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

macabresubwoofer Dec 4, 2019

#1**+2 **

V_{cone} = 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

CPhill 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

Answer: B) 6 in

macabresubwoofer
Dec 4, 2019