# What is the area of a triangle base=2 height =4, but its NOT a right triangle?

Jul 23, 2015

Area of a triangle is $\frac{1}{2} \times b a s e \times h e i g h t$
(there is no requirement that the triangle be a right triangle for this formula)
The given triangle has an area of 4

#### Explanation:

Only continue if you don't understand why
$\textcolor{w h i t e}{\text{XXXX}}$$\text{Area"_"triangle" = 1/2xx"base"xx"height}$

Consider the two triangles below:

For the Acute Angled Triangle ABC
$\triangle A B C$ is composed of $\triangle A D C$ and $\triangle D B C$

$\triangle A D C = \frac{1}{2} \square A D C P$
$\triangle D B C = \frac{1}{2} \square D B R C$

$\triangle A B C$
$\textcolor{w h i t e}{\text{XXXX}}$$= \frac{1}{2} \left(\square A D C P + \square D B R C\right)$

$\textcolor{w h i t e}{\text{XXXX}}$$= \frac{1}{2} \left(\square A B R P\right)$

and since the Area of $\square A B R P = \text{base"xx"height}$

$\textcolor{w h i t e}{\text{XXXX}}$Area of $\triangle A B C = \frac{1}{2} \times \text{base" xx "height}$

For the Obtuse Angled Triangle ABC
following a similar argument:
triangle ADC = 1/2(square ADCP) = 1/2 ("base" + x) xx "height"

$\triangle B D C = \frac{1}{2} \left(\square B D C R\right) = \frac{1}{2} x \times \text{height}$

$\triangle A B C = \triangle A D C - \triangle B D C$

Area of $\triangle A B C$
$\textcolor{w h i t e}{\text{XXXX}}$= [ 1/2("base"+x)xx "height"] - [1/2[x xx "height"]

$\textcolor{w h i t e}{\text{XXXX}}$$= \frac{1}{2} \text{base" xx "height}$