Answer
The perimeter of the figure is \[54\text{ ft}.\] and the area is\[168\text{ ft}{{.}^{2}}\].
Work Step by Step
It is required to compute the perimeter and area of the given figure. The perimeter of any figure is the sum of all the sides that makes up the boundary of the figure. The given figure consists of two triangles with same dimensions, one over the other.
In\[\vartriangle ABC\], applying Pythagoras theorem to find the side AC, we will get:
\[\begin{align}
& AC=\sqrt{A{{B}^{2}}+B{{C}^{2}}} \\
& =\sqrt{8\text{ f}{{\text{t}}^{2}}+6\text{ f}{{\text{t}}^{2}}} \\
& =10\text{ ft}.
\end{align}\]
In\[\vartriangle ABE\], applying Pythagoras theorem to find the side AE, we will get:
\[\begin{align}
& AE=\sqrt{A{{B}^{2}}+B{{E}^{2}}} \\
& =\sqrt{8\text{ f}{{\text{t}}^{2}}+15\text{ f}{{\text{t}}^{2}}} \\
& =17\text{ ft}.
\end{align}\]
Similarly, in\[\vartriangle DBC\] and\[\vartriangle DBE\], the sides DC and DE will be \[10\text{ ft}.\] and \[17\text{ ft}.\] respectively. The perimeter of the figure will be sum of all these four sides.
\[\begin{align}
& \text{Perimeter}=AC+AE+DC+DE \\
& =10\text{ ft}+17\text{ ft}+10\text{ ft}+17\text{ ft} \\
& =54\text{ ft}
\end{align}\]
The area of the figure will be the sum of the areas of \[\vartriangle AEC\]and\[\vartriangle DEC\]. The base and height of both these triangles will be \[15\text{ ft}+6\text{ ft}=21\text{ ft}.\] and \[8\text{ ft}\] respectively. So, the area of each of these triangles will be:
\[\begin{align}
& A=\frac{1}{2}bh \\
& =\frac{1}{2}\times 21\text{ ft}\times 8\text{ ft} \\
& =84\text{ f}{{\text{t}}^{2}}\text{ }
\end{align}\]
So, the total area enclosed in the figure will be twice this area, that is\[168\text{ f}{{\text{t}}^{2}}\]. Hence, the perimeter of the figure is \[54\text{ ft}\] and the area is\[168\text{ f}{{\text{t}}^{2}}\]
