AREAS OF CIRCLES AND SECTORS These regular polygons, inscribed in circles with radius r, demonstrate that as the number of sides increases, the area of the polygon approaches the value  r 2. 3-gon 4-gon 5-gon 6-gon

AREAS OF CIRCLES AND SECTORS THEOREM r THEOREM 11.7 Area of a Circle The area of a circle is  times the square of the radius, or A =  r 2

 201.06 . Find the area of P. Use r = 8 in the area formula. Using the Area of a Circle Find the area of P. . P 8 in. SOLUTION Use r = 8 in the area formula. A =  r 2 =  • 8 2 = 64   201.06 So, the area is 64, or about 201.06, square inches.

 Find the diameter of Z. The diameter is twice the radius. A =  r 2 Using the Area of a Circle Z Find the diameter of Z. • SOLUTION Area of  Z = 96 cm2 • The diameter is twice the radius. A =  r 2 96 =  r 2 = r 2 96  30.56  r 2 5.53  r Find the square roots. The diameter of the circle is about 2(5.53), or about 11.06, centimeters.

In the diagram, sector APB is bounded by AP, BP, and AB. Using the Area of a Circle The sector of a circle is the region bounded by two radii of the circle and their intercepted arc. A P r B In the diagram, sector APB is bounded by AP, BP, and AB.

Using the Area of a Circle The following theorem gives a method for finding the area of a sector. THEOREM THEOREM 11.8 Area of a Sector A B P The ratio of the area A of a sector of a circle to the area of the circle is equal to the ratio of the measure of the intercepted arc to 360°. A  r 2 = , or A = • r 2 mAB 360°

 11.17 Find the area of the sector shown at the right. Finding the Area of a Sector Find the area of the sector shown at the right. P C D 4 ft 80° SOLUTION Sector CPD intercepts an arc whose measure is 80°. The radius is 4 feet. m CD 360° A = •  r 2 Write the formula for the area of a sector. 80° 360° = •  • 4 2 Substitute known values.  11.17 Use a calculator. So, the area of the sector is about 11.17 square feet.

USING AREAS OF CIRCLES AND REGIONS Finding the Area of a Region Find the area of a the shaded region shown. 5 m 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. SOLUTION Area of shaded region = Area of circle Area of hexagon –

3 3 3  r 2  • 5 2 – USING AREAS OF CIRCLES AND REGIONS = – = – 1 2 Finding the Area of a Region 5 m Area of shaded region = Area of circle Area of hexagon – =  r 2 – 1 2 a P The apothem of a hexagon is • side length • 1 2 3 =  • 5 2 – 1 2 5 3 • • (6 • 5) = 25 – 3 75 2 or about 13.59 square meters. So, the area of the shaded region is 25  – 75 2 3 ,

Complicated shapes may involve a number of regions. Finding the Area of a Region Complicated shapes may involve a number of regions. P P Notice that the area of a portion of the ring is the difference of the areas of two sectors.

Finding the Area of a Region WOODWORKING 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? SOLUTION 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. Area = Area of rectangle + Area of sector – Area of circle

6 • 33 + •  • 9 –  • (2)2 33 +  – 4  •  34.57 = + – = + – = = Finding the Area of a Region WOODWORKING 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? Area Area of circle Area of rectangle = + Area of sector – 6 • 11 2 = + – 180° 360° •  • 32  • 1 2 • 4 2 = 33 + •  • 9 –  • (2)2 1 2 = 33 +  – 4 9 2  34.57 The area of the front of the case is about 34.57 square inches.