#### Answer

The name of the figure is a rhombus, the perimeter is $36$ ft and the area is $70$ sq ft.

#### Work Step by Step

1. Name the figure:
Rhombus
2. Find the perimeter by adding all the sides up
Let $P = $ the perimeter of the rhombus
$P = 8 + 8 + 10 + 10$
$P = 16 + 20$
$P = 36$ ft
3. Find the area of the rhombus
Note: It is made of 2 right angled triangles and a rectangle. But first you need to find the height of the triangle using the Pythagorean Formula.
$b^{2} = c^{2} - a^{2} $
$b = \sqrt (c^{2} - a^{2})$
$b = \sqrt (8^{2} - 7^{2})$
$b = \sqrt (64 - 49)$
$b = \sqrt 15$
$b = 3.87...$ sq ft
Now find the area of the triangle using $base \times height \div 2$
$= (\sqrt 15 \times 7) \div 2$
$= 13.55544...$ sq ft
Since there is two right angled triangles, multiply it by 2:
$= 13.5544 \times 2$
$= 27.1108...$ sq ft
Now, you have to find the area of the rectangle that has a width of $7$ ft and a length of $(10 - \sqrt 15) = 6.127$ ft.
$= 7 \times 6.127...$
$= 42.889...$ sq ft
Add up the area of the rectangle and the area of the triangles
$ = 42.889... + 27.110...$
$ = 70$ sq ft