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    This module is included inLens: Siyavula: Mathematics (Gr. 7-9)
    By: SiyavulaAs a part of collection: "Mathematics Grade 8"

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The characteristics of a circle

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MATHEMATICS

Grade 8

RATIONAL NUMBERS, CIRCLES AND TRIANGLES

Module 13

THE CHARACTERISTICS OF A CIRCLE

ACTIVITY 1

Discovering the characteristics of a circle

[LO 3.1, 4.2.1, 3.4]

  1. Try to copy the following design, using a pair of compasses only:
Figure 1
Figure 1 (Picture 25.png)

2. Draw a circle of any size. Refer to a textbook or any other source of information to help you indicate the following on the circle:

2.1 Centre: T

2.2 Diameter (Name it PQ.)

2.3 Radius: TS

2.4 Any arc: FG

2.5 Sector: PTW (shade this portion.)

2.6 Chord: KL

2.7 Use a coloured pencil to indicate where you would determine the circumference of the circle.

3. Refer to your sketch to answer the following questions:

3.1 What is characteristic of TW, PT, TS and TQ?

3.2 Measure PTˆWPTˆW size 12{P { hat {T}}W} {}.

3.3 What is the size of PTˆQPTˆQ size 12{P { hat {T}}Q} {}?

3.4 What do we call this type of angle?

4. Construct the following with the help of a pair of compasses:

4.1 a circle with a diameter measuring 4 cm

4.2 a circle with a radius of 1,5 cm

5. How would you go about constructing a circle of 4 m?

  • Plan:

ACTIVITY 2

Discovering the circumference of a circle and dealing with related problems

[LO 4.2.2, 4.3.1, 4.3.2, 4.3.3, 4.4, 4.5.1]

1. Make use of about four bottles / cups of different sizes. Use a length of string and measure the diameter of each of the bottles to complete the following table:

Table 1
  circumference (O) diameter (m/d) O ÷ m/d
Bottle 1      
Bottle 2      
Bottle 3      
Bottle 4      
  • What is noticeable in the last column?

circumference ÷ diameter

1.2 What is the term used for the answer in the last column?

1.3 Name two values that can be used for π: ...................... or ......................

1.4 Which formula can therefore be used to calculate the circumference of any circle?

2. We could also deduce this formula from a circle by proceeding as follows:

2.1 Draw a circle with centre P and radius 25 mm on a sheet of paper.

2.2 Cut out the circle and place a mark anywhere on the edge of the cut circle.

2.3 Draw a line (use a ruler) across the remaining area of the sheet of paper. Roll the circle (cut out disk) on its edge along this line (place the mark on the edge of the circle at the beginning of the ruled line. Mark the spot where the rotation is completed on the line when the rolled circle has completed a full rotation.

2.4 Use your ruler to measure the marked distance.

  • Distance: ......................... mm

2.5 What term would we use to describe the distance that was measured in 2.4?

2.6 Use your calculator to calculate the following:

  • circumference ÷ diameter = ..................... ÷ ..................... = ........................

2.7 What term do we use to describe the answer that you have obtained?

3. What do we actually mean when we say that the wheel of a bicycle has completed a full rotation?

4. Write the formula for calculating the circumference of a circle on the following line and answer the questions that follow:

  • Circumference = ..................................................

4.1 How would you calculate the radius of a circle when the circumference is provided?

  • Radius (R) = ..................................................

4.2 How would you calculate the diameter of a circle when the circumference is provided?

  • Diameter (d) = ..................................................

Now you should be able to answer any question dealing with the diameter, radius or circumference of a circle or wheel or any circular object.

5. Use your pocket calculator to calculate the circumference of each of the following circles:

Note this: Always write out the formula before you start.(π = 3,14).

5.1 r = 230 mm

5.2 r = 1,45 cm (answer to 2 decimal figures)

6. Determine the circumference of each of the following without the use of a pocket calculator.

Note this: Always write out the formula before you start.(π = 227227 size 12{ { {"22"} over {7} } } {})

6.1 r = 14 cm

6.2 d = 35 cm

  1. Calculate the radius of the following circle:

You may use your pocket calculator, but you have to show all the steps of the calculation. (π= 227227 size 12{ { {"22"} over {7} } } {})

7.1 circumference 242 mm

8. How many rotations will the wheel of a mountain bike complete over a distance of 7,5 m if the diameter of the wheel is 67 cm?

