Area of a Triangle

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 = \( \scriptsize 15cm\)