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## JSS1: MATHEMATICS - 2ND TERM

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#### Quizzes

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:

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

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

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

= $$\scriptsize 12cm$$

### Example 2:

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

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 = $$\scriptsize 15cm$$