2014 AMC 10A Problems/Problem 18

Problem

A square in the coordinate plane has vertices whose $y$ -coordinates are $0$ , $1$ , $4$ , and $5$ . What is the area of the square?

$\textbf{(A)}\ 16\qquad\textbf{(B)}\ 17\qquad\textbf{(C)}\ 25\qquad\textbf{(D)}\ 26\qquad\textbf{(E)}\ 27$

Solution

Let the points be $A=(x_1,0)$ , $B=(x_2,1)$ , $C=(x_3,5)$ , and $D=(x_4,4)$

Note that the difference in $y$ value of $B$ and $C$ is $4$ . By rotational symmetry of the square, the difference in $x$ value of $A$ and $B$ is also $4$ . Note that the difference in $y$ value of $A$ and $B$ is $1$ . We now know that $AB$ , the side length of the square, is equal to $\sqrt{1^2+4^2}=\sqrt{17}$ , so the area is $\boxed{\textbf{(B) }17}$ .

2014 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 17	Followed by Problem 19
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All AMC 10 Problems and Solutions

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2014 AMC 10A Problems/Problem 18

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