June 29th, 2019, 03:41 PM  #1 
Newbie Joined: Mar 2017 From: Norway Posts: 26 Thanks: 0  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.
Last edited by Farzin; June 29th, 2019 at 03:50 PM. 
June 30th, 2019, 02:45 PM  #2 
Math Team Joined: Jul 2011 From: Texas Posts: 3,044 Thanks: 1627 
I get the area of $\Delta DEF$ to be 25 square units.

July 1st, 2019, 01:24 AM  #3 
Newbie Joined: Mar 2017 From: Norway Posts: 26 Thanks: 0 
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?

July 1st, 2019, 05:45 AM  #4  
Math Team Joined: Jul 2011 From: Texas Posts: 3,044 Thanks: 1627  Quote:
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 ... $\left(\sqrt{a^2+b^2100} \right)^2 = a^2+b^2  2ab\cos(90\theta)$ $a^2+b^2100 = a^2+b^2  2ab\sin{\theta}$ $100 = 2ab\sin{\theta} \implies \sin{\theta} = \dfrac{50}{ab}$ Area of triangle FED is $A = \dfrac{1}{2}ab\sin{\theta}$ substitute $\dfrac{50}{ab}$ for $\sin{\theta}$ ... $A = \dfrac{1}{2}ab \cdot \dfrac{50}{ab} = 25$  
July 1st, 2019, 06:58 AM  #5 
Newbie Joined: Mar 2017 From: Norway Posts: 26 Thanks: 0 
Excellent job, thank you very much.


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