1)Originally Posted byConfuzzled?

Circle of radius 10cm.

Circumference, C = 2pi(10) = 20pi cm

Area, A = pi(10)^2 = 100pi sq.cm.

The minor arc has a pi/4 central angle, so the major arc has a (2pi -pi/4) = 7pi/4 central angle.

By proportion,

(maj. arc)/(7pi/4) = (Circumference)/(2pi)

(maj. arc) = [(20pi)/(2pi)]*(7pi/4)

(maj. arc) = 70pi/4 = 17.5pi cm ----------answer.

(Sector of pi/4 central angle)/(pi/4) = (area of circle)/(2pi)

sector = [(100pi)/(2pi)]*(pi/4) = 50pi/4 = 12.5pi sq.cm ---------***

Using A = (1/2)bc*sinA for area of triangle,

Area of (Isosceles triangle of pi/4 central angle) =

= (1/2)(10)(10)sin(pi/4)

= 50(1/sqrt(2)) = 50(sqrt(2) /2)

= 25sqrt(2) sq.cm. -------------------***

Therefore,

Area of minor segment = (area of minor sector) minus (25sqrt(2)

= 12.5pi -25sqrt(2)

= 3.91457 sq.cm ---------------answer.

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2)

Using Law of Cosines,

(x-2)^2 = 10^2 +x^2 -2(10)(x)cos(60deg)

x^2 -4x +4 = 100 +x^2 -10x

-4x +10x = 100 -4

6x = 96

x = 96/6 = 16 cm -----------answer.

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3)

Circle of diameter 7 cm.

Radius = 7/2 = 3.5 cm

Sector of the circle, area = 18.375 sq.cm.

Area of sector = (1/2)(radius)(arc)

18.375 = (1/2)(3.5)(arc)

arc = 2(18.375)/3.5 = 36.75/3.5 = 73.5/7 = 10.5 cm ---------answer.

No use of calculator.

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Only one problem, very detailed solution.

Three problems, (very detailed solution) divided by 3.