Water is being poured into a conical cup that is 12 cm tall.
a. When the water in the cup is 9 cm deep, what percentage of the cup is filled?
b. When the cup is 75% filled, how deep is the water?
I think it has to do with ratio, since it only gives you the height?
Along with the answer, can someone give me like a cheat sheet way of doing this, with either percentage,/ height,/ radius, etc.
you may use similar triangles there..Originally Posted by stones44
suppose that the radius of the cone is R and the radius of the cone at the hight of 9 ft is r, then the ratio is given by
$\displaystyle \frac{R}{12}=\frac{r}{9}$
which implies that
$\displaystyle r= \frac {9R}{12}$
$\displaystyle V_{cone} = \frac{1}{3}\pi R^2 H$
given th situation, the volume of the cone when full is
$\displaystyle V_{cone, full}=4\pi R^2$
while when if 9 ft full,
$\displaystyle V_{cone,9 ft}=3\pi r^2$
then, the percentage is
$\displaystyle \frac{V_{cone,9 ft}}{V_{cone, full}} = \frac{3\pi r^2}{4\pi R^2}$
$\displaystyle = \frac{3\pi (\frac {9R}{12})^2}{4\pi R^2} $
$\displaystyle = \frac{3\frac{9^2 \pi R^2}{12^2}}{4\pi R^2}$
$\displaystyle = \frac{(3)(9^2)}{(4)(12^2)}$
then multiply by 100%.. Ü
for b)
you are looking for h. so, using similar triangles again, you have
$\displaystyle \frac{r}{h}=\frac{R}{12}$
so that
$\displaystyle r=\frac{hR}{12}$
so, 75% full,
$\displaystyle \frac{V_{cone,h}}{V_{cone,full}}=\frac{3}{4} $
so,
$\displaystyle \frac{\frac{1}{3}h\pi r^2}{4\pi R^2} = \frac{\frac{1}{3}h\pi (\frac{hR}{12})^2}{4\pi R^2} = \frac{3}{4}$
can you continue? you will notice that R^2 will cancel and you can now solve for h. Ü
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