Length = x + 2 (because it is 2 cm more than x)
Width = 2x - 5 (5 cm lest than 2x)Area = 54 cm2 this is the formula to find the area Length × Width = Area (x + 2)(2x - 5) = 542x2 - x - 10 = 54 (this is you area)
Subtract 54 on both sides of equation to make the right side zero. 2x2 - x - 64 = 0 then use the quadratic formula x = (-b ± √(b2 - 4ac)) / 2a where:a = 2b = -1c = -64 Plug in these values into the formula. x = (1 ± √(1 - 4(-128))) / 4 x = (1 ± √(513)) / 4 x = (1 ± 22.65) / 4 x = (1 + 22.65) / 4 and x = (1 - 22.65) / 4 x = 5.91 and x = -5.41 Check the validity of the x values by adding them to the length and width. If the length or width should be a negative value, then that value of x is not acceptable. Now x = 5.91 Length = 5.91 + 2 (positive value.)Width = 2(5.91) - 5 ( positive value.) x = 5.91 If we look at this -- x = -5.41, Both length and width will be negative values. We reject this value of x. The answer is x = 5.91
Hope I helped and sorry it was really long
let's notice the tickmarks on the left and right sides, meaning those two sides are twins, and therefore equal, so the perimeter is simply 2.5+2.5+3.5+2.5 = 11 ft.
the trapezoid has an altitude/height of 2 ft, thus
![\bf \textit{area of a trapezoid}\\\\ A=\cfrac{h(a+b)}{2}~~ \begin{cases} a,b=\stackrel{bases}{parallel~sides}\\ h=height\\[-0.5em] \hrulefill\\ a=2.5\\ b=3.5\\ h=2 \end{cases}\implies A=\cfrac{2(2.5+3.5)}{2}\implies A=6](https://tex.z-dn.net/?f=%5Cbf%20%5Ctextit%7Barea%20of%20a%20trapezoid%7D%5C%5C%5C%5C%20A%3D%5Ccfrac%7Bh%28a%2Bb%29%7D%7B2%7D~~%20%5Cbegin%7Bcases%7D%20a%2Cb%3D%5Cstackrel%7Bbases%7D%7Bparallel~sides%7D%5C%5C%20h%3Dheight%5C%5C%5B-0.5em%5D%20%5Chrulefill%5C%5C%20a%3D2.5%5C%5C%20b%3D3.5%5C%5C%20h%3D2%20%5Cend%7Bcases%7D%5Cimplies%20A%3D%5Ccfrac%7B2%282.5%2B3.5%29%7D%7B2%7D%5Cimplies%20A%3D6)
Answer:
B. Rotation
Step-by-step explanation:
