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## JSS1: MATHEMATICS - 2ND TERM

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Lesson 8, Topic 5
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# Area of a Triangle

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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.

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

### Example 1:Â

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

Solution:

Base (b)  =  6cm, Height (h)  =  4cm

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

â‡’ $$\frac {1}{2} \scriptsize \: \times \: 6 \: \times \: 4$$

= $$\scriptsize 3 \: \times \: 4$$

= $$\scriptsize 12cm$$

### Example 2:

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

Solution

Let the base “AC” be Â  “b

Let the heightÂ  “BX” Â be Â “h”Â  = Â  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 Â  =Â  $$\frac {1}{2} \scriptsize \: \times \: b \: \times \: 16$$

120   =  $$\scriptsize 8b$$

Divide both sides by 8

= $$\frac{120}{8} = \frac{8b}{8}$$

â‡’ $$\frac{120}{8} = \frac{\not{8}b}{\not{8}}$$

move the unknown b to the left-hand side

b = $$\frac{120}{8}$$

b = $$\scriptsize 15cm$$