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avatar+163 

The water tank in the diagram below is in the shape of an inverted right circular cone. The radius of its base is 16 feet, and its height is 96 feet. The water in the tank is 25% of the tank's capacity. The height of the water in the tank can be written in the form \(a\sqrt[3]{b}\) , where  and a and b are positive integers and b is not divisible by a perfect cube greater than 1. What is a+b?

 

 ABCD is a regular tetrahedron (right triangular pyramid). If M is the midpoint of CD, then what is tan(angAMB)?

Creeperhissboom  Apr 13, 2018
edited by Creeperhissboom  Apr 13, 2018
 #1
avatar+92622 
+3

Here's the first one

 

Call the height of the water in the tank, H

 

And ......using similar triangles, we can express the radius, R, at this point thusly :

 

16 / 96  = R / H

1/6 = R/H

H/6  = R

 

Now  when the tank is 25% full, it will hold

 

(1/4) (1/3) pi 16^2 * 96   =   2048 pi  ft^3

 

So....to find the height, we have

 

2048 pi   =  (1/3) pi  (H/6)^2 * H     divide out pi, multiply both sides by 3

 

6144  =  H^3 / 36       multiply both sides by 36

 

221184  =  H^3

 

Take the cube root of both sides

 

483√2     =   H

 

So

 

a +  b   =   48 +  2   = 50

 

 

cool cool cool

CPhill  Apr 13, 2018
 #2
avatar+92622 
+2

A pic for #2  might be helpful......

 

cool cool cool

CPhill  Apr 13, 2018
edited by CPhill  Apr 13, 2018
edited by CPhill  Apr 13, 2018
 #3
avatar+163 
+2

There was no picture. It is a regular tetrahedron, so any orientation will be fine.

Creeperhissboom  Apr 13, 2018

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