A cone has a height of $16 c m$ $16 cm$ and its base has a radius of $8 c m$ $8 cm$ . 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 $8 c m$ $8 cm$ . 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 · 2017-09-17T23:05:49

$A_{s} \approx 528.233 c m^{2}$ $A_s~~528.233cm^2$

Explanation:

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, $h = 7$ $h=7$ and $R_{1} = 8$ $R_1=8$ . 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 8cm. This means the relationship between the height and the radius is equal to $\frac{16}{7}$ $16/7$ . In the frustum, we have the height, but the smaller radius is unknown, which is 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
$\frac{16}{7} = \frac{7}{R_{2}}$ $16/7 = 7/R_2$
By cross multiplying, we find that

$49 = 16 R_{2}$ $49=16R_2$

Finally we divide both sides by 16, to get

$3.0625 = R_{2}$ $3.0625=R_2$

Now we have the value 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, to get
$s = \sqrt{{(8 - 3.0625)}^{2} + 7^{2}}$ $s=sqrt((8-3.0625)^2+7^2$
$s = \sqrt{{4.9375}^{2} + 49}$ $s=sqrt(4.9375^2+49$
$s = \sqrt{73.37890625}$ $s=sqrt(73.37890625)$

Now we know $s, h, R_{1}$ $s, h, R_1$ and $R_{2}$ $R_2$ . All that is left is to calculate the Surface Area, which we do using this formula:
$A_{s} = π (s (R_{1} + R_{2}) + {(R_{1})}_{1}^{2} + {(R_{2})}_{2}^{2})$ $A_s=pi(s(R_1+R_2)+(R_1)^2+(R_2)^2)$ . How one gets to this formula can be shown with this picture:
www.ditutor.com
http://www.ditutor.com/solid_gometry/frustum_cone.html

Where the bottom circle has the larger radius, and the top circle has the smaller radius.

We have all the values needed to solve this question, so lets plug them in to get
$A_{s} = π (\sqrt{73.37890625} (8 + 3.0625) + 8^{2} + {3.0625}^{2})$ $A_s=pi(sqrt(73.37890625)(8+3.0625) + 8^2+3.0625^2)$
$A_{s} = π (94.763022 + 64 + 9.37890625)$ $A_s=pi(94.763022+64+9.37890625)$
$A_{s} = π (168.14192825$ $A_s=pi(168.14192825$
$A_{s} \approx 528.233$ $A_s~~528.233$
Therefor the surface area of the bottom segment of the cone is 528.233 $c m^{2}$ $cm^2$

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 $8 c m$ $8 cm$ . 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 $8 c m$ $8 cm$ . 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?