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 "2006 Alabama ARML TST Problems/Problem 7"

(New page: ==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...)
 
(→‎Solution)
Line 10: Line 10:
 
</asy></center>
 
</asy></center>
  
−
Since all the angles in the equilateral triangles are <math>60^\circ</math>, all the angles in the square are <math>90^\circ</math>, the angle between the sides of two adjacent equilateral triangles is <math>360^\circ - (90^\circ + 2\cdot60^\circ) = 150^\circ</math>.  
+
Since all the angles in the equilateral triangles are <math>60^\circ</math>, all the angles in the square are <math>90^\circ</math>, the anssen the sides of two adjacent equilateral triangles is <math>360^\circ - (90^\circ + 2\cdot60^\circ) = 150^\circ</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>.  
+
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>.
  
 
==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 00:57, 17 March 2021

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]

Since all the angles in the equilateral triangles are $60^\circ$, all the angles in the square are $90^\circ$, the anssen the sides of two adjacent equilateral triangles is $360^\circ - (90^\circ + 2\cdot60^\circ) = 150^\circ$.

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 $x^2 = 10^2 + 10^2 - 2\cdot10\cdot10\cdot \cos{150^\circ} = 200+100\sqrt{3}$.

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
Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2006_Alabama_ARML_TST_Problems/Problem_7&oldid=149646"