LESSON

11

VOLUMES AND SURFACE AREAS

In this lesson we will Review the formula from Grade 10 Learn the new formulae and apply them to mensuration problems Firstly, remember to think of volume as the amount of liquid or air that a 3-D shape can hold; and surface area as the exterior surface of the shape that you could paint. In Grade 10 you learnt that, in general, to work out 1. 2. Volume: we take the area of the base and multiply that by the height Surface area: we find out the area of each face seperately and then add the answers together to get a total.

Here is a reminder of formulae from Grade 10: Shape Cube Volume V=

x

Surface area

(x x) x base area height SA = 6 (x x) square = 6x2 3 =x

Rectangular Prism
height

V=

lb h SA = 2lb + 2bh + 2lh base area height = 2(lb + bh + 1h) rectangle

breadth length

Triangular Prism
base h

V=
H

H SA = two triangles + 3 rectangles 1 = 2(_b h) + (x + y + z) H 2 area of height = b h + (perimeter H) triangular base

1 (_b h) 2

z x y

Cylinder
radius

V=
height

pr2 h SA = 2pr2 + curved rectangle area of height = 2pr2 + 2prh circular base =2pr (r + h) = pr2h

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Example Here are two quick examples to refresh your skills: 1.

20 cm 30 cm 15 cm 6 cm

Example

2.
5 cm

Calculate: a. b. the volume surface area of these 3-D shapes

Solution
2 2

Solution 1.a Volume = half volume of cylinder 1 = _ (pr2 h)

1 = _ (p(3)2(20))

1.b

= 282, 74 cm3 1 1 Surface Area = 2(_ circles) + shaded rectangle + _ curved rectangle
1 1 = 2(_pr2) + (length breadth) + _(2pr H) 2 2 2 2

= 9p + 120 + 60p = 336,77 cm2 Can you see that although surface area is harder to evaluate, the best way is to break it down into the sum of its parts? 2.a Volume = x3 = Volume of the cube + volume of the cylinder = 153 + p(2,5)2(30) = 3964,05 cm3 2.b Surface Area = 6 square faces circle where handle meets (x) + end shaded circle (y) + curved rectangle = 6x2 + 2prh = 6(15) + 2 p(2,5)(30)
2

x = y, therefore they cancel each other out.

= 1821,24 cm2 You are now ready to meet the shapes we study in Grade 11. The first family of shapes are called the Right Pyramids These are 3-D shapes with a polygon as a base and triangular sides which meet at the top point, called an apex. A right pyramid has the apex directly above the middle of the base. The pyramid is named according to the shape of the base. We study pyramids with regular polygons in their base so that their slanted sides are all congruent triangles.
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Triangular Pyramid
apex

Square Pyramid

Hexagonal Pyramid

We are also required to study: Cones although we group these separately, a cone is actually a pyramid with a circular base

Now that we have introduced these shapes to you, let us look at the formulae for their volume and surface area. The volume of any pyramid is: 1 V = _ area of the base perpendicular height
3

Hopefully you will be able to see where the volume formulae come from: Look at this:

and

Can you see that by filling the cone three times over we could fill a cylinder provided they had the same radius and height. Likewise 3 fills of the square based pyramid would fill up the square based prism. Maybe you can get your teacher to help you make models of these shapes and to demonstrate this by using sand or jelly tots.

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The Surface Area of any Pyramid is: SA = the sum of the areas of the seperate faces = the area of the base + area of all the congruent triangles. The Surface Area of a Cone is: SA = the area of the circular base + area of curved surface = pr2 + pr (slant height) Surface area of a cone:

circle base

curved surface laid out flat

We now need you to study the next two shapes carefully, so that you are familiar with all the terminology we will be using in the next exercises.
A

BCDE is the base of the pyramid AG is the height of the pyramid CD is the base of ACD AF is the slant height of the pyramid, but the height of ACD ^ AFG = q is the angle at which the face is slanting GF is called the apothem. An apothem of a regular polygon is just a line segment drawn from the centre to the midpoint of one of its sides, and it is perpendicular to the side.
Apothem of a hexagon Apothem of a square

D F C

AB is the radius of the circular base AG is the height of the cone BG is called the slant height
A B

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You are also going to need to know the Theorem of Pythagoras which states that for ABG above: AB2 + AG2 = BG2 in order to work out the height, slant height or radius/apothem.
Example

Example 1 Find the volume and surface area of a pyramid _ with height 32 cm and a square base with sides all 6 cm in length.
A

Solution

Solution
1 Volume = _ area of base h 3 _ 1 = _ (6 6) 32 3 _ = 362 cm3

3 2

= 50,91 cm3 Surface Area = area of square + area 4 triangle We need slant height AC _ AC2 = 32 + (32 )2 (By Pythagoras) AC = 27 _ AC = 27
2

3 2

_ 1 SA = 6 6 + 4(_ 6 27 ) 2 = 98,35 cm2

Notice that we could work out the angle that the triangles were slanted at, using basic trigonometry. _ 32 tan q = _ Since in ABC
3

q = 54,74
3 2

How wonderful is that!

