Question

the volume of a frustrum of a right circular cone is 52pi ft^3. Its altitude is 3 ft and measure of its lower radius is three times the measure of its upper radius. Find the lateral area of the frustum.

Answer 1

@iGreen

Answer 2

\[1/3 * \Pi h(R^2 + r^2 + Rr)\] Volume of frustrum is given by (formula drawn below) where R and r are the lower and upper base radius.

Answer 3

what can we do to get the values of R and r?

Answer 4

@perl

Answer 5

\(\sf V = \dfrac{\pi h}{3} (R^2 + Rr + r^2)\)

'h' is the height, 'R' is the radius of the lower base, and 'r' is the radius of the upper base.

Answer 6

Plug in what we know:

\(\sf 52\pi = \dfrac{\pi (3)}{3} (R^2 + Rr + r^2)\)

We can cancel out pi.

\(\sf 52 = \dfrac{3}{3} (R^2 + Rr + r^2)\)

Simplify:

\(\sf 52 = (R^2 + Rr + r^2)\)

Okay, it says the lower radius is 3 times the measure of the upper radius.

So:

R = 3r

Answer 7

Plug in 3r for 'R' and solve for 'r'

Answer 8

\(\sf 52 = ((3r)^2 + (3r)r + r^2)\)

Answer 9

Can you simplify that? @~Gelmhar

Answer 10

r=2.73

Answer 11

@iGreen

Answer 12

\[7r^2=52 , r^2=52/7 ,r=2.73\]

Answer 13

No.. I got something else.

Answer 14

how?

Answer 15

\(\sf 54 = ((3r)^2 + (3r)r + r^2)\)

\(\sf (3r)^2 \rightarrow 3^2r^2 \rightarrow 9r^2\)

\(\sf (3r)r \rightarrow 3r \times r \rightarrow 3r^2\)

So we have:

\(\sf 54 = (9r^2 + 3r^2 + r^2)\)

Combine like terms.

Answer 16

i see.. wait i will solve it

Answer 17

r=2

Answer 18

can you help me again @iGreen

Answer 19

Yep, you got it.

Answer 20

\(54=(9r^2+3r^2+r^2)\)

\(54 = 13r^2\)

\(4 = r^2\)

\(r = 2\)

Answer 21

\(\sf LA=\pi(R+r)L\)

Answer 22

@~Gelmhar Do you have a picture?

Answer 23

i got the answer now, lets move to another problem can we ? :D

Answer 24

Great