ACTIVITY 3

Discovering the area of a circle and solving related problems

[LO 4.2.1, 4.5.1, 4.3]

1. Can you remember the formula for calculating the area of a rectangle?

2. Draw a circle with centre O and a radius of 60 mm on a sheet of paper. Divide the circle into 32 equal sectors. Use red for colouring 16 sectors and blue for the remaining 16 sectors.

3. Cut out all 32 sectors and arrange them in line in such a way that the segments eventually form a rectangular paving design.

Paste your triangles in the following space

4. Measure both the length and breadth of the rectangle. Use the formula from no. 1 to calculate the area of the rectangle.

5. What do you deduce with regard to the rectangle and the circle that you have drawn in no. 2?

6. Which unit of measurement is used for calculating area?

7. Provide the formula for calculating the area of any circle.

8. Calculate the area of the circle you have drawn in no. 2 with the help of the formula from no. 7.

What do you notice?

9. Calculate the area of each of the following circles without making use of a pocket calculator.

  • (π = 227227 size 12{ { {"22"} over {7} } } {})

9.1 r = 14,7 cm 9.2 d = 56,49 cm

10. Calculate the area of the shaded parts.

  • You may use your pocket calculator for this. (π = 3,14)
Figure 2
Figure 2 (Picture 27.png)

Assessment

Table 2
LO4
MeasurementThe learner will be able to use appropriate measuring units, instruments and formulae in a variety of contexts.
We know this when the learner:
4.2 solves problems involving:
4.2.1 length;
4.2.2 perimeter and area of polygonals and circles;
4.3 solves problems using a range of strategies including:
4.3.1 estimating;
4.3.2 calculating to at least two decimal positions;
4.3.3 using and converting between appropriate SI units;
4.4 describes the meaning of and uses ππ size 12{π} {} in calculations involving circles and discusses its historical development in measurement;
4.5 calculates, by selecting and using appropriate formulae:
4.5.1 perimeter of polygons and circles;
4.5.2 area of triangles, rectangles circles and polygons by decomposition into triangles and rectangles;
  • investigates (alone and / or as a member of a group or team) the relationship between the sides of a right-angled triangle to develop the Theorem of Pythagoras;
4.9 uses the Theorem of Pythagoras to calculate a missing length in a right-angled triangle leaving irrational answers in surd form (√);
4.10 describes and illustrates ways of measuring in different cultures throughout history (e.g. determining right angles using knotted string leading to the Theorem of Pythagoras).

Memorandum

ACTIVITY 2

5.1 O = ππ size 12{π} {} x d

O = ππ size 12{π} {} x 460

O = 1 444,4 mm

5.2 C = ππ size 12{π} {} x d

C = ππ size 12{π} {} x 2,9

C size 12{ approx } {} 9,11 cm

6.1 C = ππ size 12{π} {} x d

C = 22712271 size 12{ { {"22"} over { { {7}} rSub { size 8{1} } } } } {} x 28412841 size 12{ { { { {2}} { {8}} rSup { size 8{4} } } over {1} } } {}

C = 88 cm

6.2 C = ππ size 12{π} {} x d

C = 22712271 size 12{ { {"22"} over { { {7}} rSub { size 8{1} } } } } {} x 35513551 size 12{ { { { {3}} { {5}} rSup { size 8{5} } } over {1} } } {}

C = 110 cm

7.1 C = ππ size 12{π} {} x d

242 = 227227 size 12{ { {"22"} over {7} } } {} x d

24212421 size 12{ { {"242"} over {1} } } {} x 227227 size 12{ { {"22"} over {7} } } {} = d

∴∴ size 12{∴} {}d = 77 mm

8. C = ππ size 12{π} {} x d 750 ÷ 210,38 cm

= 3,14 x 67 cm = 3,6 revolutions

= 210,38 cm

ACTIVITY 3

9. A = ππ size 12{π} {} x r2

= 227227 size 12{ { {"22"} over {7} } } {} x 14,7114,71 size 12{ { {"14",7} over {1} } } {} x 14,7114,71 size 12{ { {"14",7} over {1} } } {}

= 679,14 cm2

  • r = 28,25

A = 2 505,92 cm2

10. A B

(3,14 x 152) – (3,14 x 152) (14,5)2 – (3,14 x 7,252 x 1212 size 12{ { {1} over {2} } } {})

= 706,5 – 78,5 = 210,25 – 82,52

= 628 cm2 = 127,73 cm2

11. (40 x 40) – (3,14 x 152)

= 1 600 – 706,5

= 893,5 cm2

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