2018 AMC 10B Problems/Problem 4
Contents
Problem
A three-dimensional rectangular box with dimensions , , and has faces whose surface areas are , , , , , and square units. What is + + ?
Solution 1
Let be the length of the shortest dimension and be the length of the longest dimension. Thus, , , and . Divide the first two equations to get . Then, multiply by the last equation to get , giving . Following, and .
The final answer is .
Solution 2
Simply guess and check to find that the dimensions are by by . Therefore, the answer is .
Solution 3
If you find the GCD of , , and you get your first number, . After this, do and to get and , the other 2 numbers. When you add up your numbers, you get which is .
Solution 4
Since the surface areas of the faces are the product of two of the dimensions. Therefore, , , and . You can multiply , which simplifies to which means that the volume equals. The individual dimensions, , , and can be found by doing , , and , which yields , , and . Adding this up, we have that which is - Solution by smartninja2000
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See Also
2018 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 3 |
Followed by Problem 5 | |
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