JSS1 Third Term Mathematics Lesson Note – Area of Shapes

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WEEK 3

TOPIC: AREA OF SHAPES

CONTENTS:

• Area of circles and triangles
• Area of regular quadrilaterals
• Area of irregular shapes

AREA OF CIRCLES

• Area of a circle = πr²

Where π = 22/7

Diameter = 2r, where r = radius

Example:

1. Find the area of a circle whose radius is 3½m. (Taken ∏to be 22/7)

Solution:

Area of a circle = πr²

= 22/7 × (3½)² = 22/7 × 7/2 × 7/2

=

= 77/2 m² = 38.5m²

2. The area of a circle is 126.5cm². Find its radius correct to 2 decimal places.

Solution

Area of circle = πr²

• 5 cm² = 22/7 r²

253/2 = 22/7 r²

253 X 7 = 22 X 2 X r²

r² = 161/4

r = √161/4

r = 12.689/2

r = 6.345

r = 6.35 (2 d. p.)

AREA OF TRIANGLES

Any diagonal of a rectangle divided it into two equal right- angled triangles forms a right angled triangle.

Thus:

Area of a right – angled triangle

= ½ × product of sides containing the right angle.

Area of triangle = ½ × base × height.

Example:

Calculate the area of the triangle shown below;

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