Skip to content Skip to navigation

OpenStax-CNX

You are here: Home » Content » Geometry - Grade 10

Navigation

Lenses

What is a lens?

Definition of a lens

Lenses

A lens is a custom view of the content in the repository. You can think of it as a fancy kind of list that will let you see content through the eyes of organizations and people you trust.

What is in a lens?

Lens makers point to materials (modules and collections), creating a guide that includes their own comments and descriptive tags about the content.

Who can create a lens?

Any individual member, a community, or a respected organization.

What are tags? tag icon

Tags are descriptors added by lens makers to help label content, attaching a vocabulary that is meaningful in the context of the lens.

This content is ...

Affiliated with (What does "Affiliated with" mean?)

This content is either by members of the organizations listed or about topics related to the organizations listed. Click each link to see a list of all content affiliated with the organization.
  • FETMaths display tagshide tags

    This module is included inLens: Siyavula: Mathematics (Gr. 10-12)
    By: Siyavula

    Review Status: In Review

    Click the "FETMaths" link to see all content affiliated with them.

    Click the tag icon tag icon to display tags associated with this content.

Recently Viewed

This feature requires Javascript to be enabled.

Tags

(What is a tag?)

These tags come from the endorsement, affiliation, and other lenses that include this content.
 

Introduction

Geometry (Greek: geo = earth, metria = measure) arose as the field of knowledge dealing with spatial relationships. It was one of the two fields of pre-modern mathematics, the other being the study of numbers. In modern times, geometric concepts have become very complex and abstract and are barely recognizable as the descendants of early geometry.

Research Project : History of Geometry

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. This list should serve as a guideline and provide the minimum requirement, there are many other people who contributed to the foundation of geometry.

  1. Ancient Indian geometry (c. 3000 - 500 B.C.)
    1. Harappan geometry
    2. Vedic geometry
  2. Classical Greek geometry (c. 600 - 300 B.C.)
    1. Thales and Pythagoras
    2. Plato
  3. Hellenistic geometry (c. 300 B.C - 500 C.E.)
    1. Euclid
    2. Archimedes

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. A square prism has a base that is a square and a triangular prism has a base that is a triangle.

Figure 1: Examples of a right square prism, a right triangular prism and a cylinder.
Figure 1 (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 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

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 2
Figure 2 (MG10C14_002.png)

Surface areas

  1. Calculate the surface area in each of the following:
    Figure 3
    Figure 3 (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 4
    Figure 4 (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.

Volume

  1. Write down the formula for each of the following volumes:
    Figure 5
    Figure 5 (MG10C14_005.png)
    Click here for the solution
  2. Calculate the following volumes:
    Figure 6
    Figure 6 (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 7
    Figure 7 (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 8: Rectangular prisms
Figure 8 (MG10C14_008.png)

Figure 9: Rectangular prisms 2
Figure 9 (CG10C14_1.png)

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 10
Figure 10 (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
    (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. 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 8.

Polygons

Polygons are all around us. A stop sign is in the shape of an octagon, an eight-sided polygon. The honeycomb of a beehive consists of hexagonal cells. The top of a desk is a rectangle.

In this section, you will learn about similar polygons.

Similarity of Polygons

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 11
Figure 11 (MG10C14_010.png)

  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 12
Figure 12 (MG10C14_011.png)

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.

Tip:

All squares are similar.

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 13
Figure 13 (MG10C14_012.png)

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)

Similarity of Equilateral Triangles

Working in pairs, show that all equilateral triangles are similar.

Polygons-mixed

  1. Find the values of the unknowns in each case. Give reasons.
    Figure 14
    Figure 14 (mg10c14_2.png)
    Click here for the solution
  2. Find the angles and lengths marked with letters in the following figures:
    Figure 15
    Figure 15 (MG10C14_014.png)
    Click here for the solution

Analytical Geometry

Introduction

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.

