Work in pairs or groups and investigate the history of the foundation of geometry. Describe the various stages of development and how the following cultures used geometry to improve their lives.

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 1, 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

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 2

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.

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?

          NOTE: unable to translate the contents of this figure.
          

Exercise 1: 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 8

1. a): 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. a): 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

   (1)

   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

   (2)

   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

   (3)
3. Step 3. Interpreting the above results :
   1. a): 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. b): 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

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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 Code 1.

Discussion : Similar Triangles

Fill in the table using the diagram and then answer the questions that follow.

Table 1

. AB DE . AB DE =...cm...cm=......cm...cm=...	A^A^=...∘∘	D^D^...∘∘
. BC EF . BC EF =...cm...cm=......cm...cm=...	B^B^=...∘∘	E^E^=...∘∘
. AC DF . AC DF =...cm...cm=......cm...cm=...	C^C^...∘∘	F^F^=...∘∘

Figure 9

1. What can you say about the numbers you calculated for: AB DE AB DE , BC EF BC EF , AC DF AC DF ?
2. What can you say about A^A^ and D^D^?
3. What can you say about B^B^ and E^E^?
4. What can you say about C^C^ and F^F^?

If two polygons are similar, one is an enlargement of the other. This means that the two polygons will have the same angles and their sides will be in the same proportion.

We use the symbol ≡≡ to mean is similar to.

Definition 1: Similar Polygons

Two polygons are similar if:

1. their corresponding angles are equal, and
2. the ratios of corresponding sides are equal.

Exercise 2: Similarity of Polygons

Show that the following two polygons are similar.

Figure 10

Solution

1. Step 1. Determine what is required :

   We are required to show that the pair of polygons is similar. We can do this by showing that the ratio of corresponding sides is equal and by showing that corresponding angles are equal.

2. Step 2. Corresponding angles :

   We are given the angles. So, we can show that corresponding angles are equal.

3. Step 3. Show that corresponding angles are equal :

   All angles are given to be 90∘∘ and

   A ^ = E ^ B ^ = F ^ C ^ = G ^ D ^ = H ^ A ^ = E ^ B ^ = F ^ C ^ = G ^ D ^ = H ^

   (4)
4. Step 4. Show that corresponding sides have equal ratios :

   We first need to see which sides correspond. The rectangles have two equal long sides and two equal short sides. We need to compare the ratio of the long side lengths of the two different rectangles as well as the ratio of the short side lenghts.

   Long sides, large rectangle values over small rectangle values:

   Ratio = 2 L L = 2 Ratio = 2 L L = 2

   (5)

   Short sides, large rectangle values over small rectangle values:

   Ratio = L 1 2 L = 1 1 2 = 2 Ratio = L 1 2 L = 1 1 2 = 2

   (6)

   The ratios of the corresponding sides are equal, 2 in this case.

5. Step 5. Final answer :

   Since corresponding angles are equal and the ratios of the corresponding sides are equal the polygons ABCD and EFGH are similar.

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Exercise 3: Similarity of Polygons

If two pentagons ABCDE and GHJKL are similar, determine the lengths of the sides and angles labelled with letters:

Figure 11

Solution

1. Step 1. Determine what is given :

   We are given that ABCDE and GHJKL are similar. This means that:

   AB GH = BC HJ = CD JK = DE KL = EA LG AB GH = BC HJ = CD JK = DE KL = EA LG

   (7)

   and

   A ^ = G ^ B ^ = H ^ C ^ = J ^ D ^ = K ^ E ^ = L ^ A ^ = G ^ B ^ = H ^ C ^ = J ^ D ^ = K ^ E ^ = L ^

   (8)
2. Step 2. Determine what is required :

   We are required to determine the

   1. lengths: aa, bb, cc and dd, and
   2. angles: ee, ff and gg.
3. Step 3. Decide how to approach the problem :

   The corresponding angles are equal, so no calculation is needed. We are given one pair of sides DCDC and KJKJ that correspond. DCKJ=4,53=1,5DCKJ=4,53=1,5 so we know that all sides of KJHGLKJHGL are 1,5 times smaller than ABCDEABCDE.

