Question

the volume of two similar solids are 1080cm and 1715cm .if the curved surface area of the smaller cone is 840cm .fond the curved surface area of the larger cone​

Answer 1

`A_(big) = 1143.33cm^2`

Step-by-step explanation:

The given parameters are:

`V_(small) = 1080`

`V_(big) = 1715`

`C_(small) = 840`

Required

Determine the curved surface area of the big cone

The volume of a cone is:

`V = (1)/(3)\pi r^2h`

For the big cone:

`V_(big) = (1)/(3)\pi R^2H`

Where

R = radius of the big cone and H = height of the big cone

For the small cone:

`V_(small) = (1)/(3)\pi r^2h`

Where

r = radius of the small cone and H = height of the small cone

Because both cones are similar, then:

`(H)/(h) = (R)/(r)`

and

`(V_(big))/(V_(small)) = ((1)/(3)\pi R^2H)/((1)/(3)\pi r^2h)`

`(V_(big))/(V_(small)) = (R^2H)/(r^2h)`

Substitute values for Vbig and Vsmall

`(1715)/(1080) = (R^2H)/(r^2h)`

Recall that:
`(H)/(h) = (R)/(r)`

So, we have:

`(1715)/(1080) = (R^2*R)/(r^2*r)`

`(1715)/(1080) = (R^3)/(r^3)`

Take cube roots of both sides

`\sqrt[3]{(1715)/(1080)} = (R)/(r)`

Factorize

`\sqrt[3]{(343*5)/(216*5)} = (R)/(r)`

`\sqrt[3]{(343)/(216)} = (R)/(r)`

`(7)/(6) = (R)/(r)`

The curved surface area is calculated as:

`Area = \pi rl`

Where

`l = slant\ height`

For the big cone:

`A_(big) = \pi RL`

For the small cone

`A_(small) = \pi rl`

Because both cones are similar, then:

`(L)/(l) = (R)/(r)`

and

`(A_(big))/(A_(small)) = (\pi RL)/(\pi rl)`

`(A_(big))/(A_(small)) = (RL)/(rl)`

This gives:

`(A_(big))/(A_(small)) = (R)/(r) * (L)/(l)`

Recall that:

`(L)/(l) = (R)/(r)`

So, we have:

`(A_(big))/(A_(small)) = (R)/(r) * (R)/(r)`

`(A_(big))/(A_(small)) = ((R)/(r))^2`

Make
`A_(big)` the subject

`A_(big) = ((R)/(r))^2 * A_(small)`

Substitute values for
`(R)/(r)` and
`A_(small)`

`A_(big) = ((7)/(6))^2 * 840`

`A_(big) = (49)/(36) * 840`

`A_(big) = (49* 840)/(36)`

`A_(big) = 1143.33cm^2`

Hence, the curved surface area of the big cone is 1143.33cm^2