Distance between Two Points

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 16
Figure 16 (MG10C14_015.png)

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 PQRPQR, has a right angle at RR. Therefore, the length of the side PQPQ can be obtained by using the Theorem of Pythagoras:

P Q 2 = P R 2 + Q R 2 P Q 2 = 3 2 + 4 2 P Q = 3 2 + 4 2 = 5 P Q 2 = P R 2 + Q R 2 P Q 2 = 3 2 + 4 2 P Q = 3 2 + 4 2 = 5
(12)

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 17
Figure 17 (MG10C14_016.png)

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:

A B = A C 2 + B C 2 A B = A C 2 + B C 2
(13)

However,

B C = y 2 - y 1 A C = x 2 - x 1 B C = y 2 - y 1 A C = x 2 - x 1
(14)

Therefore,

A B = A C 2 + B C 2 = ( x 1 - x 2 ) 2 + ( y 1 - y 2 ) 2 A B = A C 2 + B C 2 = ( x 1 - x 2 ) 2 + ( y 1 - y 2 ) 2
(15)

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:

Distance = ( x 1 - x 2 ) 2 + ( y 1 - y 2 ) 2 = ( 2 - ( - 2 ) ) 2 + ( 1 - ( - 2 ) ) 2 = ( 2 + 2 ) 2 + ( 1 + 2 ) 2 = 16 + 9 = 25 = 5 Distance = ( x 1 - x 2 ) 2 + ( y 1 - y 2 ) 2 = ( 2 - ( - 2 ) ) 2 + ( 1 - ( - 2 ) ) 2 = ( 2 + 2 ) 2 + ( 1 + 2 ) 2 = 16 + 9 = 25 = 5
(16)

The following video provides a summary of the distance formula.

Figure 18
Khan academy video on distance formula

Calculation of the Gradient of a Line

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 19
Figure 19 (MG10C14_017.png)

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 20
Figure 20 (MG10C14_018.png)

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

We can use the gradient of a line to determine if two lines are parallel or perpendicular. If the lines are parallel (Figure 21a) then they will have the same gradient, i.e. mAB == mCD. If the lines are perpendicular (Figure 21b) than we have: -1mAB=mCD-1mAB=mCD

Figure 21
Figure 21 (geom.png)

For example the gradient of the line between the points PP and QQ, with co-ordinates (2;1) and (-2;-2) (Figure 16) is:

Gradient = y 2 - y 1 x 2 - x 1 = - 2 - 1 - 2 - 2 = - 3 - 4 = 3 4 Gradient = y 2 - y 1 x 2 - x 1 = - 2 - 1 - 2 - 2 = - 3 - 4 = 3 4
(17)

The following video provides a summary of the gradient of a line.

Figure 22
Gradient of a line

Midpoint of a Line

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 23
Figure 23 (MG10C14_019.png)
X = x 1 + x 2 2 Y = y 1 + y 2 2 S x 1 + x 2 2 ; y 1 + y 2 2 X = x 1 + x 2 2 Y = y 1 + y 2 2 S x 1 + x 2 2 ; y 1 + y 2 2
(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:

X = x 1 + x 2 2 = - 2 + 2 2 = 0 Y = y 1 + y 2 2 = - 2 + 1 2 = - 1 2 S is at ( 0 ; - 1 2 ) X = x 1 + x 2 2 = - 2 + 2 2 = 0 Y = y 1 + y 2 2 = - 2 + 1 2 = - 1 2 S is at ( 0 ; - 1 2 )
(19)

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).

P S = ( x 1 - x 2 ) 2 + ( y 1 - y 2 ) 2 = ( 0 - 2 ) 2 + ( - 0 . 5 - 1 ) 2 = ( - 2 ) 2 + ( - 1 . 5 ) 2 = 4 + 2 . 25 = 6 . 25 P S = ( x 1 - x 2 ) 2 + ( y 1 - y 2 ) 2 = ( 0 - 2 ) 2 + ( - 0 . 5 - 1 ) 2 = ( - 2 ) 2 + ( - 1 . 5 ) 2 = 4 + 2 . 25 = 6 . 25
(20)

