Difference between revisions of "2014 AMC 10A Problems/Problem 18"

Revision as of 10:52, 13 August 2014

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 $\textbf{(B) }17$ .

2014 AMC 10A (Problems • Answer Key • Resources)
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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions.

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 <math> \textbf{(A)}\ 16\qquad\textbf{(B)}\ 17\qquad\textbf{(C)}\ 25\qquad\textbf{(D)}\ 26\qquad\textbf{(E)}\ 27 </math>
+[[Category: Introductory Geometry Problems]]
 ==Solution==

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Difference between revisions of "2014 AMC 10A Problems/Problem 18"

Revision as of 10:52, 13 August 2014

Problem

Solution

See Also