Interesting Geometry Problem...

[imhotep]

Kattiya hadala iwaraine

It's encouraging to know that there are a few people who really think....

 

[MrFrog]

mechchra marenne nathuwa ganna wena kramayk thiyanwda ? hypotenuse nm daannma ona une na..

did similar way, but had to do more steps at the beginning

once we get Theta = Pi / 6 and r = sqrt(3), what I did was...

found CE = DC . Sin(Theta) = 1
So area of DCE = 1/2 . r . 1 = sqrt(3) / 2

Area of sector = Theta. r . r / 2 = (Pi /6). (3 / 2) = Pi / 4

So the answer is (sqrt(3) / 2) - (Pi / 4)

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It's encouraging to know that there are a few people who really think....

For me, the beauty of this problem is not in the actual problem (thought process) , but the in the calculation.
------ Post added on Sep 27, 2021 at 5:57 AM

 

[imhotep]

did similar way, but had to do more steps at the beginning

once we get Theta = Pi / 6 and r = sqrt(3), what I did was...

found CE = DC . Sin(Theta) = 1
So area of DCE = 1/2 . r . 1 = sqrt(3) / 2

Area of sector = Theta. r . r / 2 = (Pi /6). (3 / 2) = Pi / 4

So the answer is (sqrt(3) / 2) - (Pi / 4)

---

For me, the beauty of this problem is not in the actual problem (thought process) , but the in the calculation.
------ Post added on Sep 27, 2021 at 5:57 AM

Correct! Initially you wouldn't think that the lengths given will simplify.
That's why I replied much earler..... "Yes.. But do the calculation... It's interesting."

 

[dilann]

found CE = DC . Sin(Theta) = 1
So area of DCE = 1/2 . r . 1 = sqrt(3) / 2

Area of sector = Theta. r . r / 2 = (Pi /6). (3 / 2) = Pi / 4

So the answer is (sqrt(3) / 2) - (Pi / 4)

ඔව් මෙහෙම ගන්න එක ලේසි