Answer:
[tex]A_{big} = 1143.33cm^2[/tex]
Explanation:
The given parameters are:
[tex]V_{small} = 1080[/tex]
[tex]V_{big} = 1715[/tex]
[tex]C_{small} = 840[/tex]
Required
Determine the curved surface area of the big cone
The volume of a cone is:
[tex]V = \frac{1}{3}\pi r^2h[/tex]
For the big cone:
[tex]V_{big} = \frac{1}{3}\pi R^2H[/tex]
Where
R = radius of the big cone and H = height of the big cone
For the small cone:
[tex]V_{small} = \frac{1}{3}\pi r^2h[/tex]
Where
r = radius of the small cone and H = height of the small cone
Because both cones are similar, then:
[tex]\frac{H}{h} = \frac{R}{r}[/tex]
and
[tex]\frac{V_{big}}{V_{small}} = \frac{\frac{1}{3}\pi R^2H}{\frac{1}{3}\pi r^2h}[/tex]
[tex]\frac{V_{big}}{V_{small}} = \frac{R^2H}{r^2h}[/tex]
Substitute values for Vbig and Vsmall
[tex]\frac{1715}{1080} = \frac{R^2H}{r^2h}[/tex]
Recall that:[tex]\frac{H}{h} = \frac{R}{r}[/tex]
So, we have:
[tex]\frac{1715}{1080} = \frac{R^2*R}{r^2*r}[/tex]
[tex]\frac{1715}{1080} = \frac{R^3}{r^3}[/tex]
Take cube roots of both sides
[tex]\sqrt[3]{\frac{1715}{1080}} = \frac{R}{r}[/tex]
Factorize
[tex]\sqrt[3]{\frac{343*5}{216*5}} = \frac{R}{r}[/tex]
[tex]\sqrt[3]{\frac{343}{216}} = \frac{R}{r}[/tex]
[tex]\frac{7}{6} = \frac{R}{r}[/tex]
The curved surface area is calculated as:
[tex]Area = \pi rl[/tex]
Where
[tex]l = slant\ height[/tex]
For the big cone:
[tex]A_{big} = \pi RL[/tex]
For the small cone
[tex]A_{small} = \pi rl[/tex]
Because both cones are similar, then:
[tex]\frac{L}{l} = \frac{R}{r}[/tex]
and
[tex]\frac{A_{big}}{A_{small}} = \frac{\pi RL}{\pi rl}[/tex]
[tex]\frac{A_{big}}{A_{small}} = \frac{RL}{rl}[/tex]
This gives:
[tex]\frac{A_{big}}{A_{small}} = \frac{R}{r} * \frac{L}{l}[/tex]
Recall that:
[tex]\frac{L}{l} = \frac{R}{r}[/tex]
So, we have:
[tex]\frac{A_{big}}{A_{small}} = \frac{R}{r} * \frac{R}{r}[/tex]
[tex]\frac{A_{big}}{A_{small}} = (\frac{R}{r})^2[/tex]
Make [tex]A_{big}[/tex] the subject
[tex]A_{big} = (\frac{R}{r})^2 * A_{small}[/tex]
Substitute values for [tex]\frac{R}{r}[/tex] and [tex]A_{small}[/tex]
[tex]A_{big} = (\frac{7}{6})^2 * 840[/tex]
[tex]A_{big} = \frac{49}{36} * 840[/tex]
[tex]A_{big} = \frac{49* 840}{36}[/tex]
[tex]A_{big} = 1143.33cm^2[/tex]
Hence, the curved surface area of the big cone is 1143.33cm^2