4. Step 4. Calculate lengths :
   a 2 = 1 , 5 ∴ a = 2 × 1 , 5 = 3 b 1 , 5 = 1 , 5 ∴ b = 1 , 5 × 1 , 5 = 2 , 25 6 c = 1 , 5 ∴ c = 6 ÷ 1 , 5 = 4 d = 3 1 , 5 ∴ d = 2 a 2 = 1 , 5 ∴ a = 2 × 1 , 5 = 3 b 1 , 5 = 1 , 5 ∴ b = 1 , 5 × 1 , 5 = 2 , 25 6 c = 1 , 5 ∴ c = 6 ÷ 1 , 5 = 4 d = 3 1 , 5 ∴ d = 2

   (9)
5. Step 5. Calculate angles :
   e = 92 ∘ ( corresponds to H ) f = 120 ∘ ( corresponds to D ) g = 40 ∘ ( corresponds to E ) e = 92 ∘ ( corresponds to H ) f = 120 ∘ ( corresponds to D ) g = 40 ∘ ( corresponds to E )

   (10)
6. Step 6. Write the final answer :
   a = 3 b = 2 , 25 c = 4 d = 2 e = 92 ∘ f = 120 ∘ g = 40 ∘ a = 3 b = 2 , 25 c = 4 d = 2 e = 92 ∘ f = 120 ∘ g = 40 ∘

   (11)

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Analytical geometry, also called co-ordinate geometry and earlier referred to as Cartesian geometry, is the study of geometry using the principles of algebra, and the Cartesian co-ordinate system. It is concerned with defining geometrical shapes in a numerical way, and extracting numerical information from that representation. Some consider that the introduction of analytic geometry was the beginning of modern mathematics.

One of the simplest things that can be done with analytical geometry is to calculate the distance between two points. Distance is a number that describes how far apart two point are. For example, point PP has co-ordinates (2,1)(2,1) and point QQ has co-ordinates (-2,-2)(-2,-2). How far apart are points PP and QQ? In the figure, this means how long is the dashed line?

Figure 14

In the figure, it can be seen that the length of the line PRPR is 3 units and the length of the line QRQR is four units. However, the ▵PQR▵PQR, has a right angle at RR. Therefore, the length of the side PQPQ can be obtained by using the Theorem of Pythagoras:

The length of PQPQ is the distance between the points PP and QQ.

In order to generalise the idea, assume AA is any point with co-ordinates (x1;y1)(x1;y1) and BB is any other point with co-ordinates (x2;y2)(x2;y2).

Figure 15

The formula for calculating the distance between two points is derived as follows. The distance between the points AA and BB is the length of the line ABAB. According to the Theorem of Pythagoras, the length of ABAB is given by:

However,

Therefore,

Therefore, for any two points, (x1;y1)(x1;y1) and (x2;y2)(x2;y2), the formula is:

Distance=(x1-x2)2+(y1-y2)2(x1-x2)2+(y1-y2)2

Using the formula, distance between the points PP and QQ with co-ordinates (2;1) and (-2;-2) is then found as follows. Let the co-ordinates of point PP be (x1;y1)(x1;y1) and the co-ordinates of point QQ be (x2;y2)(x2;y2). Then the distance is:

The gradient of a line describes how steep the line is. In the figure, line PTPT is the steepest. Line PSPS is less steep than PTPT but is steeper than PRPR, and line PRPR is steeper than PQPQ.

Figure 16

The gradient of a line is defined as the ratio of the vertical distance to the horizontal distance. This can be understood by looking at the line as the hypotenuse of a right-angled triangle. Then the gradient is the ratio of the length of the vertical side of the triangle to the horizontal side of the triangle. Consider a line between a point AA with co-ordinates (x1;y1)(x1;y1) and a point BB with co-ordinates (x2;y2)(x2;y2).

