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Surface Area Given Similar Solids

The smaller cone has a surface area of 11.74 inches[tex]$^2$[/tex]. Complete the last step to determine the surface area of the larger cone.

1. The scale factor of the larger to the smaller is [tex]$\frac{3.5}{2.1}$[/tex], or [tex]$\frac{5}{3}$[/tex].
2. The surface area will change by the square of the scale factor, which is [tex]$\frac{5^2}{3^2}$[/tex], or [tex]$\frac{25}{9}$[/tex].
3. Let the surface area of the larger cone be [tex]$x$[/tex]. Then, the proportion is [tex]$\frac{25}{9} = \frac{x}{11.74}$[/tex].
4. Solve for [tex]$x$[/tex] and round to the nearest hundredth. The surface area of the larger cone is about [tex]$\square$[/tex] inches[tex]$^2$[/tex].



Answer :

GinnyAnswer
Sure! Let's go through the solution step-by-step.

1. Understanding the Problem
- We have two similar cones: a smaller cone with a surface area of 11.74 square inches.
- The scaling factor from the smaller cone to the larger cone is given as either [tex]\(\frac{3.5}{2.1}\)[/tex] or [tex]\(\frac{5}{3}\)[/tex].

2. Determine the Scale Factor
- First, verify the scale factor:
- [tex]\(\frac{3.5}{2.1} = \frac{35}{21} = \frac{5}{3}\)[/tex]
- This confirms that the scale factor from the smaller cone to the larger cone is indeed [tex]\(\frac{5}{3}\)[/tex].

3. Calculate the Square of the Scale Factor
- Since surface areas of similar figures change by the square of the scale factor:
[tex]\[ \left(\frac{5}{3}\right)^2 = \frac{5^2}{3^2} = \frac{25}{9} \][/tex]

4. Set Up the Proportion
- Let [tex]\( x \)[/tex] be the surface area of the larger cone. From the given proportion:
[tex]\[ \frac{25}{9} = \frac{x}{11.74} \][/tex]

5. Solve for [tex]\( x \)[/tex]
- To solve for [tex]\( x \)[/tex], we multiply both sides of the proportion by the surface area of the smaller cone (11.74):
[tex]\[ x = \frac{25}{9} \times 11.74 \][/tex]
- This gives us:
[tex]\[ x = 32.6111\ldots \][/tex]

6. Round to the Nearest Hundredth
- Now round the result to the nearest hundredth:
[tex]\[ x \approx 32.61 \][/tex]

Therefore, the surface area of the larger cone is about [tex]\( 32.61 \)[/tex] square inches.