Difference between revisions of "2006 Alabama ARML TST Problems/Problem 7"
(→Solution) |
(→Solution) |
||
| Line 13: | Line 13: | ||
By the [[Law of Cosines]], the square of the length of the side of the larger square, which is also the area of the larger square, is <math>x^2 = 10^2 + 10^2 - 2\cdot10\cdot10\cdot \cos{150^\circ} = 200+100\sqrt{3}</math>. | By the [[Law of Cosines]], the square of the length of the side of the larger square, which is also the area of the larger square, is <math>x^2 = 10^2 + 10^2 - 2\cdot10\cdot10\cdot \cos{150^\circ} = 200+100\sqrt{3}</math>. | ||
| + | |||
| + | ==Solution 2== | ||
==See Also== | ==See Also== | ||
{{ARML box|year=2006|state=Alabama|num-b=6|num-a=8}} | {{ARML box|year=2006|state=Alabama|num-b=6|num-a=8}} | ||
Revision as of 22:41, 2 September 2024
Contents
Problem
Four equilateral triangles are drawn such that each one shares a different side with a square of side length 10. None of the areas of the triangles overlap with the area of the square. The four vertices of the triangles that aren’t vertices of the square are connected to form a larger square. Find the area of this larger square.
Solution
![[asy] defaultpen(linewidth(0.7)); draw((13.66,0)--(5,5)--(0,13.66)--(-5,5)--(-13.66,0)--(-5,-5)--(0,-13.66)--(5,-5)--cycle); draw((5,5)--(-5,5)--(-5,-5)--(5,-5)--cycle); draw((13.66,0)--(0,13.66)--(-13.66,0)--(0,-13.66)--cycle); [/asy]](http://latex.artofproblemsolving.com/4/9/a/49ac5640d7a3ba0ddf4e0081c1fb1d3a5450ac21.png)
Since all the angles in the equilateral triangles are 
, all the angles in the square are 
, the anssen the sides of two adjacent equilateral triangles is 
.
By the Law of Cosines, the square of the length of the side of the larger square, which is also the area of the larger square, is 
.
Solution 2
See Also
| 2006 Alabama ARML TST (Problems) | ||
| Preceded by: Problem 6 |
Followed by: Problem 8 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
