Question

Find the area Of an isosceles trapezoid with legs 5 and bases 6 and 8

Answer 1

We can draw the isosceles trapezoid as follows:

Then, the area of this isosceles trapezoid will be:

`A_{\text{trapezoid}}=(1)/(2)h(b_1+b_2)`

However, we need to find the height of the isosceles trapezoid, h, using the Pythagorean Theorem as follows:

Therefore, we can find, h, as follows:

`5^2=1^2+h^2\Rightarrow h^2=5^2-1^2\Rightarrow h^2=25-1\Rightarrow h^2=24`
`h=\sqrt[]{24}\Rightarrow24=2^2\cdot2\cdot3\Rightarrow h=\sqrt[]{2^2\cdot6}\Rightarrow h=2\cdot\sqrt[]{6}`

Then, the area of the isosceles trapezoid using the formula above as follows:

`A_{\text{trapezoid}}=(1)/(2)h(b_1+b_2)\Rightarrow\begin{cases}h=2\cdot\sqrt[]{6} \\ b_1=6 \\ b_2=8\end{cases}`

Then, we have:

`A_{\text{trapezoid}}=(1)/(2)2\cdot\sqrt[]{6}(6+8)=14\cdot\sqrt[]{6}`

In summary, the area of an isosceles trapezoid is (in square units):

`A_{\text{trapezoid}}=14\cdot\sqrt[]{6}`