I got this question from our quiz wrong I wonder how my teacher got the correct answers shown. I tried solving the first question but didn't get the right answers. I know the formula for the surface of a right circular cone is $S=πr(r+\sqrt {r^2+h^2})$ but I still followed the given formula for the sake of trying to obtain what my teacher got.
My solution:
Let $x$ be $r$ and $4x$ be $h$. Then,
$$πx\sqrt {x^2+(4x)^2}=8$$ $$πx\sqrt {17x^2}=8$$ $$17π^2x^4=8^2$$ $$x^4=\frac{8^2}{17π^2}$$ $$(x^4)^\frac{1}{4}=(\frac{8^2}{17π^2})^\frac{1}{4}$$ $$x=±\frac{2\sqrt 2}{17^\frac{1}{4}π^\frac{1}{2}}$$ $$x=\frac{2\sqrt 2}{17^\frac{1}{4}π^\frac{1}{2}}$$ $$h=\frac{8\sqrt 2}{17^\frac{1}{4}π^\frac{1}{2}}$$ $$h^2=\frac{128}{17^\frac{1}{2}π}$$ $$a=128$$ $$b=17$$
Your math is correct. Given $S = 8$, $h = 4r$, and $h^2 = \frac{a\sqrt{b}}{b\pi}$, we have:
$$ S = \pi r\sqrt{r^2 + h^2} $$
$$ = \pi \frac{\pi h}{4} \sqrt{\frac{1}{16}h^2 + h^2} $$
$$ = \pi \frac{h}{4} \sqrt{\frac{17h^2}{16}} $$
$$ = \frac{\pi h^2 \sqrt{17}}{16} $$
Substituting in $S = 8$, we have
$$ 8 = \frac{\pi h^2 \sqrt{17}}{16} $$
$$ \Rightarrow h^2 = \frac{128}{\sqrt{17}\pi} $$
$$ = \frac{128\sqrt{17}}{17\pi} $$
$$\Rightarrow a = 128, b = 17 $$
Now, if we take the "correct" solution with $a = 2$ and $b = 8$, we still have (from the algebra above):
$$ S = \frac{\pi h^2 \sqrt{17}}{16} $$
Substituting in $a = 2$ and $b = 8$, we have
$$ S = \frac{\pi \sqrt{17}}{16} \frac{2 \sqrt{8}}{8 \pi} $$
$$ = \frac{\sqrt{34}}{32} \neq 8 $$