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JSS1: MATHEMATICS - 2ND TERM

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  1. Algebraic Processes | Week 1
    4 Topics
    |
    1 Quiz
  2. Simplification of Algebraic Expressions | Week 2
    4 Topics
    |
    1 Quiz
  3. Simplification of Algebraic Expressions 2 (Use of Brackets) | Week 3
    4 Topics
    |
    1 Quiz
  4. Simple Equations | Week 4
    1 Topic
    |
    1 Quiz
  5. Simple Equations II | Week 5
    3 Topics
    |
    1 Quiz
  6. Plane Shapes I | Week 6
    5 Topics
    |
    2 Quizzes
  7. Plane Shapes II | Week 7
    7 Topics
    |
    1 Quiz
  8. Plane Shapes III | Week 8
    7 Topics
    |
    1 Quiz
  9. Decimal and Percentages I | Week 9
    2 Topics
    |
    1 Quiz
  10. Decimal and Percentages | Week 10
    3 Topics
    |
    1 Quiz



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area of circle

Area of a circle  =  \( \scriptsize \pi r^2 \)

We can also find the area given the diameter

the radius of a circle is half of its diameter

r = \( \frac{d}{2} \)

We can substitute this into Area of a circle   =  \( \scriptsize \pi r^2 \)

Therefore Area = \(\scriptsize \pi \: \times \: \normalsize \frac{d}{2} \scriptsize \: \times \: \normalsize \frac{d}{2} \\ = \frac{\pi d^2}{4} \)

semi circ e1614242421486

To calculate the Area of a Semi-circle, divide the area of a circle by 2 (as two semi-circles make a circle). We get, \( \frac{1}{2} \; \times \; \scriptsize \pi r^2 \)

 

quadrant

To calculate the Area of a Quadrant, divide the area of a circle by 4 (as four quadrants make a circle). We get, \( \frac{ \pi r^2}{4} \)

Example 1:

Find the area of a circle with a radius of 3.5cm.

\( \left ( \scriptsize \pi = \normalsize \frac{22}{7} \right) \)

Solution:

r = \( \scriptsize 3.5 \: or \: \normalsize \frac{35}{10} \)

Area of a circle  =  \( \scriptsize \pi r^2 \)

Area of a circle  =  \( \frac{22}{7} \; \times \; \frac{35}{10} \; \times \; \frac{35}{10} \)

=  \( \frac{22 \; \times \; 5 \; \times \; 35}{10 \; \times \; 10} \)

=  \( \frac{3850}{100} \)

=  \( \scriptsize 38.50 cm^2 \)

or

⇒ \( \scriptsize \pi r^2 = \normalsize \frac{22}{7} \scriptsize\; \times \; 3.5 \; \times \; 3.5 \)

= \( \frac{269.5}{7} \)

=  \( \scriptsize 38.5 cm^2 \)

Example 2:

Find the area of a semicircle with a diameter of 22mm.

\( \left ( \scriptsize \pi = \normalsize \frac{22}{7} \right) \)

Note: Radius = \( \frac{diameter}{2}\)

r = \( \frac{d}{2}\)

From the question, d  =  22 mm

r = \( \frac{d}{2}\)

r = \( \frac{22}{2}\)

r = 11 mm

Area of a semicircle   =   ½  x  area of a circle

Area of a semicircle   =   ½  x  πr2.

Area of a semicircle   =   \( \frac{1}{2} \; \times \; \frac{22}{7} \; \times \; \scriptsize (11)^2 \)

=   \( \frac{1}{2} \; \times \; \frac{22}{7} \; \times \; \scriptsize 11 \; \times \; 11 \)

= \( \frac{11\; \times \; 11 \; \times \; 11}{7} \)

= \( \frac{1331}{7} \)

Area of a semicircle = 190.1 mm2

or

Area = \( \frac{\pi d^2}{4} \)

d = 22 mm

Therefore, Area = \( \frac{\pi \; \times \; 22^2}{4} \)

= \( \frac{ \frac{22}{7} \; \times \; 22 \; \times \; 22}{4} \)

= \( \frac{ 22 \; \times \; 22 \; \times \; 22}{28} \)

Area = \( \frac{ 10648}{28} \)

Area = 380.3

Area of a semicircle   =   ½  x  area of a circle.

Area of a semicircle = \( \frac{1}{2} \; \times \; \scriptsize 380.3 \)

Area of a semicircle = 190.1 mm2

Example 3:

Find the area of a circular shape which has a circumference of 21.89m.

\( \left ( \scriptsize \pi = 3.14 \right)\)

Solution:

The circumference of a circle is given by:

C = πd

d = \( \frac{C}{\pi} \)

C  =   21.89m,

π = 3.14

d = \( \frac{21.89}{3.14} \)

d = 6.97m

So radius   =  \( \frac{d}{2} \)

=  \( \frac{6.97}{2} \)

= 3.49m

Area of a circle  = \(\scriptsize \pi r^2\)

= 3.14 × (3.49)2

= 3.14  ×  3.49  ×  3.49

= 38.2m

The area of the circle is 38.2m

Example 4:

Calculate the area of each of the following shapes:

⇒ \( \left ( \scriptsize \pi = \normalsize \frac{22}{7} \right) \)

CONSTRUCTION133 e1641480951709

(a)

Area = πr2

Area of a semi-circle  = \( \frac{\pi r^2}{2} \)

radius is half of the diameter, from the question diameter, d = 28cm

∴ r = \( \frac{d}{2} \)

r = \( \frac{28}{2} \)

r = \( \scriptsize 14 \: cm \)

Area = πr2

Area = \( \frac{22}{7} \; \times \; \scriptsize 14 \; \times \; \times \; 14 \)

= 22  ×  2  ×  14

= 616cm2

Area of semi-circle  = \( \frac{616cm^2}{2} \)

Area of semi-circle  = \( \scriptsize 308cm^2 \)

(b)

Area of circle = \(\scriptsize \pi r^2\)

Area of quadrant  =  \( \frac{\pi r^2}{4} \)

⇒ \( \scriptsize \pi = \normalsize \frac{22}{7}\scriptsize, \; r = 7 \)

Area = \( \frac{22}{7} \; \times \; \scriptsize 7 \; \times \; 7 \)

= \( \scriptsize 22 \; \times \; 7 \)

= \( \scriptsize 154cm^2 \)

Area of a quadrant   = \( \frac{154}{4} \)

= \( \scriptsize 38.5cm^2 \)

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