r h (Half a cylinder)
2
1
=      
2
=  m3
V = Rectangular prism + Half a cylinder
= 60 + 58.90486¼
= 118.90486¼
= 119 m 3
Volume of the storage shed is 119 m3 = 119000000 𝑐𝑚3
Capacity
The capacity of a container is the amount of liquid it can
hold. Some solids have both a volume and a capacity.
1 ML = 1 000 kL
1 cm3 = 1 mL
1 ML = 1 000 000 L 1 cm3 = 0.001 L
1 kL = 1000 L
1 L = 1 000 mL
1000 cm3 = 1 L
1 m3 = 1 000 000 cm3
1 m3 = 1 000 000 mL
1 m3 = 1 000 L
1 m3 = 1 kL
Example
1.
A cubic water tank has a side length of 6 m. What is
the capacity of the tank?
2. What is the capacity of a cylindrical plastic container
is 12 cm high and its circular end surfaces each have a
radius of 6 cm? Answer in litres correct to two
decimal places.
Solution
1.
V = s3
= 63
= 216 m3
Capacity = 216 1000 L
= 216 kL
2.
V =  r 2h
=   62 12
= 1357.168026... cm3
Capacity = 1357.168026  0.001 L
= 1.36 L