First, find the area of both beds!
For Heidi: 5 x 3 = 15
For Andrew: 5 x (3 x 2) = 30 ft
Added together, the beds have an area of 45 feet. To place soil with a depth of 2 feet, simply multiply by 2. You can think of this as finding the volume of the beds, which is length x width x height, or area x height. The answer is 90 cubic feet!
so we know the terminal point is at (9, -3), now, let's notice that's the IV Quadrant
![\bf (\stackrel{x}{9}~~,~~\stackrel{y}{-3})\impliedby \textit{let's find the \underline{hypotenuse}} \\\\\\ \textit{using the pythagorean theorem} \\\\ c^2=a^2+b^2\implies c=\sqrt{a^2+b^2} \qquad \begin{cases} c=hypotenuse\\ a=adjacent\\ b=opposite\\ \end{cases} \\\\\\ c=\sqrt{9^2+(-3)^2}\implies c=\sqrt{81+9}\implies c=\sqrt{90} \\\\[-0.35em] ~\dotfill](https://tex.z-dn.net/?f=%5Cbf%20%28%5Cstackrel%7Bx%7D%7B9%7D~~%2C~~%5Cstackrel%7By%7D%7B-3%7D%29%5Cimpliedby%20%5Ctextit%7Blet%27s%20find%20the%20%5Cunderline%7Bhypotenuse%7D%7D%20%5C%5C%5C%5C%5C%5C%20%5Ctextit%7Busing%20the%20pythagorean%20theorem%7D%20%5C%5C%5C%5C%20c%5E2%3Da%5E2%2Bb%5E2%5Cimplies%20c%3D%5Csqrt%7Ba%5E2%2Bb%5E2%7D%20%5Cqquad%20%5Cbegin%7Bcases%7D%20c%3Dhypotenuse%5C%5C%20a%3Dadjacent%5C%5C%20b%3Dopposite%5C%5C%20%5Cend%7Bcases%7D%20%5C%5C%5C%5C%5C%5C%20c%3D%5Csqrt%7B9%5E2%2B%28-3%29%5E2%7D%5Cimplies%20c%3D%5Csqrt%7B81%2B9%7D%5Cimplies%20c%3D%5Csqrt%7B90%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill)
1. 2+(6*(4)^2)
2+(6*(16))
3. 2+(96)
4. 98 is the answer to the question
so that means one of the relevant sides of the cylinders, namely the height, are on a 2:3 ratio, or 2/3 for that matter.
![\bf ~\hspace{5em} \textit{ratio relations of two similar shapes} \\[2em] \begin{array}{ccccllll} &\stackrel{ratio~of~the}{Sides}&\stackrel{ratio~of~the}{Areas}&\stackrel{ratio~of~the}{Volumes}\\ \cline{2-4}&\\ \cfrac{\textit{similar shape}}{\textit{similar shape}}&\cfrac{s}{s}&\cfrac{s^2}{s^2}&\cfrac{s^3}{s^3} \end{array}\\\\[-0.35em] ~\dotfill](https://tex.z-dn.net/?f=%5Cbf%20~%5Chspace%7B5em%7D%20%5Ctextit%7Bratio%20relations%20of%20two%20similar%20shapes%7D%20%5C%5C%5B2em%5D%20%5Cbegin%7Barray%7D%7Bccccllll%7D%20%26%5Cstackrel%7Bratio~of~the%7D%7BSides%7D%26%5Cstackrel%7Bratio~of~the%7D%7BAreas%7D%26%5Cstackrel%7Bratio~of~the%7D%7BVolumes%7D%5C%5C%20%5Ccline%7B2-4%7D%26%5C%5C%20%5Ccfrac%7B%5Ctextit%7Bsimilar%20shape%7D%7D%7B%5Ctextit%7Bsimilar%20shape%7D%7D%26%5Ccfrac%7Bs%7D%7Bs%7D%26%5Ccfrac%7Bs%5E2%7D%7Bs%5E2%7D%26%5Ccfrac%7Bs%5E3%7D%7Bs%5E3%7D%20%5Cend%7Barray%7D%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill)
![\bf \cfrac{\textit{similar shape}}{\textit{similar shape}}\qquad \cfrac{s}{s}=\cfrac{\sqrt{s^2}}{\sqrt{s^2}}=\cfrac{\sqrt[3]{s^3}}{\sqrt[3]{s^3}} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ \stackrel{\stackrel{\textit{ratio of the}}{sides}}{\cfrac{2}{3}}~\hspace{7em}\stackrel{\stackrel{\textit{ratio of the}}{volumes}}{\cfrac{2^3}{3^3}}\implies \cfrac{8}{27}\implies 8:27](https://tex.z-dn.net/?f=%5Cbf%20%5Ccfrac%7B%5Ctextit%7Bsimilar%20shape%7D%7D%7B%5Ctextit%7Bsimilar%20shape%7D%7D%5Cqquad%20%5Ccfrac%7Bs%7D%7Bs%7D%3D%5Ccfrac%7B%5Csqrt%7Bs%5E2%7D%7D%7B%5Csqrt%7Bs%5E2%7D%7D%3D%5Ccfrac%7B%5Csqrt%5B3%5D%7Bs%5E3%7D%7D%7B%5Csqrt%5B3%5D%7Bs%5E3%7D%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20%5Crule%7B34em%7D%7B0.25pt%7D%5C%5C%5C%5C%20%5Cstackrel%7B%5Cstackrel%7B%5Ctextit%7Bratio%20of%20the%7D%7D%7Bsides%7D%7D%7B%5Ccfrac%7B2%7D%7B3%7D%7D~%5Chspace%7B7em%7D%5Cstackrel%7B%5Cstackrel%7B%5Ctextit%7Bratio%20of%20the%7D%7D%7Bvolumes%7D%7D%7B%5Ccfrac%7B2%5E3%7D%7B3%5E3%7D%7D%5Cimplies%20%5Ccfrac%7B8%7D%7B27%7D%5Cimplies%208%3A27)
