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# ​ Circular cones A and B are congruent. If the volume of cone B is 8π in3, then its height, h =

0
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2 Circular cones A and B are congruent. If the volume of cone B is 8π in3, then its height, h =

A) 4 in

B) 6 in

C) 8 in

D) 10 in

E) 12 in

Dec 4, 2019

#1
+2

Vcone =     pi * radius^2 * height / 3    ....so we have

8 pi  =  pi * (x/2)^2 * (x + 2) / 3       divide out pi.....multiply both sides by 3

24  =  x^2 ( x + 2)/4       multiply both sides by 4

96  = x^2 ( x + 2)

96  = x^3 + 2x^2

x^3 + 2x^2 - 96   =  0

The real solution to this is  x  =  4

So....the height  of    B  =   x + 2   =    6   Dec 4, 2019
#2
+1

Cone A and cone B are congruent, so they must have the same volume. Cone A has volume of 8pi in^3 (because we're told cone B has this volume)

focus on cone A only for now
r = radius = d/2 = x/2
h = height = x+2
V = volume of cone
V = 8pi
V = (1/3)*pi*r^2*h
8pi = (1/3)*pi*(x/2)^2*(x+2)
8pi = (1/3)*pi*((x^2)/4)*(x+2)
8pi = pi*((x^2)/12)*(x+2)
8 = ((x^2)/12)*(x+2) ... divided both sides by pi
12*8 = x^2(x+2) ... multiplied both sides by 12
96 = x^3+2x^2
0 = x^3+2x^2-96
Using a graphing calculator, I found the solution to that equation to be x = 4. Basically you are looking for the x intercept of f(x) = x^3+2x^2-96

So,
h = x+2
h = 4+2
h = 6