1. ## SAT Help

Given, a paper cone for water is 6 cm across at the top and 10 cm high.

If it is filled to half of its height, what percent of it's total volume does it contain?

If it is filled to half of its volume, what is the height of the water?

This was a question given to me as a practice SAT. (It may not be exactly as written, but this is the best I can remmeber the question.) I've taken other SAT practices and this seems to be the hardest one. Could someone please go over this question and show me how to go about solving it?

Thanks.

2. Originally Posted by Mr_Green
Given, a paper cone for water is 6 cm across at the top and 10 cm high.

If it is filled to half of its height, what percent of it's total volume does it contain?

If it is filled to half of its volume, what is the height of the water?

This was a question given to me as a practice SAT. (It may not be exactly as written, but this is the best I can remmeber the question.) I've taken other SAT practices and this seems to be the hardest one. Could someone please go over this question and show me how to go about solving it?

Thanks.
The filled part of the cone is similar to the whole cone. Therefore if x1 and x2
are corresponding linear dimensions of the filled part and the whole cone
respectivly the volume of the filled part is:

V1=(x1/x2)^3 V2,

where V2 is the volume of the whole cone. If the cone is filled to half its
height, we have the heights serving for x1 and x2, so:

V1/V2=(0.5/1)^3=1/8,

Thet is the volume of the filled part of the cone is 1/8 of the volume of the
whole cone or 12.5%.

The second part works the same relation the other way around:

V1/V2=1/2=(x1/x2)^3=x1^3/1000,

so the cone is filled to a height of:

cube root(1000/2) ~=7.94 cm

RonL

3. Hello, Mr_Green!

This is a good problem.
You need to know the volume of a cone: .$\displaystyle V\:=\:\frac{\pi}{3}r^2h$
. . and be able to apply it to their questions.

Given: a paper cone for water is 6 cm across at the top and 10 cm high.

(a) If it is filled to half of its height, what percent of it's total volume does it contain?

(b) If it is filled to half of its volume, what is the height of the water?
Code:
                :   3   :
- *-------+-------*
:  \      |      /
:   \     |  r  /
:    \----+----/
10     \:::|:::/
:      \::|h:/
:       \:|:/
:        \|/
-         *

The volume of the cone is: .$\displaystyle V \:=\:\frac{\pi}{3}\cdot3^2\cdot10\:=\:30\pi$ cm³.

From the similar right triangles: .$\displaystyle \frac{r}{h} = \frac{3}{10}\quad\Rightarrow\quad r \,=\,\frac{3}{10}h$

(a) If $\displaystyle h = 5$, then $\displaystyle r = \frac{3}{10}(5) = \frac{3}{2}$

Then the volume of the water is: .$\displaystyle V \:=\:\frac{\pi}{3}\left(\frac{3}{2}\right)^2(5)\:= \:\frac{15\pi}{4}$ cm³.

The percent of volume is: .$\displaystyle \frac{\frac{15\pi}{4}}{30\pi} \;=\;\frac{1}{8}\;=\;\boxed{12.5\%}$

(b) Since $\displaystyle r = \frac{3}{10}h\!:\;\;V \:=\:\frac{\pi}{3}\!\left(\frac{3}{100}h\right)^2\ !\!h\quad\Rightarrow\quad V\:=\:\frac{3\pi}{100}h^3$

If the cone is half full, the volume of water is $\displaystyle 15\pi$ cm³.

We have: .$\displaystyle \frac{3\pi}{100}h^3\:=\:15\pi\quad\Rightarrow\quad h^3\:=\:500\quad\Rightarrow\quad h \,=\,\sqrt[3]{500}$

Therefore: .$\displaystyle h \;=\;5\sqrt[3]{4}\;\approx\;\boxed{7.937\text{ cm}}$