# How do I find the area of an isosceles triangle with.a base that is 20 m long and each of the other two sides is 26 m?

Dec 17, 2015

$166 {m}^{2}$

#### Explanation:

Use the formula $\frac{1}{2} \cdot b a s e \cdot h e i g h t$

You need to find the height of the triangle, so split the isosceles into 2 right angled triangles.

So, one newly- created right angled triangle will have a base length of half of the original triangle, which is $20$

$\frac{20}{2} = 10$

The new base length of the right- angled triangle is $10$.

You know the length of the hypotenuse is $26$, so you can use the Pythagoras theorem to solve for the height of the triangle.

${c}^{2} = {a}^{2} + {b}^{2}$
${26}^{2} = {a}^{2} + {20}^{2}$
$676 = {a}^{2} + 400$
$276 = {a}^{2}$
$a \approx 16.6132477258$
$a = 16.6 \implies$ (correct to 3 s.f)

So the height of the triangle is $\approx 16.6 m$

Substitute everything in the formula again.

$\frac{1}{2} \cdot 20 \cdot 16.6$
= $166 {m}^{2}$