Difference between revisions of "2018 AMC 10B Problems/Problem 4"
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Let <math>X</math> be the length of the shortest dimension and <math>Z</math> be the length of the longest dimension. Thus, <math>XY = 24</math>, <math>YZ = 72</math>, and <math>XZ = 48</math>. | Let <math>X</math> be the length of the shortest dimension and <math>Z</math> be the length of the longest dimension. Thus, <math>XY = 24</math>, <math>YZ = 72</math>, and <math>XZ = 48</math>. | ||
− | Divide the first to equations to get <math>\frac{Z}{X} = 3</math>. Then, multiply by the last equation to get <math>Z^2 = 144</math>, giving <math>Z = 12</math>. Following, <math>X = 4</math> and <math>Y = 6</math>. The final answer <math>4 + 6 + 12 = 22</math>. <math>\boxed{B}</math> | + | Divide the first to equations to get <math>\frac{Z}{X} = 3</math>. Then, multiply by the last equation to get <math>Z^2 = 144</math>, giving <math>Z = 12</math>. Following, <math>X = 4</math> and <math>Y = 6</math>. |
+ | The final answer <math>4 + 6 + 12 = 22</math>. <math>\boxed{B}</math> |
Revision as of 15:48, 16 February 2018
Problem
A three-dimensional rectangular box with dimensions ,
, and
has faces whose surface areas are
,
,
,
,
, and
square units. What is
+
+
?
Solution
Let be the length of the shortest dimension and
be the length of the longest dimension. Thus,
,
, and
.
Divide the first to equations to get
. Then, multiply by the last equation to get
, giving
. Following,
and
.
The final answer
.