Respuesta :
Our aim is to calculate the Radius so that to use the formula related to the area of a segment of a circle, that is: Aire of segment = Ф.R²/2
Let o be the center of the circle, AB the chord of 8 in subtending the arc f120°
Let OH be the altitude of triangle AOB. We know that a chord perpendicular to a radius bisects the chord in the middle. Hence AH = HB = 4 in
The triangle HOB is a semi equilateral triangle, so OH (facing 30°)=1/2 R. Now Pythagoras: OB² = OH² + 4²==> R² = (R/2)² + 16
R² = R²/4 +16. Solve for R ==> R =8/√3
OB² = OH² +
Let o be the center of the circle, AB the chord of 8 in subtending the arc f120°
Let OH be the altitude of triangle AOB. We know that a chord perpendicular to a radius bisects the chord in the middle. Hence AH = HB = 4 in
The triangle HOB is a semi equilateral triangle, so OH (facing 30°)=1/2 R. Now Pythagoras: OB² = OH² + 4²==> R² = (R/2)² + 16
R² = R²/4 +16. Solve for R ==> R =8/√3
OB² = OH² +
Answer:
Step-by-step explanation:
Given that a segment of a circle has a 120 arc and a chord of 8in.
Consider the triangle formed by two radii and the chord of 8 inches. This triangle is isosceles with two sides as equal radii. Hence each base angle is 30 degrees.
By sine formula for triangles
[tex]\frac{r}{sin30} =\frac{8}{sin120} \\r = \frac{8\sqrt{3} }{3}[/tex]
Area of segment = area of sector - area of triangle
Area of sector = [tex]\frac{120}{360} (\pi 8^2) = \frac{64\pi}{3}[/tex] ... I
Area of triangle = [tex]\frac{1}{2} r^2 sin 120 =16\sqrt{3}[/tex] ...II
Hence area of segment = I-II = [tex]\frac{64\pi}{3}-16\sqrt{3}[/tex]