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.

Guest Aug 28, 2019

#1**+1 **

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

CPhill Aug 28, 2019

#2**+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

__D2__^{2} = __BD__^{2} - __B2__^{2}

__D2__ = 1.5

Area of a triangle ABD = __D2__ * __B2__ = 3

(AD)^{2} * pi = circle area Area = 19.635 n^{2}

{(19.635 / 360 ) * 106.26} - 3 = 2.795 n^{2}^{ }

Guest Aug 29, 2019