3. Objectives
At the end of the lesson the students
should be able;
To find the surface area of a
cylinder ..
4. What is a cylinder?
The term Cylinder refers to a right
circular cylinder. Like a right prism, its
altitude is perpendicular to the bases
and has an endpoint in each base.
6. What will happen if we
removed the end of the
cylinder and unrolled the
body?
Lets find out
!!!!
7. This will happen if we unrolled
and removed the end of a
cylinder….
Circumference
of the base
h
2Πr2
8. Notice that we had formed 2
circles and a 1 rectangle….
The 2 circles serves as our bases of
our cylinder and the rectangular
region represent the body
9. How can we solved the surface
area of a Cylinder?
To solve the surface area of a
cylinder, add the areas of the
circular bases and the area of
the rectangular region which is
the body of the cylinder.
10. This is the formula in order to
solved the surface area of a
cylinder.
SA= area of 2 circular bases
+ area of a rectangle
11. We derived at this formula..!!
SA=2Πr2 +2Πrh
Or
SA=2Πr (r + h)
12. Find the surface area of a
cylindrical water tank given the
height of 20m and the radius of
5m? (Use π as 3.14)
Given:
h=20m
r=5m
SA=2πr2 +2πrh
=2(3.14)(5m)2 +
2[(3.14)(5m)(20m)
=157m2 + 628m
SA =785m2
14. To find the surface area of a shape, we calculate the
total area of all of the faces.
A cuboid has 6 faces.
The top and the bottom of the
cuboid have the same area.
Surface area of a cuboid
15. To find the surface area of a shape, we calculate the
total area of all of the faces.
A cuboid has 6 faces.
The front and the back of the
cuboid have the same area.
Surface area of a cuboid
16. To find the surface area of a shape, we calculate the
total area of all of the faces.
A cuboid has 6 faces.
The left hand side and the right
hand side of the cuboid have
the same area.
Surface area of a cuboid
17. To find the surface area of a shape, we calculate the
total area of all of the faces.
Can you work out the
surface area of this cubiod?
Surface area of a cuboid
7 cm
8 cm 5 cm
The area of the top = 8 × 5
= 40 cm2
The area of the front = 7 × 5
= 35 cm2
The area of the side = 7 × 8
= 56 cm2
18. To find the surface area of a shape, we calculate the
total area of all of the faces.
So the total surface area =
Surface area of a cuboid
7 cm
8 cm 5 cm
2 × 40 cm2
+ 2 × 35 cm2
+ 2 × 56 cm2
Top and bottom
Front and back
Left and right side
= 80 + 70 + 112 = 262 cm2
19. We can find the formula for the surface area of a cuboid
as follows.
Surface area of a cuboid =
Formula for the surface area of a cuboid
h
l
w
2 × lw Top and bottom
+ 2 × hw Front and back
+ 2 × lh Left and right side
= 2lw + 2hw + 2lh
20. How can we find the surface area of a cube of length x?
Surface area of a cube
x
All six faces of a cube have the
same area.
The area of each face is x × x = x2
Therefore,
Surface area of a cube = 6x2
21. This cuboid is made from alternate purple and green
centimetre cubes.
Checkered cuboid problem
What is its surface area?
Surface area
= 2 × 3 × 4 + 2 × 3 × 5 + 2 × 4 × 5
= 24 + 30 + 40
= 94 cm2
How much of the
surface area is green?
48 cm2
22. What is the surface area of this L-shaped prism?
Surface area of a prism
6 cm
5 cm
3 cm
4 cm
3 cm
To find the surface area of
this shape we need to add
together the area of the two
L-shapes and the area of the
6 rectangles that make up
the surface of the shape.
Total surface area
= 2 × 22 + 18 + 9 + 12 + 6
+ 6 + 15
= 110 cm2
23. 5 cm
6 cm
3 cm
6 cm
3 cm
3 cm
3 cm
It can be helpful to use the net of a 3-D shape to calculate its
surface area.
Using nets to find surface area
Here is the net of a 3 cm by 5 cm by 6 cm cubiod.
Write down the
area of each
face.
15 cm2 15 cm2
18 cm2
30 cm2 30 cm2
18 cm2
Then add the
areas together
to find the
surface area.
Surface Area = 126 cm2
24. Here is the net of a regular tetrahedron.
Using nets to find surface area
What is its surface area?
6 cm
5.2 cm
Area of each face = ½bh
= ½ × 6 × 5.2
= 15.6 cm2
Surface area = 4 × 15.6
= 62.4 cm2
25. 3-Warm up: Finding the Area of a
Lateral Face
Architecture. The lateral faces of the
Pyramid Arena in Memphis, Tennessee,
are covered with steal panels. Use the
diagram of the arena to find the area of
each lateral face of this regular pyramid.
29. A cone has a circular base and a vertex that is not in the
same plane as a base.
In a right cone, the height meets the base at its center.
The height of a cone is the perpendicular distance between
the vertex and the base.
The slant height of a cone is the distance between the vertex
and a point on the base edge.
