# What is the area of an isosceles triangle with two equal sides of 10 cm and a base of 12 cm?

Nov 21, 2015

Area $= 48$ $c {m}^{2}$

#### Explanation:

Since an isosceles triangle has two equal sides, if the triangle is split in half vertically, the length of the base on each side is:

$12$ $c m$$\div 2 =$$6$ $c m$

We can then use the Pythagorean theorem to find the height of the triangle.

The formula for the Pythagorean theorem is:

${a}^{2} + {b}^{2} = {c}^{2}$

To solve for the height, substitute your known values into the equation and solve for $a$:

where:
$a$ = height
$b$ = base
$c$ = hypotenuse

${a}^{2} + {b}^{2} = {c}^{2}$
${a}^{2} = {c}^{2} - {b}^{2}$
${a}^{2} = {\left(10\right)}^{2} - {\left(6\right)}^{2}$
${a}^{2} = \left(100\right) - \left(36\right)$
${a}^{2} = 64$
$a = \sqrt{64}$
$a = 8$

Now that we have our known values, substitute the following into the formula for area of a triangle:

$b a s e = 12$ $c m$
$h e i g h t = 8$ $c m$

$A r e a = \frac{b a s e \cdot h e i g h t}{2}$

$A r e a = \frac{\left(12\right) \cdot \left(8\right)}{2}$

$A r e a = \frac{96}{2}$

$A r e a = 48$

$\therefore$, the area is $48$ $c {m}^{2}$.