Question

Express answer in exact form. Show all work for full credit.

Answer 1

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 2

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

$\frac{r}{sin30} =\frac{8}{sin120} \\r = \frac{8\sqrt{3} }{3}$

Area of segment = area of sector - area of triangle

Area of sector = $\frac{120}{360} (\pi 8^2) = \frac{64\pi}{3}$  ... I

Area of triangle = $\frac{1}{2} r^2 sin 120 =16\sqrt{3}$  ...II

Hence area of segment = I-II = $\frac{64\pi}{3}-16\sqrt{3}$