Figure 17

Gradient=y2-y1x2-x1=y2-y1x2-x1

For example the gradient of the line between the points PP and QQ, with co-ordinates (2;1) and (-2;-2) (Figure 14) is:

Sometimes, knowing the co-ordinates of the middle point or midpoint of a line is useful. For example, what is the midpoint of the line between point PP with co-ordinates (2;1)(2;1) and point QQ with co-ordinates (-2;-2)(-2;-2).

The co-ordinates of the midpoint of any line between any two points AA and BB with co-ordinates (x1;y1)(x1;y1) and (x2;y2)(x2;y2), is generally calculated as follows. Let the midpoint of ABAB be at point SS with co-ordinates (X;Y)(X;Y). The aim is to calculate XX and YY in terms of (x1;y1)(x1;y1) and (x2;y2)(x2;y2).

Figure 18

Then the co-ordinates of the midpoint (SS) of the line between point PP with co-ordinates (2;1)(2;1) and point QQ with co-ordinates (-2;-2)(-2;-2) is:

It can be confirmed that the distance from each end point to the midpoint is equal. The co-ordinate of the midpoint SS is (0;-0,5)(0;-0,5).

and

It can be seen that PS=QSPS=QS as expected.

Figure 19

When something is moved in a straight line, we say that it is translated. What happens to the co-ordinates of a point that is translated horizontally or vertically?

Discussion : Translation of a Point Vertically

Complete the table, by filling in the co-ordinates of the points shown in the figure.

Figure 21

Table 2

Point	xx co-ordinate	yy co-ordinate
A
B
C
D
E
F
G

What do you notice about the xx co-ordinates? What do you notice about the yy co-ordinates? What would happen to the co-ordinates of point A, if it was moved to the position of point G?

When a point is moved vertically up or down on the Cartesian plane, the xx co-ordinate of the point remains the same, but the yy co-ordinate changes by the amount that the point was moved up or down.

For example, in Figure 22 Point A is moved 4 units upwards to the position marked by G. The new xx co-ordinate of point A is the same (xx=1), but the new yy co-ordinate is shifted in the positive yy direction 4 units and becomes yy=-2+4=2. The new co-ordinates of point A are therefore G(1;2). Similarly, for point B that is moved downwards by 5 units, the xx co-ordinate is the same (x=-2,5x=-2,5), but the yy co-ordinate is shifted in the negative yy-direction by 5 units. The new yy co-ordinate is therefore yy=2,5 -5=-2,5.

Figure 22: Point A is moved 4 units upwards to the position marked by G. Point B is moved 5 units downwards to the position marked by H.

Discussion : Translation of a Point Horizontally

Complete the table, by filling in the co-ordinates of the points shown in the figure.

Figure 23

Table 3

Point	xx co-ordinate	yy co-ordinate
A
B
C
D
E
F
G

What do you notice about the xx co-ordinates? What do you notice about the yy co-ordinates?

What would happen to the co-ordinates of point A, if it was moved to the position of point G?

When a point is moved horizontally left or right on the Cartesian plane, the yy co-ordinate of the point remains the same, but the xx co-ordinate changes by the amount that the point was moved left or right.

For example, in Figure 24 Point A is moved 4 units right to the position marked by G. The new yy co-ordinate of point A is the same (yy=1), but the new xx co-ordinate is shifted in the positive xx direction 4 units and becomes xx=-2+4=2. The new co-ordinate of point A at G is therefore (2;1). Similarly, for point B that is moved left by 5 units, the yy co-ordinate is the same (y=-2,5y=-2,5), but the xx co-ordinate is shifted in the negative xx-direction by 5 units. The new xx co-ordinate is therefore xx=2,5 -5=-2,5. The new co-ordinates of point B at H is therefore (-2,5;1).

Figure 24: Point A is moved 4 units to the right to the position marked by G. Point B is moved 5 units to the left to the position marked by H.

When you stand in front of a mirror your reflection is located the same distance (dd) behind the mirror as you are standing in front of the mirror.

Figure 25

We can apply the same idea to a point that is reflected on the xx-axis, the yy-axis and the line y=xy=x.

Reflection on the xx-axis

If a point is reflected on the xx-axis, then the reflection must be the same distance below the xx-axis as the point is above the xx-axis and vice-versa, as though it were a mirror image.

