# geometry problem: circle inscribed in a sector?

a circle having a diameter of 8 cm is inscribed in a sector of a circle whose angle is 80°. find the area of the sector.

Relevance

Since the two straight sides bordering the sector

are tangent to the circle, we know that the right

triangle formed by the centers of the two circles

and the point of tangency of the smaller circle

and either of the sector sides has the radius of

the smaller circle as one leg and the angle opposite

this leg being (1/2)*80 = 40 degrees. The hypotenuse

of this triangle is then the distance between the two

centers, and the radius of the larger circle will then be

this hypotenuse length plus the radius of the smaller

circle. So with the smaller circle having a radius of 4 cm,

the hypotenuse has a length of 4*csc(40), and thus the

radius of the larger circle is 4*csc(40) + 4 = 4*(1 + csc(40)) cm.

The area of the sector is therefore

(1/2)*r^2*(theta) = (1/2)*[4*(1 + csc(40))]^2 * (80 * (pi/180)) =

8*(1 + csc(40))^2 * (4/9) * pi = (32/9)*pi*(1 + csc(40))^2 cm^2,

which is 72.96 cm^2 to 2 decimal places.

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• If you draw a sketch you'll find that the inscribed circle has its center on the bisector of the

80 degree angle of the sector. If you draw perpendiculars from the center to the two lines defining

the sector then you can express the distance between the center of the inscribed circle and the

tip of the sector as

r / x = tan(80/2) = tan (40)

Therefroe, x = r / tan(40)

The radius of the sector is x + r = r (1/tan(40) + 1) = 8 ( 1/ tan(40) + 1) = 17.534 cm

Now we can find the area of the sector

Area = 1/2 R^2 (theta), where theta is expressed in radians.

Area = (1/2) (17.534)^2 ( 80 * pi /180 ) = 214.634 cm^2

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• area of sector is given by 80 pie R^2 / 360 where R = 4(1 + csc 40) ANSWER

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• Area of sector = (x/360)πr^2

where x is central angle of sector

=> A = (80/360)π(8)^2

= 44.68 sq.cm

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