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

The area of a triangle is $E = \frac{1}{2} b \cdot h$ where b is the base and h is the height.
The height is $h = \sqrt{{a}^{2} - {\left(\frac{b}{2}\right)}^{2}} = \sqrt{{4}^{2} - {3}^{2}} = \sqrt{16 - 9} = \sqrt{7}$

So we have that $E = \frac{1}{2} 6 \sqrt{7} = 3 \cdot \sqrt{7} = 7.94$

Sep 8, 2015

I found $A = 8$

#### Explanation:

You can use Phytagoras Theorem to find $h$ as:
${4}^{2} = {3}^{2} + {h}^{2}$
$h = 2.6$
$A = \frac{1}{2} \left(b a s e \times h e i g h t\right) = \frac{1}{2} \left(6 \times 2.6\right) = 7.8 \approx 8$