Answer:
I guess it is the Answer C
from the diagram, we can see that the height or line perpendicular to the parallel sides is 8.5.
likewise we can see that the parallel sides or "bases" are 24.3 and 9.7, so
![\textit{area of a trapezoid}\\\\ A=\cfrac{h(a+b)}{2}~~ \begin{cases} h=height\\ a,b=\stackrel{parallel~sides}{bases}\\[-0.5em] \hrulefill\\ h=8.5\\ a=24.3\\ b=9.7 \end{cases}\implies \begin{array}{llll} A=\cfrac{8.5(24.3+9.7)}{2}\\\\ A=\cfrac{8.5(34)}{2}\implies A=144.5~in^2 \end{array}](https://tex.z-dn.net/?f=%5Ctextit%7Barea%20of%20a%20trapezoid%7D%5C%5C%5C%5C%20A%3D%5Ccfrac%7Bh%28a%2Bb%29%7D%7B2%7D~~%20%5Cbegin%7Bcases%7D%20h%3Dheight%5C%5C%20a%2Cb%3D%5Cstackrel%7Bparallel~sides%7D%7Bbases%7D%5C%5C%5B-0.5em%5D%20%5Chrulefill%5C%5C%20h%3D8.5%5C%5C%20a%3D24.3%5C%5C%20b%3D9.7%20%5Cend%7Bcases%7D%5Cimplies%20%5Cbegin%7Barray%7D%7Bllll%7D%20A%3D%5Ccfrac%7B8.5%2824.3%2B9.7%29%7D%7B2%7D%5C%5C%5C%5C%20A%3D%5Ccfrac%7B8.5%2834%29%7D%7B2%7D%5Cimplies%20A%3D144.5~in%5E2%20%5Cend%7Barray%7D)
Here you have 2 linear equations and are to solve this system. Both equations have already been solved for y, so you can set one of them = to the other one:
-2x+11 = -3x+21.
Then 3x-2x = 21 - 11, or x = 10 (answer)
Find y by subst. x = 10 into either of the given equations.
Solution is then (10, ? )
$10.50 X 35 would give you your answer which would equal $367.5
Area = side * side
Area of the square = 6 1/9 * 6/1/9
= 36 1/81