Figure 26: Points A and B are reflected on the xx-axis. The original points are shown with •• and the reflected points are shown with ∘∘.

Tip:

When a point is reflected about the xx-axis, only the yy co-ordinate of the point changes.

Exercise 4: Reflection on the xx-axis

Find the co-ordinates of the reflection of the point P, if P is reflected on the xx-axis. The co-ordinates of P are (5;10).

Solution

1. Step 1. Determine what is given and what is required :

   We are given the point P with co-ordinates (5;10) and need to find the co-ordinates of the point if it is reflected on the xx-axis.

2. Step 2. Determine how to approach the problem :

   The point P is above the xx-axis, therefore its reflection will be the same distance below the xx-axis as the point P is above the xx-axis. Therefore, yy=-10.

   For a reflection on the xx-axis, the xx co-ordinate remains unchanged. Therefore, xx=5.

3. Step 3. Write the final answer :

   The co-ordinates of the reflected point are (5;-10).

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Reflection on the yy-axis

If a point is reflected on the yy-axis, then the reflection must be the same distance to the left of the yy-axis as the point is to the right of the yy-axis and vice-versa.

Figure 27: Points A and B are reflected on the yy-axis. The original points are shown with •• and the reflected points are shown with ∘∘.

Tip:

When a point is reflected on the yy-axis, only the xx co-ordinate of the point changes. The yy co-ordinate remains unchanged.

Exercise 5: Reflection on the yy-axis

Find the co-ordinates of the reflection of the point Q, if Q is reflected on the yy-axis. The co-ordinates of Q are (15;5).

Solution

1. Step 1. Determine what is given and what is required :

   We are given the point Q with co-ordinates (15;5) and need to find the co-ordinates of the point if it is reflected on the yy-axis.

2. Step 2. Determine how to approach the problem :

   The point Q is to the right of the yy-axis, therefore its reflection will be the same distance to the left of the yy-axis as the point Q is to the right of the yy-axis. Therefore, xx=-15.

   For a reflection on the yy-axis, the yy co-ordinate remains unchanged. Therefore, yy=5.

3. Step 3. Write the final answer :

   The co-ordinates of the reflected point are (-15;5).

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Reflection on the line y=xy=x

The final type of reflection you will learn about is the reflection of a point on the line y=xy=x.

Casestudy : Reflection of a point on the line y=xy=x

Figure 28

Study the information given and complete the following table:

Table 4

	Point	Reflection
A	(2;1)	(1;2)
B	(-112112;-2)	(-2;-11212)
C	(-1;1)
D	(2;-3)

What can you deduce about the co-ordinates of points that are reflected about the line y=xy=x?

The xx and yy co-ordinates of points that are reflected on the line y=xy=x are swapped around, or interchanged. This means that the xx co-ordinate of the original point becomes the yy co-ordinate of the reflected point and the yy co-ordinate of the original point becomes the xx co-ordinate of the reflected point.

Figure 29: Points A and B are reflected on the line y=xy=x. The original points are shown with •• and the reflected points are shown with ∘∘.

Tip:

The xx and yy co-ordinates of points that are reflected on the line y=xy=x are interchanged.

Exercise 6: Reflection on the line y=xy=x

Find the co-ordinates of the reflection of the point R, if R is reflected on the line y=xy=x. The co-ordinates of R are (-5;5).

Solution

1. Step 1. Determine what is given and what is required :

   We are given the point R with co-ordinates (-5;5) and need to find the co-ordinates of the point if it is reflected on the line y=xy=x.

2. Step 2. Determine how to approach the problem :

   The xx co-ordinate of the reflected point is the yy co-ordinate of the original point. Therefore, xx=5.

   The yy co-ordinate of the reflected point is the xx co-ordinate of the original point. Therefore, yy=-5.

3. Step 3. Write the final answer :

   The co-ordinates of the reflected point are (5;-5).

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Rules of Translation

A quick way to write a translation is to use a 'rule of translation'. For example (x;y)→(x+a;y+b)(x;y)→(x+a;y+b) means translate point (x;y) by moving a units horizontally and b units vertically.

So if we translate (1;2) by the rule (x;y)→(x+3;y-1)(x;y)→(x+3;y-1) it becomes (4;1). We have moved 3 units right and 1 unit down.

Translating a Region

To translate a region, we translate each point in the region.

Example

Region A has been translated to region B by the rule: (x;y)→(x+4;y+2)(x;y)→(x+4;y+2)

Figure 30

Discussion : Rules of Transformations

Work with a friend and decide which item from column 1 matches each description in column 2.

Table 5

Column 1	Column 2
.(x;y)→(x;y-3).(x;y)→(x;y-3)	a reflection on x-y line
. ( x ; y ) → ( x - 3 ; y ) . ( x ; y ) → ( x - 3 ; y )	a reflection on the x axis
. ( x ; y ) → ( x ; - y ) . ( x ; y ) → ( x ; - y )	a shift of 3 units left
. ( x ; y ) → ( - x ; y ) . ( x ; y ) → ( - x ; y )	a shift of 3 units down
. ( x ; y ) → ( y ; x ) . ( x ; y ) → ( y ; x )	a reflection on the y-axis

Transformations

1. Find the co-ordinates of each of the points ( S - Z) if they are reflected about the given lines:
   1. a): y-axis (x=0)
   2. b): x-axis (y=0)
   3. c): y=-x
   4. d): y=x

   Figure 31

2. Write down the rule used for each of the following reflections:
   1. a): Z(7;3), Z'(3;7)
   2. b): Y(-1;-8), Y'(1;-8)
   3. c): X(5;9), X'(-5;9)
   4. d): W(4;6), W'(4;6)
   5. e): V(-37-37;5353), V'(5353;-37-37)
3. 1. a): Reflect the given points using the rules that are given.
   2. b): Identify the line of reflection in each case (some may not exist):
      1. (i): H(-4;3); (x;y)→→ (-x;y)
      2. (ii): H(-4;3); (x;y) →→ (-y;-x)
      3. (iii): H(-4;3); (x;y) →→ (y;x)
      4. (iv): H(-4;3); (x;y) →→ (-x;-y)
      5. (v): H(-4;3); (x;y) →→ (x;-y)

4. Figure 32

   Using squared paper, copy the diagram given. Let -10≤≤x≤≤10, -10≤≤y≤≤10.
   1. i): Identify the transformation.
   2. ii): Draw the image of the figure according the rules given.
   1. a): (x;y)→(-x;y)(x;y)→(-x;y)
   2. b): (x;y)→(y;x)(x;y)→(y;x)
   3. c): (x;y)→(x;y-3)(x;y)→(x;y-3)
   4. d): (x;y)→(x+5;y)(x;y)→(x+5;y)
   5. e): (x;y)→(x;-y)(x;y)→(x;-y)

Investigation : Calculation of Volume, Surface Area and scale factors of objects

1. Look around the house or school and find a can or a tin of any kind (e.g. beans, soup, cooldrink, paint etc.)
2. Measure the height of the tin and the diameter of its top or bottom.
3. Write down the values you measured on the diagram below:

   Figure 33

4. Using your measurements, calculate the following (in cm22, rounded off to 2 decimal places):
   1. the area of the side of the tin (i.e. the rectangle)
   2. the area of the top and bottom of the tin (i.e. the circles)
   3. the total surface area of the tin
5. If the tin metal costs 0,17 cents/cm22, how much does it cost to make the tin?
6. Find the volume of your tin (in cm33, rounded off to 2 decimal places).
7. What is the volume of the tin given on its label?
8. Compare the volume you calculated with the value given on the label. How much air is contained in the tin when it contains the product (i.e. cooldrink, soup etc.)
9. Why do you think space is left for air in the tin?
10. If you wanted to double the volume of the tin, but keep the radius the same, by how much would you need to increase the height?
11. If the height of the tin is kept the same, but now the radius is doubled, by what scale factor will the:
    1. area of the side surface of the tin increase?
    2. area of the bottom/top of the tin increase?