given the similar cylinder below,determine the volume of the larger
cylinder if the ratio of total areas is 769/64.

larger cylinder: r=10x-6.5

smaller cylinder: r=2x h=x

a)100.53

b)3864.75

c)1717.67

d)1455.91

e)1145.11

how can I solve the equations of the radius of both cylinder?
thanks

The ratio of areas is 769 : 64, then the ratio of lengths will be 27 : 8, the square roots of the given numbers. Can you take it from here?

so folks I asked this question and I answer back.despite many viewer there was no answer,so ı assume this could help if anybody wonder how to do this calculation,I just find out but got many help though. Thanks

The principle we are supposed to know here is that if two similar solids
have similarity ratio k, then their areas have ratio k^2 and their volumes
have ratio k^3.

x is the height of the small cylinder
2x is the radius of the small cylinder
10x - 6.5 is the radius of the larger cylinder.

Now, since we know that the cylinders are similar, you know that the
height of the large cylinder is half its radius, so the height of the large
cylinder must be

5x - 3.25

the area formula would be A = pi*r^2 + 2*pi*r*h, and so

area of smaller cylinder = pi*(2x)^2+ 2*pi*(2x)*x = pi*4*x^2 + pi*4*x^2 =
8*pi*x^2
area of larger cylinder = pi*(10x - 6.5)^2 + 2*pi*(10x-6.5)(5x-3.25)

Before multiplying all this out, note that 2(5x-3.25) = 10x - 6.5, so what
you have here is really pi*(10x - 6.5)^2 + pi*(10x - 6.5)^2 = 2*pi*(10x -
6.5)^2

You know the area ratio, so you know that

(2*pi*(10x-6.5)^2 / (8*pi*x^2) = 729/64

You can solve this equation for x. here how we do that.

GIVEN

2*pi*(10x-6.5)^2 / (8*pi*x^2) = 729/64

The above reduces to

(10x - 6.5)^2 / (4x^2) = 729/64


Multiplying both sides by 4x^2

(10x - 6.5)^2 = 729x^2/16

Extracting the square root of both sides,

10x - 6.5 = 27x/4

Multiplying both sides by 4,

40x - 26 = 27x

Combining terms and rearranging,

40x - 27x = 26

13x = 26

and solving for "x"

x = 2

now we know the x ,so radius and diameters basically we can calculate any thing from there.