A REAS OF C IRCLES AND S ECTORS 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-gon4-gon5-gon6-gon

A REAS OF C IRCLES AND S ECTORS THEOREM THEOREM 11.7 Area of a Circle The area of a circle is  times the square of the radius, or A =  r 2 r

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

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

P 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 B r In the diagram, sector APB is bounded by AP, BP, and AB.

THEOREM THEOREM 11.8 Area of a Sector 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°. Using the Area of a Circle The following theorem gives a method for finding the area of a sector. A r 2A r 2 =, or A = r 2 r 2 m AB 360° m AB 360° A A B P

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

U SING A REAS OF C IRCLES AND R EGIONS Finding the Area of a Region Find the area of a the shaded region shown. 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 – S OLUTION 5 m

U SING A REAS OF C IRCLES AND R EGIONS Finding the Area of a Region Area of shaded region = Area of circle Area of hexagon – =  5 2 – (6 5) The apothem of a hexagon is side length =  r 2 r 2 – 1212 a Pa P or about square meters. So, the area of the shaded region is 25  – , 5 m = 25  –

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

Area of circle Finding the Area of a Region W OODWORKING 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? S OLUTION 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 of rectangle Area =+ Area of sector –

Finding the Area of a Region Area Area of circle Area of rectangle =+ Area of sector – = +– 180° 360°  32  32  = 33 +  9 –  (2) = 33 +  – 4  9292  W OODWORKING 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? The area of the front of the case is about square inches.

U SING I NSCRIBED A NGLES An inscribed angle is an angle whose vertex is on a circle and whose sides contain chords of the circle. The arc that lies in the interior of an inscribed angle and has endpoints on the angle is called the intercepted arc of the angle. inscribed angle intercepted arc

THEOREM U SING I NSCRIBED A NGLES THEOREM 10.8 Measure of an Inscribed Angle If an angle is inscribed in a circle, then its measure is half the measure of its intercepted arc. m ADB = m AB 1212 mAB = 2m ADB C A B D

Find the measure of the blue arc or angle. Finding Measures of Arcs and Inscribed Angles S OLUTION N P M 100° C W X Y Z 115° C R S T Q C mQTS = 2m QRS = 2(90°) = 180° mZWX = 2m ZYX = 2(115°) = 230° M NMP = mNP = (100°) = 50°

THEOREM A D C B U SING I NSCRIBED A NGLES THEOREM 10.9 If two inscribed angles of a circle intercept the same arc, then the angles are congruent. C D

Using the Measure of an Inscribed Angle You decide that the middle of the sixth row has the best viewing angle. If someone is sitting there, where else can you sit to have the same viewing angle? T HEATER D ESIGN When you go to the movies, you want to be close to the movie screen, but you don’t want to have to move your eyes too much to see the edges of the picture. If E and G are the ends of the screen and you are at F, m EFG is called your viewing angle. movie screen EG F

S OLUTION Using the Measure of an Inscribed Angle Draw the circle that is determined by the endpoints of the screen and the sixth row center seat. Any other location on the circle will have the same viewing angle.

U SING P ROPERTIES OF I NSCRIBED P OLYGONS If all of the vertices of a polygon lie on a circle, the polygon is inscribed in the circle and the circle is circumscribed about the polygon. The polygon is an inscribed polygon and the circle is a circumscribed circle.

U SING P ROPERTIES OF I NSCRIBED P OLYGONS THEOREMS ABOUT INSCRIBED POLYGONS THEOREM If a right triangle is inscribed in a circle, then the hypotenuse is a diameter of the circle. Conversely, if one side of an inscribed triangle is a diameter of the circle, then the triangle is a right triangle and the angle opposite the diameter is the right angle. C A B B is a right angle if and only if AC is a diameter of the circle.

U SING P ROPERTIES OF I NSCRIBED P LOYGONS THEOREMS ABOUT INSCRIBED POLYGONS THEOREM A quadrilateral can be inscribed in a circle if and only if its opposite angles are supplementary. D, E, F, and G lie on some circle, C, if and only if m D + m F = 180° and m E + m G = 180°.. C E D F G

Using an Inscribed Quadrilateral In the diagram, ABCD is inscribed in P. Find the measure of each angle.. P B C D A 2y ° 3y ° 5x ° 3x °

P B C D A 2y ° 3y ° 5x ° 3x ° Using an Inscribed Quadrilateral S OLUTION ABCD is inscribed in a circle, so opposite angles are supplementary. 3x + 3y = 1805x + 2y = 180

Using an Inscribed Quadrilateral 3x + 3y = 1805x + 2y = 180 To solve this system of linear equations, you can solve the first equation for y to get y = 60 – x. Substitute this expression into the second equation. y = 60 – 20 = 40 Substitute and solve for y. 5x + 2y = 180 Write second equation. 5x + 2(60 – x) = 180 Substitute 60 – x for y. 5x – 2x = 180 Distributive property 3x = 60 Subtract 120 from each side. x = 20 Divide each side by 3. P B C D A 2y ° 3y ° 5x ° 3x °

Using an Inscribed Quadrilateral P B C D A 2y ° 3y ° 5x ° 3x ° x = 20 and y = 40, so m A = 80°, m B = 60°, m C = 100°, and m D = 120°.