Question

the total surface area is 774 in.²
(round to the nearest integer as needed.)
the volume is □ in.³
(round to the nearest integer as needed.)
find the total surface area and volume of the frustum of a cone.

Explanation:

Step1: Identify the formula for the volume of a frustum of a cone

The formula for the volume \( V \) of a frustum of a cone is \( V=\frac{1}{3}\pi h(R^{2}+Rr + r^{2}) \), where \( h \) is the height of the frustum, \( R \) is the radius of the larger base, and \( r \) is the radius of the smaller base.

From the diagram, the height \( h = 13 \) inches. The diameter of the larger base is \( 15 \) inches, so \( R=\frac{15}{2}=7.5 \) inches. The diameter of the smaller base: let's find the radius \( r \). Wait, maybe we can get \( r \) from the slant height or other? Wait, the surface area is given as 774, but we can also use the dimensions. Wait, the top base: the length from the top is 10? Wait, no, the diagram has 13.2, 10, 13, 15. Wait, maybe the height of the frustum is 13, the larger diameter is 15 (so \( R = 7.5 \)), the smaller diameter: let's see, the slant height? Wait, maybe the smaller radius \( r \): let's check the surface area formula for frustum of cone: \( SA=\pi(R + r)l+\pi R^{2}+\pi r^{2} \), where \( l \) is the slant height. But maybe we can find \( r \) first. Wait, the problem is to find the volume. Let's re - check the dimensions.

Wait, the height \( h = 13 \) in, larger radius \( R=\frac{15}{2}=7.5 \) in. Let's find the smaller radius \( r \). Let's assume that the frustum is part of a cone. Let the height of the smaller cone (the one that is cut off) be \( x \), then the height of the larger cone is \( x + 13 \). The radii are proportional, so \( \frac{r}{R}=\frac{x}{x + 13} \). But maybe we can get \( r \) from the slant height? Wait, the surface area is given as 774. But maybe we can use the given surface area to find \( r \), but that might be complicated. Wait, maybe the smaller diameter is 10? Wait, the diagram has a 10 - unit measurement. Wait, if the smaller diameter is 10, then \( r=\frac{10}{2}=5 \) inches. Let's test this.

Step2: Substitute the values into the volume formula

Now, \( h = 13 \), \( R = 7.5 \), \( r = 5 \).

Substitute into \( V=\frac{1}{3}\pi h(R^{2}+Rr + r^{2}) \)

\( V=\frac{1}{3}\pi\times13\times(7.5^{2}+7.5\times5 + 5^{2}) \)

First, calculate the terms inside the parentheses:

\( 7.5^{2}=56.25 \), \( 7.5\times5 = 37.5 \), \( 5^{2}=25 \)

Sum: \( 56.25+37.5 + 25=118.75 \)

Then, \( V=\frac{1}{3}\pi\times13\times118.75 \)

\( V=\frac{13\times118.75}{3}\pi \)

\( 13\times118.75 = 1543.75 \)

\( V=\frac{1543.75}{3}\pi\approx\frac{1543.75}{3}\times3.1416 \)

\( \frac{1543.75}{3}\approx514.5833 \)

\( 514.5833\times3.1416\approx1616.5 \) (Wait, that can't be right. Wait, maybe I made a mistake in \( r \). Wait, maybe the smaller diameter is not 10. Wait, let's re - examine the diagram. The top has a length of 10? Wait, maybe the height of the frustum is 13, the larger radius \( R = 7.5 \), and the smaller radius \( r \): let's use the surface area formula to find \( r \).

The surface area of frustum \( SA=\pi(R + r)l+\pi R^{2}+\pi r^{2} \), where \( l \) is the slant height. The slant height \( l \) can be found by \( l=\sqrt{h^{2}+(R - r)^{2}} \). We know \( SA = 774 \), \( R = 7.5 \), \( h = 13 \).

Let \( r \) be the radius of the smaller base.

\( SA=\pi(7.5 + r)l+\pi(7.5)^{2}+\pi r^{2}=774 \)

And \( l=\sqrt{13^{2}+(7.5 - r)^{2}}=\sqrt{169+(7.5 - r)^{2}} \)

This is a bit complex. Wait, maybe the initial assumption of \( r = 5 \) is wrong. Wait, let's check the volume again with correct \( r \). Wait, maybe the smaller diameter is 10, so \( r = 5 \), let's recalculate the volume:

\( V=\frac{1}{3}\times\pi\times13\times(7.5^{2}+7.5\times5 + 5^{2})=\frac{1}{3}\times\pi\times13\times(56.25 + 37.5+25)=\frac{1}{3}\times\pi\times13\times118.75 \)

\( 13\times118.75 = 1543.75 \)

\( \frac{1543.75}{3}\times\pi\approx514.5833\times3.1416\approx1616 \). But this seems large. Wait, maybe the height is 13, and the radii are \( R = 7.5 \) and \( r = 2 \)? No, that doesn't make sense. Wait, maybe I misread the diagram. Wait, the height of the frustum is 13, the larger radius \( R = 7.5 \), the smaller radius \( r \): let's look at the 10 - unit measurement. Maybe the 10 is the diameter of the smaller base, so \( r = 5 \). Then the volume is:

\( V=\frac{1}{3}\pi\times13\times(7.5^{2}+7.5\times5 + 5^{2})=\frac{1}{3}\pi\times13\times(56.25 + 37.5+25)=\frac{1}{3}\pi\times13\times118.75 \)

\( 13\times118.75 = 1543.75 \)

\( \frac{1543.75}{3}\pi\approx\frac{1543.75\times3.1416}{3}\approx\frac{4849.5}{3}=1616.5\approx1617 \)

Wait, but let's check with the surface area. The surface area of frustum is \( \pi(R + r)l+\pi R^{2}+\pi r^{2} \). If \( R = 7.5 \), \( r = 5 \), \( l=\sqrt{13^{2}+(7.5 - 5)^{2}}=\sqrt{169 + 6.25}=\sqrt{175.25}\approx13.24 \). Then \( \pi(7.5 + 5)\times13.24+\pi\times7.5^{2}+\pi\times5^{2}=\pi\times12.5\times13.24+\pi\times56.25+\pi\times25 \)

\( 12.5\times13.24 = 165.5 \), so \( 165.5\pi+56.25\pi + 25\pi=(165.5 + 56.25+25)\pi=246.75\pi\approx246.75\times3.1416\approx775 \), which is close to the given 774 (due to rounding). So our values of \( R = 7.5 \), \( r = 5 \), \( h = 13 \) are correct.

So the volume \( V=\frac{1}{3}\pi\times13\times(7.5^{2}+7.5\times5 + 5^{2})=\frac{1}{3}\pi\times13\times118.75\approx1617 \)

Answer:

\( 1617 \)