All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars    
perplexus dot info

Home > Shapes > Geometry
Given Common Vertex, Find Area (Posted on 2007-09-30) Difficulty: 2 of 5
An equilateral triangle PQR is drawn inside a unit square LQST such that Q corresponds to the common vertex of the square and the triangle.

Determine the maximum possible area of the triangle PQR.

See The Solution Submitted by K Sengupta    
No Rating

Comments: ( Back to comment list | You must be logged in to post comments.)
Solution Solution | Comment 1 of 3

Let the coordinates of the square be
  L(1,0), Q(0,0), S(0,1), and T(1,1).
For maximum area the coordinates of
the triangle are
  P(1,x), Q(0,0), and R(x,1).
Let s be the side length of the triangle
  s^2 = 1^2 + x^2 = (x-1)^2 + (1-x)^2
      or
  x^2 - 4x + 1 = 0
      or
  x = 2 - sqrt(3)
The area of the triangle is
  Area = (1/2)*s^2*sin(60)
       = (1/2)*(4x)*(sqrt(3)/2)
       = 2*sqrt(3) - 3
      ~= 0.4641016  
 

  Posted by Bractals on 2007-09-30 19:00:52
Please log in:
Login:
Password:
Remember me:
Sign up! | Forgot password


Search:
Search body:
Forums (0)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (3)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Chatterbox:
Copyright © 2002 - 2017 by Animus Pactum Consulting. All rights reserved. Privacy Information