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.
+128407 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
