Concept explainers
To find: which cardboard box does not meet the requirement.
Answer to Problem 23STP
The correct choice is (D).
Explanation of Solution
Given:
The dimension of box whose surface area is more is 6 in. by 14 in. by 20 in.
A) 7 in. by 10 in. by 24 in
B) 7 in. by 12 in. by 20 in
C) 10 in. by 12 in. by 14 in
D) 5 in. by 16 in. by 21 in
Calculation:
Consider two cardboard has same volume but one has less surface area than the other.
The dimension of box whose surface area is more is 6 in. by 14 in. by 20 in.
The objective is to find the size of box from the given box which does not meet the above criteria.
Consider l represents the length, w represents the width and h represents the height of a rectangular prism.
Find the surface area of the rectangular prism.
The lateral area L of a rectangular prism is, L = Ph.
Here, P is the perimeter of the base and h is the height the rectangular prism.
The surface area S of a rectangular prism is, S = L + 2B .
Here, L is the lateral area and B is the area of the base.
The perimeter P of the base of the prism is,
Here, l is the length and w is the width of the rectangular base.
Thus,
Find the surface area of the original prism as,
First find the area B of the base of the prism.
The area B of the base is,
Here, l is the length and w is the width of the rectangular base.
Thus, the surface area S of the original rectangular prism is
Therefore, the surface area of the rectangular prism is
The volume V of the rectangular box with dimension 6 in. by 14 in. by 20 in. is,
The surface area S of the rectangular box with dimension 6 in. by 14 in. by 20 in. is,
The volume
The surface area
Thus,
The volume
The surface area of the rectangular box with dimension 7 in. by 12 in. by 20 in. is,
Thus,
The volume
The surface area
Thus,
The volume
The surface area
Thus,
Conclusion:
Therefore, the correct choice is (D).
Chapter 12 Solutions
Pre-Algebra Student Edition
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