Example

Example 2 Find the volume and surface area of a pyramid with a triangular base with each edge 6 cm and a height of 10 cm.
10 cm

Solution

Solution
1 Volume = _ area of base h 3 To work out area of base triangle _ AB2 = 62 32 93 _ AB = 27 _ 1 1 Volume = _ (_ 6 27 ) 10 3 _ 2 = 303
A

A C 6 cm

_
B

= 51,96 cm3
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Or to work out the area of the base we could use the area rule from 1 Trigonometry so area = _(6)(6)sin 60 (equilateral has int. angle = 60) 2 _ = 93 Area base + 3 other s _ _ 1 93 + 3 (_ 6 103 ) = 106,93 cm2 2 _ Slant height = 103 = 62,35 cm3 Example 3 Find the volume and surface area of a cone with diameter 8 cm and height 8 cm. Solution
1 Volume = _ pr2h 3 1 = _ p(4)2(8) 3
A

10
3

_
Example

Solution

= 134,04 cm3 Remember if diameter = 8 cm, we use radius = 4 cm

8 cm

4 cm

Surface Area = area of base circle + area curved surface = pr2 + pr (slant height) Now we need to use Pythagoras to get the slant heights: so now s2 = 82 + 42 s2 = 80 _ s = 80 _ SA = p(4)2 + p(4)(80 ) = 162,66 cm2
4 cm 8 cm

Spheres this is a body bounded by a surface whose every point is the same distance from a centre point. For example: a soccer ball, a tennis ball. Remember that half a sphere is called a HEMISPHERE We now just need to give you the formulae for measuring the volume and surface area of a sphere. For a sphere: 4 Volume = _ pr3 (where r is the spheres radius)
3

Surface Area = 4pr2 Example 1 An ice cream cone has a radius of 3 cm and a height of 12 cm. A half scoop of ice cream is paced on the cone. If the ice cream melts, will it fit into the cone?
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Lesson 1 | Algebra

Example

Solution

Solution
1 Volume cone = _ pr2h 3 1 = _ p(3)2(12) 3
3

= 36 p = 113,1 cm3 1 4 Volume scoop = _(_pr3)

1 4 = _(_ p(3)3) 2 3 2 3
12

= 18 p = 56,55 cm3 So yes, it will fit and will actually only half fill the cone. All you now need to do is practise applying these formulae:
Activity

Activity 1
Find the volume of these shapes (correct to 2 decimal digits) 1.
6 cm

12 cm

2.

10 cm 3,5 cm

3.
20 cm

30 cm 15 cm 30 cm
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4.

89

_
C

20 cm

Activity 2
Find the surface area of the shapes 1 4 in Exercise 1.

Activity

Activity 3
To finish off this module we are going to challenge you to use your knowledge to solve real life problems. 1. If we take a cone and we remove the top of it by making a cut parallel to the circular base we get a small cone and a shape called a frustum.

Activity

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small cone

frustum

Ayanda wants to make a lamp shade. She first makes a cone with dimensions as shown and then cuts off the top. She covers the frustum with material. How much material does she need?
10 cm

30 cm

25 cm 10 30 cm 20

20 cm

2.

Paul wants to put 1 200 steel ball bearings (which are spherical in shape) with a radius of 5 mm, into a cylindrical container which is 30 cm high and 18 cm in diameter. Will all the bearings fit into the container? Show all your working.

5 mm

30 cm

18 cm

3.

Abigail buys a new salt cellar in the shape of a cylinder topped by a hemisphere, as shown below. The cylinder has a diameter of 6 cm and a height of 10 cm. She pours the salt into the salt cellar, so that it takes up half the total volume of the pot. Find the depth of the salt, marked with x in the diagram.
10 cm x

4.

The pyramid of the sun of Teotihuacan in mexico is the third largest pyramid in the world behind the great Pyramid of Cholula and the Great Pyramid of Giza. It is a right pyramid with a base that is approximately square.

