Difference between revisions of "2014 AMC 10A Problems/Problem 18"
Isabelchen (talk | contribs) (→Solution 2) |
Erics son07 (talk | contribs) (→Solution 3) |
||
Line 97: | Line 97: | ||
~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen] | ~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen] | ||
+ | |||
+ | ==Solution 4 (not rigorous)== | ||
+ | Draw it out, by inspection the coordinates are <math>(-1, 4)</math>, <math>(0, 0)</math>, <math>(4, 1)</math>, and <math>(3, 5)</math>. The side length is <math>\sqrt{17}, [ABCD]=\boxed{\textbf{(B) }17}</math> | ||
+ | ~JH. L | ||
== Video Solution== | == Video Solution== |
Latest revision as of 03:18, 20 June 2022
Contents
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
.
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
Solution 4 (not rigorous)
Draw it out, by inspection the coordinates are ,
,
, and
. The side length is
~JH. L
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.