QUESTION IMAGE
Question
ma.912.gr.4.3
volume, surface area, and dilations
use the table below to answer the following questions.
original cylinder
dilation with
scale factor k
new
surface
area
new
volume
radius of the base is 5 inches
height of the cylinder is 12
inches
surface area =______in²
inches volume =______in³
k = 2
k = 3
k = 1/2
part a. determine the surface area and volume of the given
cylinder.
part b. given the three different dilations determine the new
surface areas and volumes.
Part A
Surface Area
- Step1: Recall the formula for the surface area of a cylinder
The formula for the surface area of a cylinder is \(S = 2\pi r(r + h)\), where \(r\) is the radius and \(h\) is the height.
- Step2: Substitute the given values
Given \(r = 5\) inches and \(h = 12\) inches.
Volume
- Step1: Recall the formula for the volume of a cylinder
The formula for the volume of a cylinder is \(V=\pi r^{2}h\).
- Step2: Substitute the given values
Given \(r = 5\) inches and \(h = 12\) inches.
Part B
For \(k = 2\)
- Surface Area
The formula for the surface area of a dilated cylinder is \(S_{new}=k^{2}S_{original}\).
Since \(S_{original}=170\pi\approx533.8\) and \(k = 2\), then \(S_{new}=2^{2}\times533.8 = 4\times533.8=2135.2\)
- Volume
The formula for the volume of a dilated cylinder is \(V_{new}=k^{3}V_{original}\).
Since \(V_{original}=300\pi\approx942\) and \(k = 2\), then \(V_{new}=2^{3}\times942=8\times942 = 7536\)
For \(k = 3\)
- Surface Area
Using \(S_{new}=k^{2}S_{original}\), with \(S_{original}=170\pi\approx533.8\) and \(k = 3\)
\(S_{new}=3^{2}\times533.8=9\times533.8 = 4804.2\)
- Volume
Using \(V_{new}=k^{3}V_{original}\), with \(V_{original}=300\pi\approx942\) and \(k = 3\)
\(V_{new}=3^{3}\times942=27\times942=25434\)
For \(k=\frac{1}{2}\)
- Surface Area
Using \(S_{new}=k^{2}S_{original}\), with \(S_{original}=170\pi\approx533.8\) and \(k=\frac{1}{2}\)
\(S_{new}=(\frac{1}{2})^{2}\times533.8=\frac{1}{4}\times533.8 = 133.45\)
- Volume
Using \(V_{new}=k^{3}V_{original}\), with \(V_{original}=300\pi\approx942\) and \(k=\frac{1}{2}\)
\(V_{new}=(\frac{1}{2})^{3}\times942=\frac{1}{8}\times942=117.75\)
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- Part A
- Surface Area: \(533.8\space in^{2}\)
- Volume: \(942\space in^{3}\)
- Part B
- For \(k = 2\):
- New Surface Area: \(2135.2\space in^{2}\)
- New Volume: \(7536\space in^{3}\)
- For \(k = 3\):
- New Surface Area: \(4804.2\space in^{2}\)
- New Volume: \(25434\space in^{3}\)
- For \(k=\frac{1}{2}\):
- New Surface Area: \(133.45\space in^{2}\)
- New Volume: \(117.75\space in^{3}\)