A cone has a height of $16 c m$ $16 cm$ and its base has a radius of $3 c m$ $3 cm$ . If the cone is horizontally cut into two segments $7 c m$ $7 cm$ from the base, what would the surface area of the bottom segment be?

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A cone has a height of $16 c m$ $16 cm$ and its base has a radius of $3 c m$ $3 cm$ . If the cone is horizontally cut into two segments $7 c m$ $7 cm$ from the base, what would the surface area of the bottom segment be?

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Answer 1 · 2017-09-21T00:10:32

By cutting off a segment of a cone parallel to the base, you create what is known as a frustum. There are four important pieces of information in a fustrum: The height $(h)$ $(h)$ , the larger radius $(R_{1})$ $(R_1)$ , the smaller radius $(R_{2})$ $(R_2)$ and the slant $(s)$ $(s)$ . In the question asked we know two of these variables, h=7 and R1=3. The first thing we need to do is find $R_{2}$ $R_2$ .

By looking at the question, we see the original cone has a height of 16cm and a radius of 3cm. This means the relationship between the height and the radius is equal to $\frac{16}{3}$ $16/3$ . In the frustum, we have the height, but the smaller radius is unknown, which makes the relationship of height to radius equal to $\frac{7}{R_{2}}$ $7/R_2$ . Since the ratios of the cone have been unchanged while making it a fustrum, we can safely say that the height-radius ratio of the cone is the same in the fustrum, so
$\frac{16}{3} = \frac{7}{R_{2}}$ $16/3=7/R_2$ . By cross-multiplying, we find that
$21 = 16 R_{2}$ $21=16R_2$ . Divide both sides by 16, and we get
$\frac{21}{16} = R_{2}$ $21/16=R_2$
$1.3125 = R_{2}$ $1.3125=R_2$

Now we have the values of $h, R_{1} and R_{2}$ $h, R_1 and R_2$ , all that is left is $s$ $s$ . The formula for finding $s$ $s$ is as follows:
$s = \sqrt{{(R_{1} - R_{2})}_{1}^{2} + h^{2}}$ $s=sqrt((R_1-R_2)^2+h^2)$
By looking at the image below, you should get an understanding of how this works using Pythagoras' Theorem.
www.ditutor.com
http://www.ditutor.com/solid_gometry/frustum_cone.html
Now we simply plug in the values we have into the formula, to get
$s = \sqrt{{(3 - 1.3125)}^{2} + 7^{2}}$ $s=sqrt((3-1.3125)^2+7^2)$
$s = \sqrt{2.8476 + 49}$ $s=sqrt(2.8476+49)$
$s = \sqrt{51.8476}$ $s=sqrt(51.8476)$

Now we know all the values necessary to calculate the Surface Area, the formula for which is as follows:
$A_{s} = π (s (R_{1} + R_{2}) + {(R_{1})}_{1}^{2} + (R_{2}))$ $A_s=pi(s(R_1+R_2)+(R_1)^2+(R_2))$ . This formula is derived from showing the 2-D map of a fustrum, such as below, where the bottom circle has $R_{1}$ $R_1$ , and the top circle has $R_{2}$ $R_2$ .
www.ditutor.com
http://www.ditutor.com/solid_gometry/frustum_cone.html

Lastly, we plug in all our known values into the equation to get
$A_{s} = π (\sqrt{51.8476} (3 - 1.3125) + 3^{2} + {(1.3125)}^{2})$ $A_s=pi(sqrt(51.8476)(3-1.3125)+3^2+(1.3125)^2)$
$A_{s} = π (12.151 + 9 + 1.7226)$ $A_s=pi(12.151+9+1.7226)$
$A_{s} = π (22.874)$ $A_s=pi(22.874)$
$A_{s} \approx 71.859 c m^{2}$ $A_s~~71.859cm^2$
Therefore the surface area of the bottom segment of the cone is roughly 71.859 $c m^{2}$ $cm^2$ .

I hope I helped

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A cone has a height of $16 c m$ $16 cm$ and its base has a radius of $3 c m$ $3 cm$ . If the cone is horizontally cut into two segments $7 c m$ $7 cm$ from the base, what would the surface area of the bottom segment be?

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A cone has a height of $16 c m$ $16 cm$ and its base has a radius of $3 c m$ $3 cm$ . If the cone is horizontally cut into two segments $7 c m$ $7 cm$ from the base, what would the surface area of the bottom segment be?