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November 26th, 2014, 05:44 AM   #1
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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?
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Last edited by skipjack; November 26th, 2014 at 11:51 AM.
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November 26th, 2014, 07:39 AM   #2
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i) Let y = AF; and
ii) Create a new point G which lies on CD vertically above F.

Since A'E = x,

$\displaystyle DA' = \sqrt{x^2 - (10-x)^2} = \sqrt{(x-(10-x))(x+(10-x))} = \sqrt{20(x-5)} = 2\sqrt{5x-25}$

$\displaystyle Area(AEF) = Area(A'EF) = A$

$\displaystyle 2Area(AEF) + Area(A'ED) + Area(A'FG) = Area(AFGD)$
$\displaystyle 2A + \frac{1}{2}(10-x)(2\sqrt{5x-25}) + \frac{1}{2}\cdot 10(y-2\sqrt{5x-25}) = 10y$
$\displaystyle 2A = (x-10)\sqrt{5x-25} + 5y + 10\sqrt{5x-25}$
$\displaystyle 2A = 5y + x\sqrt{5x-25}$

$\displaystyle Area(AEF) = A = \frac{xy}{2}$

Therefore,
$\displaystyle y = \frac{2A}{x}$.

Consequently,
$\displaystyle 2A = \frac{10A}{x} + x\sqrt{5x-25}$
$\displaystyle 2A\left(1 - \frac{5}{x}\right) = x\sqrt{5x-25}$
$\displaystyle A = \frac{x\sqrt{5x-25}}{2\left(1 - \frac{5}{x}\right)}$
$\displaystyle A = \frac{x^2\sqrt{5x-25}}{2(x - 5)}$

Multiplying top and bottom by $\displaystyle \sqrt{5x-25}$ gives

$\displaystyle A = \frac{x^2 \cdot 5(x-5)}{2(x - 5)\sqrt{5x-25}}$
$\displaystyle A = \frac{5x^2}{2\sqrt{5x-25}}$

Last edited by Benit13; November 26th, 2014 at 07:42 AM.
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November 26th, 2014, 12:40 PM   #3
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Quote:
Originally Posted by 9vswt View Post
Where have I gone wrong?
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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