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Geometry: Measurement (Grade 10)

Module by: Free High School Science Texts Project. E-mail the author

Measurement

Areas of Polygons

  1. Area of triangle: 12×12× base ×× perpendicular height
    Figure 1
    Figure 1 (MG10C13_047.png)
  2. Area of trapezium: 12×12× (sum of ∥∥ (parallel) sides) ×× perpendicular height
    Figure 2
    Figure 2 (MG10C13_048.png)
  3. Area of parallelogram and rhombus: base ×× perpendicular height
    Figure 3
    Figure 3 (MG10C13_049.png)
  4. Area of rectangle: length ×× breadth
    Figure 4
    Figure 4 (MG10C13_050.png)
  5. Area of square: length of side ×× length of side
    Figure 5
    Figure 5 (MG10C13_051.png)
  6. Area of circle: ππ x radius22
    Figure 6
    Figure 6 (MG10C13_052.png)

Figure 7
Khan Academy video on area and perimeter

Figure 8
Khan Academy video on area of a circle

Exercise 1: Finding areas

Find the area of the following figure:

Figure 9
Figure 9 (area1.png)

Solution
  1. Step 1. Find the height: We first need to find the height, BE, of the parallelogram. We can use Pythagoras to do this:
    BE2 = AB2−AE2 BE2 = 52−32 BE2 = 16 BE = 4BE2 = AB2−AE2 BE2 = 52−32 BE2 = 16 BE = 4
    (1)
  2. Step 2. Apply the formula: We apply the formula for the area of a parallelogram to find the area:
    Area = h × b = 4 × 7 = 28 Area = h × b = 4 × 7 = 28
    (2)

Polygons

  1. For each case below, say whether the statement is true or false. For false statements, give a counter-example to prove it:
    1. All squares are rectangles
    2. All rectangles are squares
    3. All pentagons are similar
    4. All equilateral triangles are similar
    5. All pentagons are congruent
    6. All equilateral triangles are congruent
    Click here for the solution
  2. Find the areas of each of the given figures - remember area is measured in square units (cm22, m22, mm22).
    Figure 10
    Figure 10 (MG10C13_064.png)
    Click here for the solution

Right prisms and cylinders

In this section we study how to calculate the surface areas and volumes of right prisms and cylinders. A right prism is a polygon that has been stretched out into a tube so that the height of the tube is perpendicular to the base (the definition is motivated by the fact that the angle between base and side form a right angle). A square prism has a base that is a square and a triangular prism has a base that is a triangle.

Figure 11: Examples of a right square prism, a right triangular prism and a cylinder.
Figure 11 (MG10C14_001.png)

It is relatively simple to calculate the surface areas and volumes of prisms.

Surface Area

The term surface area refers to the total area of the exposed or outside surfaces of a prism. This is easier to understand if you imagine the prism as a solid object.

If you examine the prisms in Figure 11, you will see that each face of a prism is a simple polygon. For example, the triangular prism has two faces that are triangles and three faces that are rectangles. Therefore, in order to calculate the surface area of a prism you simply have to calculate the area of each face and add it up. In the case of a cylinder the top and bottom faces are circles, while the curved surface flattens into a rectangle.

Surface Area of Prisms

Calculate the area of each face and add the areas together to get the surface area. To do this you need to determine the correct shape of each and every face of the prism and then for each one determine the surface area. The sum of the surface areas of all the faces will give you the total surface area of the prism.

Discussion : surface areas

In pairs, study the following prisms and the adjacent image showing the various surfaces that make up the prism. Explain to your partner, how each relates to the other.

Figure 12
Figure 12 (MG10C14_002.png)

Activity: Surface areas

Find (or take one yourself) a picture of a building that does not have a well defined shape (i.e. is not simply a rectangle). For example a castle with towers, or a house with gable windows or a porch. Assume you have to paint the outside of the building. How much paint would you need? Think about what you have learnt about surface area and the area of polygons. Can you find regular polygons on your picture and use those to find the surface area?

Surface areas

  1. Calculate the surface area in each of the following:
    Figure 13
    Figure 13 (MG10C14_003.png)
    Click here for the solution
  2. If a litre of paint covers an area of 2m22m2, how much paint does a painter need to cover:
    1. A rectangular swimming pool with dimensions 4m×3m×2,5m4m×3m×2,5m, inside walls and floor only.
    2. The inside walls and floor of a circular reservoir with diameter 4m4m and height 2,5m2,5m
    Figure 14
    Figure 14 (MG10C14_004.png)
    Click here for the solution

Volume

The volume of a right prism is calculated by multiplying the area of the base by the height. So, for a square prism of side length aa and height hh the volume is a×a×h=a2ha×a×h=a2h.

Volume of Prisms

Calculate the area of the base and multiply by the height to get the volume of a prism.

Exercise 2

Find the surface area and volume for the a square prism of height 4cm4cm and base length, 3cm3cm.

