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December 15th, 2013, 12:04 AM   #1
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Geometry-Triangle

[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)$a/b$

(b)$ab/(a+b)$

(c)$(a+b)/ab$

(d)$[ab/(a+b)]^2$

Ansd). Now can this question be solved algebraically without using trigonometric ratios? Request a non-trigonometric solution.
Thanks in advance!
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 December 15th, 2013, 12:28 AM #2 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,209 Thanks: 517 Math Focus: Calculus/ODEs Re: Geometry-Triangle Let the triangle above the square, which is similar to the larger triangle in which the square is inscribed have base $a_1$ and height $b_1$. The area of the square is then: $A=a_1^2$ By similarity, we have: $a_1=b_1\frac{a}{b}$ We also have: $b_1=b-a_1$ Hence: $a_1=$$b-a_1$$\frac{a}{b}$ $a_1$$1+\frac{a}{b}$$=a$ $a_1=\frac{ab}{a+b}$ And so: $A=$$\frac{ab}{a+b}$$^2$
 December 15th, 2013, 01:40 AM #3 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 13,795 Thanks: 970 Re: Geometry-Triangle BC = a, AC = b, square sides = x; then AD = b - x and BF = a-x : Code: B a-x F x E x x C x D b-x A Since triangle BEF is similar to triangle EAD: (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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