and

Q S = ( x 1 - x 2 ) 2 + ( y 1 - y 2 ) 2 = ( 0 - ( - 2 ) ) 2 + ( - 0 . 5 - ( - 2 ) ) 2 = ( 0 + 2 ) ) 2 + ( - 0 . 5 + 2 ) ) 2 = ( 2 ) ) 2 + ( - 1 . 5 ) ) 2 = 4 + 2 . 25 = 6 . 25 Q S = ( x 1 - x 2 ) 2 + ( y 1 - y 2 ) 2 = ( 0 - ( - 2 ) ) 2 + ( - 0 . 5 - ( - 2 ) ) 2 = ( 0 + 2 ) ) 2 + ( - 0 . 5 + 2 ) ) 2 = ( 2 ) ) 2 + ( - 1 . 5 ) ) 2 = 4 + 2 . 25 = 6 . 25
(21)

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

Figure 24
Figure 24 (MG10C14_020.png)

The following video provides a summary of the midpoint of a line.

Figure 25
Khan academy video on midpoint of a line

Co-ordinate Geometry

  1. In the diagram given the vertices of a quadrilateral are F(2;0), G(1;5), H(3;7) and I(7;2).
    Figure 26
    Figure 26 (MG10C14_021.png)
    1. What are the lengths of the opposite sides of FGHI?
    2. Are the opposite sides of FGHI parallel?
    3. Do the diagonals of FGHI bisect each other?
    4. Can you state what type of quadrilateral FGHI is? Give reasons for your answer.
    Click here for the solution
  2. A quadrialteral ABCD with vertices A(3;2), B(1;7), C(4;5) and D(1;3) is given.
    1. Draw the quadrilateral.
    2. Find the lengths of the sides of the quadrilateral.
    Click here for the solution
  3. ABCD is a quadrilateral with verticies A(0;3), B(4;3), C(5;-1) and D(-1;-1).
    1. Show that:
      1. AD = BC
      2. AB DC
    2. What name would you give to ABCD?
    3. Show that the diagonals AC and BD do not bisect each other.
    Click here for the solution
  4. P, Q, R and S are the points (-2;0), (2;3), (5;3), (-3;-3) respectively.
    1. Show that:
      1. SR = 2PQ
      2. SR PQ
    2. Calculate:
      1. PS
      2. QR
    3. What kind of a quadrilateral is PQRS? Give reasons for your answers.
  5. EFGH is a parallelogram with verticies E(-1;2), F(-2;-1) and G(2;0). Find the co-ordinates of H by using the fact that the diagonals of a parallelogram bisect each other.
    Click here for the solution

Transformations

In this section you will learn about how the co-ordinates of a point change when the point is moved horizontally and vertically on the Cartesian plane. You will also learn about what happens to the co-ordinates of a point when it is reflected on the xx-axis, yy-axis and the line y=xy=x.

Translation of a Point

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 27
Figure 27 (MG10C14_022.png)

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 28 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 28: 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.
Figure 28 (MG10C14_023.png)

Tip:

If a point is shifted upwards, the new yy co-ordinate is given by adding the shift to the old yy co-ordinate. If a point is shifted downwards, the new yy co-ordinate is given by subtracting the shift from the old yy co-ordinate.

Discussion : Translation of a Point Horizontally

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

Figure 29
Figure 29 (MG10C14_024.png)

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 30 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 30: 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.
Figure 30 (MG10C14_025.png)

Tip:

If a point is shifted to the right, the new xx co-ordinate is given by adding the shift to the old xx co-ordinate. If a point is shifted to the left, the new xx co-ordinate is given by subtracting the shift from the old xx co-ordinate.

Reflection of a Point

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 31
Figure 31 (MG10C14_026.png)

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 32: Points A and B are reflected on the xx-axis. The original points are shown with and the reflected points are shown with .
Figure 32 (MG10C14_027.png)
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).

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 33: Points A and B are reflected on the yy-axis. The original points are shown with and the reflected points are shown with .
Figure 33 (MG10C14_028.png)
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).

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 34
Figure 34 (MG10C14_029.png)

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 35: 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 .
Figure 35 (MG10C14_030.png)
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).