Figure 15
Figure 15 (squareprism.png)

Solution
  1. Step 1. Find the surface area: We use the formula for the surface area of a prism:
    S.A. = 2 2 L × b + b × h = 2 2 3 × 4 + 3 × 4 = 72cm 2 S.A. = 2 2 L × b + b × h = 2 2 3 × 4 + 3 × 4 = 72cm 2
    (3)
  2. Step 2. Find the volume: To find the volume of the prism, we find the area of the base and multiply it by the height:
    V = l 2 × h = 3 2 × 4 = 36cm 3 V = l 2 × h = 3 2 × 4 = 36cm 3
    (4)

Volume

  1. Write down the formula for each of the following volumes:
    Figure 16
    Figure 16 (MG10C14_005.png)
    Click here for the solution
  2. Calculate the following volumes:
    Figure 17
    Figure 17 (MG10C14_006.png)
    Click here for the solution
  3. A cube is a special prism that has all edges equal. This means that each face is a square. An example of a cube is a die. Show that for a cube with side length aa, the surface area is 6a26a2 and the volume is a3a3.
    Figure 18
    Figure 18 (MG10C14_007.png)
    Click here for the solution

Now, what happens to the surface area if one dimension is multiplied by a constant? For example, how does the surface area change when the height of a rectangular prism is divided by 2?

Figure 19: Rectangular prisms
Figure 19 (MG10C14_008.png)

Table 1
Surface area=2(l×h+l×b+b×h)Surface area=2(l×h+l×b+b×h) Surface area=2(l×12h+l×b+b×12h)Surface area=2(l×12h+l×b+b×12h)
Volume=l×b×hVolume=l×b×h Volume=l×b×12h =12×l×b×hVolume=l×b×12h =12×l×b×h

Exercise 3: Scaling the dimensions of a prism

The size of a prism is specified by the length of its sides. The prism in the diagram has sides of lengths LL, bb and hh.

Figure 20
Figure 20 (MG10C14_009.png)

  1. Consider enlarging all sides of the prism by a constant factor xx, where x>1x>1. Calculate the volume and surface area of the enlarged prism as a function of the factor xx and the volume of the original volume.
  2. In the same way as above now consider the case, where 0<x<10<x<1. Now calculate the reduction factor in the volume and the surface area.
Solution
  1. Step 1. Identify :

    The volume of a prism is given by: V=L×b×hV=L×b×h

    The surface area of the prism is given by: A=2×(L×b+L×h+b×h)A=2×(L×b+L×h+b×h)

  2. Step 2. Rescale :

    If all the sides of the prism get rescaled, the new sides will be:

    L ' = x × L b ' = x × b h ' = x × h L ' = x × L b ' = x × b h ' = x × h
    (5)

    The new volume will then be given by:

    V ' = L ' × b ' × h ' = x × L × x × b × x × h = x 3 × L × b × h = x 3 × V V ' = L ' × b ' × h ' = x × L × x × b × x × h = x 3 × L × b × h = x 3 × V
    (6)

    The new surface area of the prism will be given by:

    A ' = 2 × ( L ' × b ' + L ' × h ' + b ' × h ' ) = 2 × ( x × L × x × b + x × L × x × h + x × b × x × h ) = x 2 × 2 × ( L × b + L × h + b × h ) = x 2 × A A ' = 2 × ( L ' × b ' + L ' × h ' + b ' × h ' ) = 2 × ( x × L × x × b + x × L × x × h + x × b × x × h ) = x 2 × 2 × ( L × b + L × h + b × h ) = x 2 × A
    (7)
  3. Step 3. Interpreting the above results :
    1. We found above that the new volume is given by: V'=x3×VV'=x3×V Since x>1x>1, the volume of the prism will be increased by a factor of x3x3. The surface area of the rescaled prism was given by: A'=x2×AA'=x2×A Again, since x>1x>1, the surface area will be increased by a factor of x2x2. Surface areas which are two dimensional increase with the square of the factor while volumes, which are three dimensional, increase with the cube of the factor.
    2. The answer here is based on the same ideas as above. In analogy, since here 0<x<10<x<1, the volume will be reduced by a factor of x3x3 and the surface area will be decreased by a factor of x2x2

When the length of one of the sides is multiplied by a constant the effect is to multiply the original volume by that constant, as for the example in Figure 19.

Right Pyramids, Right Cones and Spheres

A pyramid is a geometric solid that has a polygon base and the base is joined to a point, called the apex. Two examples of pyramids are shown in the left-most and centre figures in (Reference). The right-most figure has an apex which is joined to a circular base and this type of geometric solid is called a cone. Cones are similar to pyramids except that their bases are circles instead of polygons.

Figure 21: Examples of a square pyramid, a triangular pyramid and a cone.
Figure 21 (MG11C16_001.png)

Surface Area of a Pyramid

Figure 22
Khan academy video on solid geometry volumes

The surface area of a pyramid is calculated by adding the area of each face together.

Exercise 4: Surface Area

If a cone has a height of hh and a base of radius rr, show that the surface area is πr2+πrr2+h2πr2+πrr2+h2.

