Basic College Mathematics (10th Edition)

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