Answer
the perimeter is $\left( 12\sqrt{3}+13\sqrt{7} \right) \text{ meters}$;
the area is $39\sqrt{21} \text{ square meters}$
Work Step by Step
Adding all the given sides of the trapezoid, then its perimeter, $P,$ is
\begin{array}{l}\require{cancel}
P=2\sqrt{27}+2\sqrt{63}+6\sqrt{3}+7\sqrt{7}
.\end{array}
Extracting the factor of the radicand that is a perfect power of the index, the equation above is equivalent to
\begin{array}{l}\require{cancel}
P=2\sqrt{9\cdot3}+2\sqrt{9\cdot7}+6\sqrt{3}+7\sqrt{7}
\\
P=2\sqrt{(3)^2\cdot3}+2\sqrt{(3)^2\cdot7}+6\sqrt{3}+7\sqrt{7}
\\
P=2(3)\sqrt{3}+2(3)\sqrt{7}+6\sqrt{3}+7\sqrt{7}
\\
P=6\sqrt{3}+6\sqrt{7}+6\sqrt{3}+7\sqrt{7}
\\
P=\left( 6\sqrt{3}+6\sqrt{3}\right) +\left(6\sqrt{7}+7\sqrt{7}\right)
\\
P=12\sqrt{3}+13\sqrt{7}
.\end{array}
Hence, the perimeter, $P,$ is $
\left( 12\sqrt{3}+13\sqrt{7} \right) \text{ meters}
.$
Using $A=\dfrac{h(b_1+b_2)}{2}$ or the formula for the area of a trapezoid, then
\begin{array}{l}\require{cancel}
A=\dfrac{6\sqrt{3}(2\sqrt{63}+7\sqrt{7})}{2}
.\end{array}
Extracting the factor of the radicand that is a perfect power of the index, the equation above is equivalent to
\begin{array}{l}\require{cancel}
A=\dfrac{6\sqrt{3}(2\sqrt{9\cdot7}+7\sqrt{7})}{2}
\\
A=\dfrac{6\sqrt{3}(2\sqrt{(3)^2\cdot7}+7\sqrt{7})}{2}
\\
A=\dfrac{6\sqrt{3}(2(3)\sqrt{7}+7\sqrt{7})}{2}
\\
A=\dfrac{6\sqrt{3}(6\sqrt{7}+7\sqrt{7})}{2}
\\
A=\dfrac{6\sqrt{3}(13\sqrt{7})}{2}
\\
A=\dfrac{\cancel2\cdot3\sqrt{3}(13\sqrt{7})}{\cancel2}
\\
A=3\sqrt{3}(13\sqrt{7})
\\
A=39\sqrt{21}
.\end{array}
Hence, the area, $A,$ is $
39\sqrt{21} \text{ square meters}
.$
