+0  
 
+2
102
4
avatar+186 

In triangle ABC,  

 Jun 6, 2020
edited by Caffeine  Jun 6, 2020
 #1
avatar+186 
0

I'm sorry, it seems that I cannot write the rest of the question...

 Jun 6, 2020
 #2
avatar+186 
+1

Web calc isn't working for me at the moment.

Caffeine  Jun 6, 2020
edited by Caffeine  Jun 6, 2020
 #3
avatar+902 
+1

Hi, caffeine. 

 

Sometimes the site has a problem and limits your question to the first three words.

 

If you would take a screenshot of the question that would be extremely helpful!

 

Happy coffee/tea drinking(no decaf)!

 Jun 6, 2020
 #4
avatar+902 
0

I think I might be able to guess the problem

 

SO we are looking at the shaded part's area?

 

K.

 

Area of a circle is... πr^2.

 

The larger circle(with the unshaded) is π(BC)^2 /2, and the smaller one is π(AB)^2 /2.

 

And then we calc. the unshaded.

 

That's π(AC)^2 /2, minus the area of triangle BCA, is BC * BA/2 .

 

Thus the shaded area is,

 

π(BC^2 + AB^2) /2 - (π(AC^2) /2 -BC * BA/2)

 

You can review this in case my small brain messed something up and plug it in and potentially simplify. 

 

Cheers.

 

(If you don't understand anything feel free to ask.)

 

Don't forget to drink Coffee.

 

Lol

 

Idk why I'm saying drink coffee.

 

Huh.

hugomimihu  Jun 6, 2020

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