November 26th, 2014, 05:44 AM  #1 
Newbie Joined: Nov 2014 From: Singapore Posts: 1 Thanks: 0  Find the area of the triangle
Hello, I'd like to know the how to derive the area of the triangle. I have tried the steps shown in the attachments. Where have I gone wrong? Last edited by skipjack; November 26th, 2014 at 11:51 AM. 
November 26th, 2014, 07:39 AM  #2 
Senior Member Joined: Apr 2014 From: Glasgow Posts: 2,074 Thanks: 695 Math Focus: Physics, mathematical modelling, numerical and computational solutions 
i) Let y = AF; and ii) Create a new point G which lies on CD vertically above F. Since A'E = x, $\displaystyle DA' = \sqrt{x^2  (10x)^2} = \sqrt{(x(10x))(x+(10x))} = \sqrt{20(x5)} = 2\sqrt{5x25}$ $\displaystyle Area(AEF) = Area(A'EF) = A$ $\displaystyle 2Area(AEF) + Area(A'ED) + Area(A'FG) = Area(AFGD)$ $\displaystyle 2A + \frac{1}{2}(10x)(2\sqrt{5x25}) + \frac{1}{2}\cdot 10(y2\sqrt{5x25}) = 10y$ $\displaystyle 2A = (x10)\sqrt{5x25} + 5y + 10\sqrt{5x25}$ $\displaystyle 2A = 5y + x\sqrt{5x25}$ $\displaystyle Area(AEF) = A = \frac{xy}{2}$ Therefore, $\displaystyle y = \frac{2A}{x}$. Consequently, $\displaystyle 2A = \frac{10A}{x} + x\sqrt{5x25}$ $\displaystyle 2A\left(1  \frac{5}{x}\right) = x\sqrt{5x25}$ $\displaystyle A = \frac{x\sqrt{5x25}}{2\left(1  \frac{5}{x}\right)}$ $\displaystyle A = \frac{x^2\sqrt{5x25}}{2(x  5)}$ Multiplying top and bottom by $\displaystyle \sqrt{5x25}$ gives $\displaystyle A = \frac{x^2 \cdot 5(x5)}{2(x  5)\sqrt{5x25}}$ $\displaystyle A = \frac{5x^2}{2\sqrt{5x25}}$ Last edited by Benit13; November 26th, 2014 at 07:42 AM. 
November 26th, 2014, 12:40 PM  #3 
Global Moderator Joined: Dec 2006 Posts: 18,154 Thanks: 1421  You incorrectly identified the triangle similar to triangle EDA'. I'll use Benit13's notation and ignore the inappropriate "cm" in the diagram. By similar triangles, DA'/GF = EA'/A'F = x/y, and so DA' = 10x/y. As (10x/y)² + (10  x)² = x², y² = 100x²/(20x  100). Hence area(A'EF) = area(AEF) = xy/2 = 5x²/(2√(5x  25)) (since x > 5). Can you make any progress with part (b) of the question? 

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