Area Of A Triangle

Theorem: Area of a triangle is equal to one half of the product of the length of any side and an altitude drawn over it by the vertex opposite to it.

Prerequisites:
Area of a right triangle (proof)

Proof:

In the given triangle ABC, draw an altitude from A to join BC at D. As a result, $\triangle ABC$ is divided into two right triangles, i.e., $\triangle ABD$ and $\triangle ACD$.

Since, these are right triangles, hence there area is given by:

$\qquad\quad\text{Area of $\triangle ABD$} = \dfrac{1}{2} (AD).(BD)\\ \text{And}\quad\text{Area of $\triangle ACD$} = \dfrac{1}{2}  (AD).(CD)$

$\begin{align}\therefore\;\;\quad\text{Area of $\triangle ABC$} &= \text{Area of $\triangle ABD$} + \text{Area of $\triangle ACD$}\\ &= \frac{1}{2}  (AD).(BD) + \frac{1}{2}  (AD).(CD)\\ &= \frac{1}{2}  (AD).(BD + CD)\\ &= \frac{1}{2}  (AD).(BC)\end{align}$

Hence the result.


Recommended:
Area of rectangle
Area of rhombus
Isosceles triangle theorem