Question

Circular cones A and B are congruent. If the area of the base of cone B is 16π in2, then the volume of cone A is

Answer 1

if the area of the base of cone B is 16π, since both cones are congruent, then the area of the base of cone A is also 16π.

well, the base of cone A is a circle, and we know its diameter is "x", so its radius is half that, or x/2, and we also know that the area is 16π, thus

$\bf \textit{area of a circle}\\\\ A=\pi r^2~~ \begin{cases} r=radius\\ -----\\ r=\frac{x}{2}\\ A=16\pi \end{cases}\implies 16\pi =\pi \left( \frac{x}{2} \right)^2\implies 16\pi =\pi \cdot \cfrac{x^2}{2^2} \\\\\\ \cfrac{16\pi }{\pi }=\cfrac{x^2}{4}\implies 16=\cfrac{x^2}{4}\implies 64=x^2\implies \sqrt{64}=x\implies \boxed{8=x}$

and since we know what "x" is, then the radius of it is (8)/2, or just 4, and its height is (8) + 2, or just 10, then

$\bf \textit{volume of a cone}\\\\ V=\cfrac{\pi r^2 h}{3}~~ \begin{cases} r=radius\\ h=height\\ -----\\ r=4\\ h=10 \end{cases}\implies V=\cfrac{\pi (4)^2(10)}{3}\implies V=\cfrac{160\pi }{3}$

Answer 2

Answer: 6 in

Step-by-step explanation: 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