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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.
Attached Images photo_2019-06-29_20-43-34.jpg (56.8 KB, 17 views)

Last edited by Farzin; June 29th, 2019 at 03:50 PM. June 30th, 2019, 02:45 PM   #2
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I get the area of $\Delta DEF$ to be 25 square units.
Attached Images area_prob.jpg (65.2 KB, 11 views) 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
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Quote:
 Originally Posted by Farzin 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 $\sin{\theta}$ ...

$A = \dfrac{1}{2}ab \cdot \dfrac{50}{ab} = 25$ July 1st, 2019, 06:58 AM #5 Newbie   Joined: Mar 2017 From: Norway Posts: 26 Thanks: 0 Excellent job, thank you very much. Tags area, shaded Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post yeoky Algebra 4 November 22nd, 2012 01:48 AM Tutu Algebra 3 May 8th, 2012 07:14 AM tatertots49 Algebra 2 May 18th, 2011 03:59 PM ILoveMaths Algebra 3 December 5th, 2010 11:51 AM roehm12 Calculus 1 March 12th, 2009 07:57 PM

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