Difference between revisions of "2014 AMC 10A Problems/Problem 18"
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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>y</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>. | + | Note that the difference in <math>y</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>\boxed{\textbf{(B) }17}</math>. |
==See Also== | ==See Also== |
Revision as of 07:17, 7 July 2015
Problem
A square in the coordinate plane has vertices whose -coordinates are , , , and . What is the area of the square?
Solution
Let the points be , , , and
Note that the difference in value of and is . By rotational symmetry of the square, the difference in value of and is also . Note that the difference in value of and is . We now know that , the side length of the square, is equal to , so the area is .
See Also
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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