Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Solution)
(→‎Solution)
Line 8: Line 8:
 
Let the points be <math>A=(x_1,0)</math>, <math>B=(x_2,1)</math>, <math>C=(x_3,5)</math>, and <math>D=(x_4,4)</math>
 
Let the points be <math>A=(x_1,0)</math>, <math>B=(x_2,1)</math>, <math>C=(x_3,5)</math>, and <math>D=(x_4,4)</math>
  
−
Note that the difference in <math>x</math> value of <math>B</math> and <math>C</math> is <math>4</math>. By rotational symmetry of the square, the difference in <math>y</math> value of <math>A</math> and <math>B</math> is also <math>4</math>. Note that the difference in <math>x</math> value of <math>A</math> and <math>B</math> is <math>1</math>. We now know that <math>AB</math>, the side length of the square, is equal to <math>\sqrt{1^2+4^2}=\sqrt{17}</math>, so the area is <math>\textbf{(B) }17</math>.
+
Note that the difference in <math>x</math> value of <math>B</math> and <math>C</math> is <math>4</math>. By rotational symmetry of the square, the difference in <math>x</math> value of <math>A</math> and <math>B</math> is also <math>4</math>. Note that the difference in <math>y</math> value of <math>A</math> and <math>B</math> is <math>1</math>. We now know that <math>AB</math>, the side length of the square, is equal to <math>\sqrt{1^2+4^2}=\sqrt{17}</math>, so the area is <math>\textbf{(B) }17</math>.
  
 
==See Also==
 
==See Also==

Revision as of 14:44, 10 February 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 $x$ 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 $\textbf{(B) }17$.

See Also

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

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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2014_AMC_10A_Problems/Problem_18&oldid=59562"