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Geometry and Trigonometry / Area and volume Difficulty: Hard

Rectangle $ABCD$ is similar to rectangle $EFGH$ . The area of rectangle $ABCD$ is $648$ square inches, and the area of rectangle $EFGH$ is $72$ square inches. The length of the longest side of rectangle $ABCD$ is $36$ inches. What is the length, in inches, of the longest side of rectangle $EFGH$ ?

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Explanation

Choice C is correct. It's given that rectangle $A B C D$ is similar to rectangle $E F G H$ . Therefore, if the length of each side of rectangle $A B C D$ is $k$ times the length of the corresponding side of rectangle $E F G H$ , then the area of rectangle $A B C D$ is $k^{2}$ times the area of rectangle $E F G H$ . It’s given that the area of rectangle $A B C D$ is $648$ square inches and the area of rectangle $E F G H$ is $72$ square inches. It follows that $k^{2} = \frac{648}{72}$ , or $k squared equals 9$ . Taking the square root of each side of this equation yields $k equals StartRoot 9 EndRoot$ , or $k equals 3$ . It follows that the length of each side of rectangle $A B C D$ is $3$ times the length of the corresponding side of rectangle $E F G H$ . It’s given that the length of the longest side of rectangle $A B C D$ is $36$ inches. Therefore, $36$ inches is $3$ times the length of the longest side of rectangle $E F G H$ , and the longest side of rectangle $E F G H$ is equal to $StartFraction 36 Over 3 EndFraction$ , or $12$ , inches.

Choice A is incorrect. This is the length, in inches, of the longest side of a rectangle with side lengths that are $\frac{1}{9}$ the corresponding side lengths of rectangle $A B C D$ , rather than a rectangle with an area that is $\frac{1}{9}$ the area of rectangle $A B C D$ .

Choice B is incorrect. This is the factor by which the area of rectangle $A B C D$ is larger than the area of rectangle $E F G H$ , not the length, in inches, of the longest side of rectangle $E F G H$ .

Choice D is incorrect. This is the length, in inches, of the longest side of rectangle $A B C D$ , not rectangle $E F G H$ .