Solution

  1. Step 1. Draw a picture :

    Figure 23
    Figure 23 (MG11C16_002.png)

  2. Step 2. Identify the faces that make up the cone :

    The cone has two faces: the base and the walls. The base is a circle of radius rr and the walls can be opened out to a sector of a circle.

    Figure 24
    Figure 24 (MG11C16_003.png)

    This curved surface can be cut into many thin triangles with height close to aa (aa is called the slant height). The area of these triangles will add up to 12×12×base××height(of a small triangle) which is 12×2πr×a=πra12×2πr×a=πra

  3. Step 3. Calculate aa :

    aa can be calculated by using the Theorem of Pythagoras. Therefore:

    a = r 2 + h 2 a = r 2 + h 2
    (8)
  4. Step 4. Calculate the area of the circular base :
    A b = π r 2 A b = π r 2
    (9)
  5. Step 5. Calculate the area of the curved walls :
    A w = π r a = π r r 2 + h 2 A w = π r a = π r r 2 + h 2
    (10)
  6. Step 6. Calculate surface area A :
    A = A b + A w = π r 2 + π r r 2 + h 2 A = A b + A w = π r 2 + π r r 2 + h 2
    (11)

Volume of a Pyramid: The volume of a pyramid is found by:

V = 1 3 A · h V = 1 3 A · h
(12)

where AA is the area of the base and hh is the height.

A cone is like a pyramid, so the volume of a cone is given by:

V = 1 3 π r 2 h . V = 1 3 π r 2 h .
(13)

A square pyramid has volume

V = 1 3 a 2 h V = 1 3 a 2 h
(14)

where aa is the side length of the square base.

Exercise 5: Volume of a Pyramid

What is the volume of a square pyramid, 3cm high with a side length of 2cm?

Figure 25
Figure 25 (MG11C16_004.png)

Solution

  1. Step 1. Determine the correct formula :

    The volume of a pyramid is

    V = 1 3 A · h , V = 1 3 A · h ,
    (15)

    where AA is the area of the base and hh is the height of the pyramid. For a square base this means

    V = 1 3 a · a · h V = 1 3 a · a · h
    (16)

    where aa is the length of the side of the square base.

  2. Step 2. Substitute the given values :
    = 1 3 · 2 · 2 · 3 = 1 3 · 12 = 4 c m 3 = 1 3 · 2 · 2 · 3 = 1 3 · 12 = 4 c m 3
    (17)

We accept the following formulae for volume and surface area of a sphere (ball).

Surface area = 4 π r 2 Volume = 4 3 π r 3 Surface area = 4 π r 2 Volume = 4 3 π r 3
(18)

Exercise 6

A triangular pyramid is placed on top of a triangular prism. The prism has an equilateral triangle of side length 20 cm as a base, and has a height of 42 cm. The pyramid has a height of 12 cm.

  1. Find the total volume of the object.
  2. Find the area of each face of the pyramid.
  3. Find the total surface area of the object.
Figure 26
Figure 26 (MG11C16_006.png)

Solution

  1. Step 1. Find the volume of the triangular prism: We use the formula for the volume of a triangular prism:
    V = 12bh2 = 1220422 = 17640 V = 12bh2 = 1220422 = 17640
    (19)
  2. Step 2. Find the volume of the triangular pyramid: We use the formula for the volume of a triangular pyramid:
    V = 16bh2 = 1620422 = 5880 V = 16bh2 = 1620422 = 5880
    (20)
  3. Step 3. Find the volume: We note that we can simply add the volumes of each of the two shapes. So we obtain: 17640+5880=2352017640+5880=23520. This is the answer to part a.
  4. Step 4. Find the area of the faces of the pyramid: We note that we have four triangles that make up the pyramid. So the area of each face is simply:
    Area = 12bh = 122042 = 420 Area = 12bh = 122042 = 420
    (21)
    This is the answer to part b.
  5. Step 5. Find the area of the pyramid: The total area of the pyramid is simply 4×420=16804×420=1680
  6. Step 6. Find the area of the prism: The surface area of the prism is:
    Surface area = b×h+2×H×S+H×b = 20×20+2×12×20+12×20 = 1120 Surface area = b×h+2×H×S+H×b = 20×20+2×12×20+12×20 = 1120
    (22)
  7. Step 7. Find the total surface area: To find the total surface area, we must subtract the area of one face of the pyramid from the area of the prism. We must also subtract the area of one of the triangular faces of the prism. Doing this gives the total surface area as: 1120-420+1680-420=19601120-420+1680-420=1960 This is the answer to part c.

Surface Area and Volume

  1. Calculate the volumes and surface areas of the following solids: *Hint for (e): find the perpendicular height using Pythagoras.
    Figure 27
    Figure 27 (MG11C16_005.png)
    Click here for the solution
  2. Water covers approximately 71% of the Earth's surface. Taking the radius of the Earth to be 6378 km, what is the total area of land (area not covered by water)?
    Click here for the solution

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