Question

Good-afternoon everyone! This past week I've managed to finish up some math work. However, I want to make sure I didn't make any errors in the process of solving and if I did, I want to know how to fix them and make sure I don't make those errors next time. If anyone would be willing to check my work, I would deeply appreciate it! Basically, the question is asking to give the error that the student made and to find the correct area & volume of the cone. For the surface area, I got an answer of 1244.1 meters squared. For the volume, I got an answer of 2787.6 meters squared. Thank you so much for your time and help! :) ​

Answer 1

you're correct, Jan used the height instead of the slant-height, which is needed for the SA, so hmm well, we really don't know the slant-height, hmmm let's find it, keeping in mind the slant-height is just the hypotenuse of either triangular cross-section

`\begin{array}{llll} \textit{using the pythagorean theorem} \\\\ c^2=a^2+o^2\implies c=√(a^2 + o^2) \end{array} \qquad \begin{cases} c=\stackrel{hypotenuse}{L}\\ a=\stackrel{adjacent}{11}\\ o=\stackrel{opposite}{22} \end{cases} \\\\\\ L=√( 11^2 + 22^2)\implies L=√( 121 + 484 ) \implies L=√( 605 )\implies L=11√(5) \\\\[-0.35em] ~\dotfill`

`\textit{surface area of a cone}\\\\ SA=\pi rL + \pi r^2~~ \begin{cases} r=radius\\ L=slant~height\\[-0.5em] \hrulefill\\ r=11\\ L=11√(5) \end{cases}\implies SA=\pi (11)(11√(5))+\pi (11)^2 \\\\\\ SA=121\pi √(5)+121\pi \implies \boxed{SA\approx 1230.14~m^2} \\\\[-0.35em] ~\dotfill\\\\ \textit{volume of a cone}\\\\ V=\cfrac{\pi r^2 h}{3}~~ \begin{cases} r=radius\\ h=height\\[-0.5em] \hrulefill\\ r=11\\ h=22 \end{cases}\implies V=\cfrac{\pi (11)^2(22)}{3}\implies \boxed{V\approx 2787.64~m^3}`