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 $\boxed{\textbf{(B) }17}$.

Video Solution

https://www.youtube.com/watch?v=iPPQUrNE4RE

~ naren_pr

See Also

2014 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 17
Followed by
Problem 19
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 10 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Invalid username
Login to AoPS