Cups A and B are cone shaped and have heights of $25 c m$ $25 cm$ and $26 c m$ $26 cm$ and openings with radii of $9 c m$ $9 cm$ and $7 c m$ $7 cm$ , respectively. If cup B is full and its contents are poured into cup A, will cup A overflow? If not how high will cup A be filled?

Question

1693 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-06-14T07:28:53

See below for explanation

Cone Volume is given by $V = \frac{1}{3} π r^{2} h$ $V=1/3pir^2h$

We have cone $A$ $A$ which volume is $V_{A} = \frac{1}{3} π 9^{2} \cdot 25 = 675 π$ $V_A=1/3pi9^2·25=675pi$

For cone $B$ $B$ : $V_{B} = \frac{1}{3} π 7^{2} \cdot 26 = \frac{1274}{3} π$ $V_B=1/3pi7^2·26=1274/3pi$

Comparing both volume it is obvius that $\frac{1274}{3} < 675$ $1274/3<675$ then the cup B do not overfilled cup A

In this case, what will be the height of cup A filled?. Lets see.

With volume B into cone A, we have a situation like this
enter image source here

By Thales theorem we know that $\frac{9}{r} = \frac{25}{h}$ $9/r=25/h$ then $h = \frac{25 r}{9}$ $h=(25r)/9$ (1)

Revolving DEC around CA axis, we have the cone produced by volume B in cup A, so his volume is known. Then

$\frac{1274}{3} π = \frac{1}{3} π r^{2} \cdot 25 \frac{r}{9}$ $1274/3pi=1/3pir^2·25r/9$ we can remove $π$ $pi$ and $\frac{1}{3}$ $1/3$

$1274 = 25 \frac{r^{3}}{9}$ $1274=25r^3/9$ from here we have $r^{3} = \frac{1274}{25} \cdot 9 = 458.64$ $r^3=1274/25·9=458.64$

Then $r = \sqrt[3]{458.64} = 7.71182$ $r=root(3)458.64=7.71182$ cm

In (1), we have $h = 25 \cdot \frac{7.71182}{9} = 21.42$ $h=25·7.71182/9=21.42$ cm

Cups A and B are cone shaped and have heights of 25 cm25cm25 cm and 26 cm26cm26 cm and openings with radii of 9 cm9cm9 cm and 7 cm7cm7 cm, respectively. If cup B is full and its contents are poured into cup A, will cup A overflow? If not how high will cup A be filled?