Cups A and B are cone shaped and have heights of $25 cm$ and $26 cm$ and openings with radii of $9 cm$ and $7 cm$ , 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?

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Answer 1 · 2018-06-14T07:28:53

See below for explanation

Cone Volume is given by $V=1/3pir^2h$

We have cone $A$ which volume is $V_A=1/3pi9^2·25=675pi$

For cone $B$ : $V_B=1/3pi7^2·26=1274/3pi$

Comparing both volume it is obvius that $1274/3<675$ 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 $9/r=25/h$ then $h=(25r)/9$ (1)

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

$1274/3pi=1/3pir^2·25r/9$ we can remove $pi$ and $1/3$

$1274=25r^3/9$ from here we have $r^3=1274/25·9=458.64$

Then $r=root(3)458.64=7.71182$ cm

In (1), we have $h=25·7.71182/9=21.42$ cm

Cups A and B are cone shaped and have heights of 25 cm25 cm and 26 cm26 cm and openings with radii of 9 cm9 cm and 7 cm7 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?