Question

Find the area of the trapezoid in the figure below ground your final answer to the nearest 10th

Answer 1

First, find the length of side CD.

`\begin{gathered} CD=2CE+13.5 \\ \\ \text{By Pythagorean theorem} \\ CA^2=CE^2+AE^2 \\ CE^2=CA^2-AE^2 \\ CE^2=(30.5)^2-(26)^2 \\ CE^2=930.25-676 \\ CE^2=254.25 \\ CE=\sqrt[]{254.25} \end{gathered}`
`\begin{gathered} CD=2CE+13.5 \\ CD=2(\sqrt[]{254.25})+13.5 \\ CD=31.89+13.5 \\ CD=45.39 \end{gathered}`

Now that we have length CD, recall that the area of the trapezoid is

`\begin{gathered} A=(a+b)/(2)h \\ \text{where} \\ a\text{ is the top base} \\ b\text{ is the bottom base} \\ h\text{ is the height} \end{gathered}`
`\begin{gathered} \text{Given} \\ h=AE=26 \\ a=CD=45.39 \\ b=AB=13.5 \\ \\ \text{Substitute these values to the formula and we have} \\ A=(a+b)/(2)h \\ A=(45.39+13.5)/(2)(26) \\ A=(58.89)/(2)(26) \\ A=(29.445)(26) \\ A=765.57 \end{gathered}`

Rounding the area to the nearest tenth, the area of the given trapezoid is equal to 765.6 square units.