6 cm

The length of one side is approximately 223,5 m and the height is about 71,2 m

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Calculate: 4.1 4.2 4.3 4.4 The angle at which the slanted face is sloping By how much would the volume of material increase if the Pyramid was 1 m Higher/taller? The volume of material in a pyramid that has dimensions that are half the original ie: Length of square base = 111,75 m Height = 35,6 m The ratio of the pyramid volume calculated in 4.3 above, to the volume of the original pyramid.

Solutions to Activities
Activity 1 1.
1 Volume = _ (volume of cone) 2 1 1 = _(_ pr2h) 1 = _ (p)(3)2(12) 6 2 3

= 18p = 56,55 cm3 2.

1 Volume = volume cone + _ sphere 2 1 1 4 = _ pr2h + _ (_ pr3) 1 2 = _ p(3,5)2(10) + _ p(3,5)3 3 3 3 2 3 343 245 =_p+_p 12 6 833 =_p 12

= 218,08 cm3 3. Volume = volume prism + volume pyramid 1 = 30 30 15 + (_ 30 30 20)

3

= 13 500 + 6 000 = 19 500 cm3 4. Volume = volume cylinder + volume cone 1 = pr2h + _pr2h
1 = p(5)2(20) + _p 52(AB) 3 3

Need to find AB : AB2 = AC2 BC2 AB2 = 89 25 AB = 8 1 = (5)2 20 + _ (5)28

200 = 500 + _ 1700 =_ 3 = 1780,24 cm3 3 3

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Activity 2 1.
1 1 Surface area = _ circle + _ curved area + triangle 2 2 1 1 1 = _ (pr2) + _ (pr (s)) + _ base h 2 2 2 __ 1 1 1 = _ p (3)2 + _ p(3)(122 + 32 ) + _ (6)(12) 9 = _ p+ 58,29 + 36 2 2 2 2
s (slanted height) 3

2.

= 108,43 cm2 1 Surface area = _ (4pr2) + curved area

2 1 = _ (4p(3,5)2) + pr (s) 2 __ 49 = _ p+ p(3,5)(100 + (3,5)2 ) 2

= 193,47 cm2 3. Surface area = area of 5 surfaces of prism + area 4 triangles We need to know the height of each triangle, so we calculate it by the theorem of Pythagoras. Note: We can get the apothem as half the side length So in the triangle S2 = 202 + 152 S2 = 400 + 225 S2 = 625 S = 25
1 So surface area becomes = (30 30) + 4(30 15) + 4(_ 30 25) 2
15 20 25

= 900 + 1800 + 1500 = 4 200cm2 4. Surface area = base circle + curved bit + curved cone area _ = p(5)2 + 2p(5)(20) + p(5)(89 ) = 855,05 cm2 Activity 3 We will get the surface area of the frustum 1. We work out the surface area of the whole cone and then subtract the surface area of the small one. We do not need the circular base. Big Cone SA = prs __ = p(20)(55)2 + (20)2 = 3 677,14 cm2
55 (slant height)

20

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Little Cone SA = prs

__ = p(10)(25)2 + (10)2 = 845,90 cm2

25 (slant height)

\ Frustum/ lamp shade needs 3 677, 14 845, 90 = 2 831, 24 cm2 of material

10

2.

We are going to work in cm so first we change 1 5 mm 0,5 cm = _ cm

4 4 1 4 1 Each ball bearing has volume = _ p(r)3 = _ p(_)3 = _ p(_) 3 3 2 3 8 4 1 =_._p p 1 = _ p = _ @ 0,52cm3 6 6 3 8 2

Cylinder has volume = pr2h = p(9)2(30) = 2 430p = 7 634,07 cm3

p Now 1 200 bearings will have volume 1200 x _ = 200p @ 628,32 cm3 so 6 they will be able to fit in with lots of room to spare. 1 4 Volume of salt cellar = pr2h + _ (_ pr3) 1 4 = p(3) (10) + _ _ p(3)3 2 3
2

3.

= 108p @ 339,29 cm3 So half the total volume = 54p or 169,65 cm3 We want to find x so that pr2x = 169,65 p(3)2x = 169,65 x = 6 cm So the salt will be 6 cm deep. 4.1

71,2

71,2

111,75

233,5

apothem is half the side length

71,2 Now tan q = _ 111,75

q = 32,5

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4.2

1 Current volume = _ base area h 3 1 = _(223,5)2 72,2 3

= 1185533,4 m3 1 If increased height by 1 m = _ (223,5)2 72,2

3

= 1202184,15 m3 4.3 4.4 \ an extra 16650,75 m3 1 Volume = _ (111,75)2 35,6

3

= 148191,675 m3 148191,675 1 __ = 0,125 = _

1185533,4 8

ratio 1: 8

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