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June 29th, 2019, 03:41 PM   #1
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What is the area of the shaded region?

I need help to solve this problem. I tried different measures and found that there is a fixed pattern but I don't know how it can be calculated. Any help is highly appreciated.
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Last edited by Farzin; June 29th, 2019 at 03:50 PM.
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June 30th, 2019, 02:45 PM   #2
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I get the area of $\Delta DEF$ to be 25 square units.
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July 1st, 2019, 01:24 AM   #3
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I drew it by a geometry software and got 25 too. How about you, did you also use a software? How did you calculate it?
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July 1st, 2019, 05:45 AM   #4
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Quote:
Originally Posted by Farzin View Post
I drew it by a geometry software and got 25 too. How about you, did you also use a software? How did you calculate it?
I just copied your original and pasted to MS Paint to add labels.

I labeled AC = CF = FE as $\color{red}{\text{a}}$ and FD = DB as $\color{red}{\text{b}}$

Used Pythagoras to determine the lengths of BC and DC.

Used the cosine law on triangle FDC ...

$\left(\sqrt{a^2+b^2-100} \right)^2 = a^2+b^2 - 2ab\cos(90-\theta)$

$a^2+b^2-100 = a^2+b^2 - 2ab\sin{\theta}$

$-100 = -2ab\sin{\theta} \implies \sin{\theta} = \dfrac{50}{ab}$

Area of triangle FED is $A = \dfrac{1}{2}ab\sin{\theta}$

substitute $\dfrac{50}{ab}$ for $\sin{\theta}$ ...

$A = \dfrac{1}{2}ab \cdot \dfrac{50}{ab} = 25$
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July 1st, 2019, 06:58 AM   #5
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Excellent job, thank you very much.
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