Height
Lateral Surface
The vertex is directly
above the center of
the circle.
Base
r
Slant Height
r
30. Surface Area of a Cone
Surface Area = area of base + area of sector
= area of base + π(radius of base)(slant height)
S B r
2
r r
2
B r
r
31. Lateral Area of a Cone
Since Lateral Area = Surface Area – area of the
base
2
r r
L.A. =
32. Example 1:
Find the surface area of the cone to the nearest
whole number.
a. r = 4 slant height = 6
4 in.
6 in.
2
S r r
2
(4) (4)(6)
16 24
40
40(3.14)
2
126 .
in
33. Example 2:
Find the surface area of the cone to the nearest whole
number.
b.
First, find the slant height. Next, r = 12,
12 ft.
5 ft.
2 2 2
r h
2 2
(12) (5)
144 25 169
169 13
13.
2
S r r
2
(12) (12)(13)
144 156
300
2
942 .
ft
34. On your own #1
Calculate the surface
area of:
•S = (7)2 + (7)(11.40)
•S = 49 + 79.80
•S = 128.8
2
S r r
35. On your own #2
Calculate the lateral area of:
•L.A. = (5)(13)
•L.A. = 65
2
S r r
L.A. =
46. 10 cm
4 cm
6 cm
Look at this cuboid
Now imagine it
is full of cubic
centimetres
Can you see that there are 10 4 = 40 cubic centimetres
on the bottom layer?
There are 6 layers of 40 cubes making 40 6 = 240 cm3
1 cm3
47. 10 cm
4 cm
6 cm
Let us go back and look at what we did here
length
breadth
height
When we worked out the volume we multiplied the length by the
breadth and then by the height
Volume of a cuboid = length breadth height
or
V = l b h
48. 10 cm
4 cm
6 cm
V = l b h
= 10 4 6 cm3
= 240 cm3
Lets us look again
at the same
cuboid and this
time try the
formula
You will see that this is the same answer as we got before
55. Pieces Missing
Find the volume of concrete used to
make this pipe
Volume of Concrete = Volume of Big
Cylinder – Volume of Small Cylinder
(hole)
62. Conversions of Units
1 cm2 = 10 mm x 10 mm =100 mm2
1 m2 = 100 cm x 100 cm = 10 000 cm2
1 m2 = 1000 mm x 1000 mm = 1 000 000
mm2
1 ha = 100 m x 100 m = 10 000 m2
1 km2 = 100 ha
63. What about when cubic units?
1 cm3
= 1cm x 1cm x 1cm
= 10 mm × 10 mm × 10 mm
= 1000 mm3
1 m3
= 1m x 1m x 1m
= 100 cm × 100 cm × 100 cm
= 1 000 000 cm3
64. Capacity
Volume - The volume of a three-
dimensional figure is the amount of space
within it.
Measured in Units Cubed (e.g. cm3)
Volume and capacity are related.
Capacity is the amount of material
(usually liquid) that a container can hold.
Capacity is measured in millilitres, litres
and kilolitres.
74. Compare Cone and Cylinder
Use plastic space figures.
Fill cone with water.
Pour water into cylinder.
Repeat until cylinder is full.
r r
h
75. Volume of Cone?
3 cones fill the cylinder, so…
Volume = ⅓ Base x height
=
76. Volume of Cone
3 cones fill the cylinder
Volume = ⅓ Base x height
V = ⅓ Bh
Base area = r2
V = ⅓ ( . 2.5 2) . 7
V = ⅓ 3.14 . 6.25 . 7
V = 45.79 cm3
r =2.5 cm
h = 7 cm
78. Volume of a Sphere
Using relational solids and pouring material we noted
that the volume of a cone is the same as the volume of a
hemisphere (with corresponding dimensions)
Using “math language” Volume (cone) = ½ Volume (sphere)
Therefore 2(Volume (cone)) = Volume (sphere)
=
OR
+
79. Volume of a Sphere
We already know the formula for the volume of a
cone.
3
cylinder
cone
Volume
Volume
= ÷ 3
OR
80. AND we know the formula for the volume of a cylinder
Volume of a Sphere
)
(
)
( Height
X
Base
of
Area
Volumecylinder
BASE
Height
81. SUMMARIZING:
Volume (cylinder) = (Area Base) (height)
Volume (cone) = Volume (cylinder) /3
Volume (cone) = (Area Base) (height)/3
AND 2(Volume (cone)) = Volume (sphere)
Volume of a Sphere
= ÷ 3
2 X =
82. 2(Volume (cone)) = Volume (sphere)
2( ) (height) /3= Volume (sphere)
2( )(h)/3= Volume (sphere)
BUT h = 2r
2(r2)(2r)/3 = Volume(sphere)
4(r3)/3 = Volume(sphere)
Volume of a Sphere
Area of Base
r2
2 X =
h
r
r
83. Volume of a Sphere
3
4 3
r
Volumesphere
3
4 3
r
3
4 3
r
3
4 3
r
3
4 3
r
