# What is the area of an isosceles triangle with a base of 6 and sides of 4? units are in meters?

Jun 1, 2015

Area of a triangle is $\frac{1}{2} \times b a s e \times h e i g h t$

If we divide the isosceles triangle down the middle into two triangles, then they will be right angled triangles with sides $3$, $4$, and the height of the isosceles triangle.

Unfortunately, this isn't a $3$, $4$, $5$, triangle as it's the hypotenuse that is of length $4$. We can still use Pythagoras to calculate:

height $= \sqrt{{4}^{2} - {3}^{2}} = \sqrt{7}$

So the area of the triangle is

$\frac{1}{2} \times b a s e \times h e i g h t = \frac{1}{2} \cdot 6 \cdot \sqrt{7} = 3 \sqrt{7} \cong 7.94 {m}^{2}$