December 14th, 2013, 11:04 PM  #1 
Member Joined: Dec 2013 Posts: 86 Thanks: 1  GeometryTriangle
[attachment=0:1a0w2j4t]untitled.JPG[/attachment:1a0w2j4t] Q: In a right triangle ABC,what is the maximum possible area of a square that can be inscribed when one of its vertices coincide with the vertex of the right angle of the triangle ? (a) (b) (c) (d) Ansd). Now can this question be solved algebraically without using trigonometric ratios? Request a nontrigonometric solution. Thanks in advance! 
December 14th, 2013, 11:28 PM  #2 
Senior Member Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,204 Thanks: 511 Math Focus: Calculus/ODEs  Re: GeometryTriangle
Let the triangle above the square, which is similar to the larger triangle in which the square is inscribed have base and height . The area of the square is then: By similarity, we have: We also have: Hence: And so: 
December 15th, 2013, 12:40 AM  #3 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 13,296 Thanks: 934  Re: GeometryTriangle
BC = a, AC = b, square sides = x; then AD = b  x and BF = ax : Code: B ax F x E x x C x D bx A (a  x) / x = x / (b  x) x^2 = (a  x)(b  x) x^2 = ab  ax  bx + x^2 ax + bx = ab x = ab / (a + b) ........area = x^2 

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