# How do you find the area of an isosceles triangle if the two equal sides are 10cm and the base is 12cm?

Jul 22, 2015

I found: $48 {\text{cm}}^{2}$

#### Explanation:

Considering:

applying Pythagoras on half triangle you get:
${h}^{2} + {6}^{2} = {10}^{2}$
$h = 8 c m$
So $A r e a = \frac{b a s e \times h e i g h t}{2} = \frac{12 \times 8}{2} = 48 {\text{cm}}^{2}$

Jul 22, 2015

Area $= 48$ sq. cm.
$\textcolor{w h i t e}{\text{XXXX}}$$\textcolor{w h i t e}{\text{XXXX}}$(Using Heron's formula)

#### Explanation:

As an alternate solution method:

Heron's Formula for the area of a triangle with sides $a , b , c$
$\textcolor{w h i t e}{\text{XXXX}}$$A = \sqrt{s \left(s - a\right) \left(s - b\right) \left(s - c\right)}$
$\textcolor{w h i t e}{\text{XXXX}}$$\textcolor{w h i t e}{\text{XXXX}}$where $s$ is the semi-perimeter (i.e. $s = \frac{a + b + c}{2}$

In this case:
$\textcolor{w h i t e}{\text{XXXX}}$$s = 16$

$\textcolor{w h i t e}{\text{XXXX}}$$A = \sqrt{16 \left(6\right) \left(6\right) \left(4\right)}$
$\textcolor{w h i t e}{\text{XXXX}}$$\textcolor{w h i t e}{\text{XXXX}}$$= \sqrt{{4}^{2} \cdot {6}^{2} \cdot {2}^{2}}$
$\textcolor{w h i t e}{\text{XXXX}}$$\textcolor{w h i t e}{\text{XXXX}}$$= 48$