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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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If you divide a rectangle into two equal parts by drawing any of its diagonals, each half of the rectangle is a right-angled triangle. See the diagram below. We can then use this to find the area of a triangle.

Screenshot 2022 12 09 at 15.27.46

Area of a right-angled \( \scriptsize \Delta = \normalsize \frac {1}{2} \scriptsize \; \times \; base \; \times \; height \)

Example 1: 

Calculate the area of the triangle with a base of 6cm and a height of 4cm

Solution:

Screenshot 2022 12 09 at 15.35.12

Base (b)  =  6cm, Height (h)  =  4cm

Area of \( \scriptsize \Delta = \normalsize \frac {1}{2} \scriptsize \: \times \: base \: \times \: height \)

⇒ \( \frac {1}{2} \scriptsize \: \times \: 6 \: \times \: 4 \)

= \( \scriptsize 3 \: \times \: 4 \)

= \( \scriptsize 12cm \)

Example 2:

Given that the area of triangle ABC is 120cm2 and its height BX is 16cm. Find the length AC.

Screenshot 2022 12 09 at 15.44.18

Solution

Let the base “AC” be   “b

Let the height  “BX”  be  “h”  =   16cm

Area of \( \scriptsize \Delta ABC = \normalsize \frac {1}{2} \scriptsize \: \times \: base \: \times \: height \)

120   =  \( \frac {1}{2} \scriptsize \: \times \: base \: \times \: 16 \)

120   =  \( \frac {1}{2} \scriptsize \: \times \: b \: \times \: 16 \)

120   =  \( \scriptsize 8b\)

Divide both sides by 8

= \( \frac{120}{8} = \frac{8b}{8} \)

⇒ \( \frac{120}{8} = \frac{\not{8}b}{\not{8}} \)

move the unknown b to the left-hand side

b = \( \frac{120}{8} \)

b = \( \scriptsize 15cm\)

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