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

 Given Common Vertex, Find Area (Posted on 2007-09-30)
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 | 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 ofthe 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

 Search: Search body:
Forums (0)