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 36
Figure 36 (MG10C14_031.png)

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. Describe the translations in each of the following using the rule (x;y) (...;...)
    Figure 37
    Figure 37 (mg10c14_4.png)
    1. From A to B
    2. From C to J
    3. From F to H
    4. From I to J
    5. From K to L
    6. From J to E
    7. From G to H
    Click here for the solution
  2. A is the point (4;1). Plot each of the following points under the given transformations. Give the co-ordinates of the points you have plotted.
    1. B is the reflection of A in the x-axis.
    2. C is the reflection of A in the y-axis.
    3. D is the reflection of B in the line x=0.
    4. E is the reflection of C is the line y=0.
    5. F is the reflection of A in the line y= x
    Click here for the solution
  3. In the diagram, B, C and D are images of polygon A. In each case, the transformation that has been applied to obtain the image involves a reflection and a translation of A. Write down the letter of each image and describe the transformation applied to A in order to obtain the image.
    Figure 38
    Figure 38 (mg10c14_5.png)
    Click here for the solution
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 39
    Figure 39 (MG10C14_034.png)
  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?

End of Chapter Exercises

  1. Using the rules given, identify the type of transformation and draw the image of the shapes.
    1. (x;y)(x+3;y-3)
      Figure 40
      Figure 40 (MG10C14_037.png)
    2. (x;y)(x-4;y)
      Figure 41
      Figure 41 (MG10C14_038.png)
    3. (x;y)(y;x)
      Figure 42
      Figure 42 (MG10C14_039.png)
    4. (x;y)(-x;-y)
      Figure 43
      Figure 43 (MG10C14_040.png)
    Click here for the solution
  2. PQRS is a quadrilateral with points P(0; −3) ; Q(−2;5) ; R(3;2) and S(3;–2) in the Cartesian plane.
    1. Find the length of QR.
    2. Find the gradient of PS.
    3. Find the midpoint of PR.
    4. Is PQRS a parallelogram? Give reasons for your answer.
    Click here for the solution
  3. A(–2;3) and B(2;6) are points in the Cartesian plane. C(a;b) is the midpoint of AB. Find the values of a and b.
    Click here for the solution
  4. Consider: Triangle ABC with vertices A (1; 3) B (4; 1) and C (6; 4):
    1. Sketch triangle ABC on the Cartesian plane.
    2. Show that ABC is an isoceles triangle.
    3. Determine the co-ordinates of M, the midpoint of AC.
    4. Determine the gradient of AB.
    5. Show that the following points are collinear: A, B and D(7;-1)
    Click here for the solution
  5. In the diagram, A is the point (-6;1) and B is the point (0;3)
    Figure 44
    Figure 44 (MG10C14_5.png)
    1. Find the equation of line AB
    2. Calculate the length of AB
    3. A’ is the image of A and B’ is the image of B. Both these images are obtain by applying the transformation: (x;y)(x-4;y-1). Give the coordinates of both A’ and B’
    4. Find the equation of A’B’
    5. Calculate the length of A’B’
    6. Can you state with certainty that AA'B'B is a parallelogram? Justify your answer.
    Click here for the solution
  6. The vertices of triangle PQR have co-ordinates as shown in the diagram.
    Figure 45
    Figure 45 (mg10c14_6.png)
    1. Give the co-ordinates of P', Q' and R', the images of P, Q and R when P, Q and R are reflected in the line y=x.
    2. Determine the area of triangle PQR.
    Click here for the solution

Content actions

Download module as:

Add module to:

My Favorites (?)

'My Favorites' is a special kind of lens which you can use to bookmark modules and collections. 'My Favorites' can only be seen by you, and collections saved in 'My Favorites' can remember the last module you were on. You need an account to use 'My Favorites'.

| A lens I own (?)

Definition of a lens

Lenses

A lens is a custom view of the content in the repository. You can think of it as a fancy kind of list that will let you see content through the eyes of organizations and people you trust.

What is in a lens?

Lens makers point to materials (modules and collections), creating a guide that includes their own comments and descriptive tags about the content.

Who can create a lens?

Any individual member, a community, or a respected organization.

What are tags? tag icon

Tags are descriptors added by lens makers to help label content, attaching a vocabulary that is meaningful in the context of the lens.

| External bookmarks