# AREAS OF CIRCLES AND SECTORS WORKSHEET

Problem 1 :

Find the area of the circle shown below. Problem 2 :

If the area of a circle is 96 square centimeters, find its diameter.

Problem 3 :

Find the area of the sector shown at the right. Problem 4 :

A and B are two points on a P with radius 9 inches and APB = 60°. Find the areas of the sectors formed by APB.

Problem 5 :

Find the area of the shaded region shown below. Problem 6 :

You are cutting the front face of a clock out of wood, as shown in the diagram. What is the area of the front of the case ?   Formula area of a circle is given by

A = πr2

Substitute r = 8.

A = π(8)2

A = 64π

Use calculator.

≈ 201.06

So, the area is 64π, or about 201.06, square inches.

Formula area of a circle is given by

A = πr2

Substitute A = 96.

96 = πr2

Divide each side π.

96/π πr2/π

96/π = r2

Use calculator.

30.56  r2

Take square root on each side.

5.53  r

So, the diameter of the circle is about 2(5.53), or about 11.06, centimeters. Sector CPD intercepts an arc whose measure is 80°. The radius is 4 feet.

Formula for area of a sector is given by

A = [m∠arc CD/360°] ⋅ πr2

Substitute m∠arc CD = 80° and r = 4.

A = [80°/360°] ⋅ π(4)2

A = (2/9) ⋅ 16π

Use calculator.

A ≈ 11.17

So, the area of the sector is about 11.17 square feet.

Draw a diagram of ⊙P and APB. Shade the sectors. Label a point Q on the major arc.

Find the measures of the minor and major arcs. Because m∠APB = 60°, we have

m∠arc AB = 60°

and

m∠AQB = 360° - 60° = 300°

Use the formula for the area of a sector.

A = [m∠arc CD/360°] ⋅ πr2

Substitute m∠arc CD = 80° and r = 4.

A = [80°/360°] ⋅ π(4)2

A = (2/9) ⋅ 16π

Use calculator.

A ≈ 11.17

So, the area of the sector is about 11.17 square feet.

 Area of Smaller SectorA = 60°/360° ⋅ π(9)2A = 1/6 ⋅ π ⋅ 81A ≈ 42.41 square inches Area of Larger SectorA = 300°/360° ⋅ π(9)2A = 5/6 ⋅ π ⋅ 81A ≈ 212.06 square inches The diagram shows a regular hexagon inscribed in a circle with radius 5 meters. The shaded region is the part of the circle that is outside of the hexagon.

Area of shaded region  = Area of circle - Area of hexagon

Area of shaded region  =  πr- 1/2 ⋅ a ⋅ p

Radius of the circle is 5 and the apothem of a hexagon is

=  1/2 ⋅ side length ⋅ √3

=  1/2 ⋅ 5 ⋅ √3

=  5√3/2

So, the area of the shaded region is

=  [π ⋅ 52]  -  [1/2 ⋅ (5√3/2) ⋅ (6 ⋅ 5)]

=  25π  -  75√3/2

Use calculator.

≈  13.59

So, the area of the shaded region is about 13.59 square meters. The front of the case is formed by a rectangle and a sector, with a circle removed. Note that the intercepted arc of the sector is a semicircle.

So, the required area is

= Area of rectangle + Area of sector - Area of circle

= [6 ⋅ 11/2] + [180°/360° ⋅ π ⋅ 32] -  [π ⋅ (1/2  4)2]

= 33 + 9/2 ⋅ π -  4π

Use calculator.

≈ 34.57

The area of the front of the case is about 34.57 square inches.

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