Chapter Closure
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Exercise 150 Page 702
a The surface area of a cylinder equals 2 π r h + 2 π r^2, where r is the radius of the cylinder's base and h is the cylinder's height. The surface area of a rectangular prism equals 2 l w + 2 l h + 2wh, where l is the length, w is the width, and h is the height of the prism.
b Examine the given diagrams. Divide the ribbon into parts whose length is easier to determine.
a The cylindrical box requires less wrapping paper.
b The rectangular box requires less ribbon.
a We consider two boxes: a cylinder with diameter of 6 inches and height of 10 inches, and a rectangular box with dimensions 5 inches by 6 inches by 9 inches. We want to find which of the two boxes requires less wrapping paper. This means we want to find which box has less surface area. Let's find these areas one box at a time.
Cylinder
Let's recall the formula for the surface area of a cylinder. S = 2 π r h + 2 π r^2S = 2 π r h + 2 π r^2
SubstituteII
r= 3, h= 10
S = 2 π ( 3) ( 10) + 2 π ( 3)^2
CalcPow
Calculate power
S = 2 π(3)(10) + 2π(9)
Multiply
Multiply
S = 60π + 18π
AddTerms
Add terms
S = 78π
UseCalc
Use a calculator
S = 245.044226...
RoundInt
Round to nearest integer
S ≈ 254
Rectangular Box
The rectangular box is a rectangular prism. Let's recall the formula for the surface area of a rectangular prism. S = 2 l w + 2 l h + 2 w h Our prism has dimensions 5 inches by 6 inches by 9 inches. Let's substitute 5 for l, 6 for w, and 9 for h in the formula for the surface area and simplify the result.S = 2 l w + 2 l h + 2 wh
SubstituteValues
Substitute values
S = 2( 5)( 6) + 2( 5)( 9) + 2( 6)( 9)
Multiply
Multiply
S = 60 + 90 + 108
AddTerms
Add terms
S = 258
Conclusion
The surface area of the rectangular box is 258 square inches which is more then 254 square inches, the approximate surface area of the cylindrical box. Therefore, the cylindrical box requires less wrapping paper.
b Talila wants to tie two loops of ribbon about the package. We want to know which package requires less ribbon. Let's consider each box one at a time.
Cylinder
Let's take a look at how Talila wants to tie the ribbon around the cylindrical box.
Now, let's think how long is each part of the ribbon. Each of the 2 crosses consists of 2 segments that are the diameters of the base. We have 4 vertical segments, one for each end of the top cross. These segments have the length of cylinder's height. The loop around the cylinder is as long as the base's circumference.
| Part | Times of Appearing | Length |
|---|---|---|
| Cross | 2(2)=4 | 2*diameter |
| Vertical segment | 4 | Cylinder's height |
| Loop | 1 | Base's circumference |
We know that the diameter is 6 inches, the height is 10 inches. The circumference of a circle is π times the diameter. Therefore, the base's circumference is 6π. Let's find the total lengths of ribbon by summing the lengths of all parts. 4(2*6) + 4(10) + 1(6π) ≈ 106.9 The ribbon's length for the cylinder is about 106.9 inches.
Rectangular Box
Let's take a look at how Talila wants to tie the ribbon around the rectangular box.
Looking at the given diagram, we see that she wants to make a cross on each face of the box. Each segment creating the cross is as long as the edges that are parallel to that segment. Each such segment appears exactly 4 times — once per each face that have an edge with the same length as the segment.
The rectangular box has dimensions 5 inches by 6 inches by 9 inches. We have 4 parts of the ribbon that have the length of each of these dimensions. Let's use this fact to find the total length of ribbon needed to wrap the box. 4(5) + 4(6) + 4(9) = 80 The ribbon's length for the rectangular box is 80 inches.
Conclusion
The amount of ribbon needed for the cylindrical box is about 106.9 inches. This is more than the 80 inches required for the rectangular box. Therefore, the rectangular box requires less ribbon.
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