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The triangle with the sides of 3, 4, and 5 is inscribed in a circle. Find the area between the side with number 4 and the circumference.

 Aug 28, 2019
 #1
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See the following image :

 

 

Since this is a right triangle....the center of the circle will lie  on the midpoint of  hypotenuse BC

 

The midpoint of the hypotenuse  is   (2, 1.5)

 

The radius   of the circumscribing circle  is  sqrt [  ( 4-2)^2 + (3 - 1.5)^2 ]  =  

sqrt [ 2^2  + 1.5^2 ]   =  sqrt [ 4 + 2.25]  =  sqrt [ 6.25]  = 2.5

 

So  the equation of the circumscribing circle is  (x - 2)^2 + (y - 1.5)^2  = 2.5^2

 

And we can find central angle ADB  using the Law of Cosines

 

AB^2  = AD^2 + BD^2 - 2(AD)(BD) cosADB

4^2  = 6.25 + 6.25 - 2 (2.5)(2.5)cosADB

[16  - 2(6.25)]  / [ -2(2.5)(2.5)) ]  =  cos ADB

 

And  using the arccos, we can find the measure of ADB

 

arccos [ (16 - 2(6.25)^2) / ( -2(2.5)(2.5) ]  = ADB ≈ 106.26°

 

So  the  area of the  sector DAEB  = pi * (2.5)^2 * (106.26/ 360)      (1)

 

And the area of triangle ADB  =  (1/2)(2.5)^2 sin (106,26°)       (2)  

 

So  the area  between the side of 4 and the circumference  =  (1)  - (2)  =

 

(2.5)^2 [ pi * (106.26/360)  - sin (106.26°) / 2 ]  ≈  2.796 units^2

 

 

cool cool cool

 Aug 28, 2019
 #2
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+1

I've don it differently, but got the same answer.

I connected D and 2 in order to form a right triangle.

BD = 2.5      B2 = 2

D22 = BD2 - B22

D2 = 1.5

Area of a triangle ABD =  D2 * B2 = 3

(AD)2 * pi = circle area       Area = 19.635 n2

{(19.635 / 360 ) * 106.26} - 3 = 2.795 n2     smiley

Guest Aug 29, 2019

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