Difference between revisions of "2014 AMC 10A Problems/Problem 18"
Isabelchen (talk | contribs) (→Solution 3) |
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Because <math>AD=AB,</math> | Because <math>AD=AB,</math> | ||
+ | <math>\sqrt{x_3^2 +4^2} = \sqrt{(x_1)^2+1^2}</math> | ||
<math>\sqrt{(- \frac{4}{x_1})^2 +4^2} = \sqrt{(x_1)^2+1^2}</math> | <math>\sqrt{(- \frac{4}{x_1})^2 +4^2} = \sqrt{(x_1)^2+1^2}</math> | ||
<math>\frac{16}{x_1^2}+16=x_1^2+1</math> | <math>\frac{16}{x_1^2}+16=x_1^2+1</math> | ||
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~isabelchen | ~isabelchen | ||
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==Solution 3== | ==Solution 3== |
Revision as of 11:34, 3 October 2021
Problem
A square in the coordinate plane has vertices whose -coordinates are , , , and . What is the area of the square?
Solution 1
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 .
Solution 2
By translation, we can move the square with point at the origin. Then, . We will use the relationship among the 4 sides of being perpendicular and equal.
The slope of is .
Because is perpendicular to , the slope of . From the information above we could have the equation:
Because is perpendicular to , the slope of . From the information above we could have the equation:
Because
Note that the square with is just the reflection of square with over the origin. I will use .
~isabelchen
Solution 3
In this solution, we will use the fact that the diagonals of a square bisect each other, they are perpendicular to each other, and they are equal in length.
Using the fact that the diagonals bisect each other, we get the equation:
Now we use the fact that the diagonals are perpendicular to each other:
Using the fact that the diagonals are equal in length, we get the equation:
Now we have 3 equations with 3 variables:
We substitute into the 2 other equations:
Now we have 2 equations of and :
This is the same equation as solution . So
~isabelchen
Video Solution
https://www.youtube.com/watch?v=iPPQUrNE4RE
~ naren_pr
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 |
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