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A segment of a circle has a 120 arc and a chord of 8in. Find the area of the segment.

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 +

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]

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