Difference between revisions of "2021 JMC 10 Problems/Problem 7"

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Latest revision as of 16:10, 1 April 2021

Problem

For some real $x,$ the area of a square equals $3x+1$ and the product of the lengths of its diagonals equals $5x+7.$ What is the perimeter of this square?

$\textbf{(A) } 16 \qquad\textbf{(B) } 17 \qquad\textbf{(C) } 18 \qquad\textbf{(D) } 19 \qquad\textbf{(E) } 20$

Solution

Let the square have side length $s.$ The diagonals have length $s\sqrt{2}.$ So, the product of the length of both diagonals is $2s^2$ and the square's area is $s^2,$ which is half of the product of the diagonals. We have $5x+7 = 2(3x+1) \implies x=5.$ The area is $3\cdot5+1 = 16,$ implying that the side length is $\sqrt{16}=4$ and the perimeter is $4\cdot 4 = 16.$