AREA AND PERIMETER OF COMPLEX PLANE FIGURES

Transcription

1 AREA AND PERIMETER OF OMPLEX PLANE FIGURES AREA AND PERIMETER OF POLYGONAL FIGURES DISSETION PRINIPLE: Every polygon can be dissected (or broken up) into triangles (or rectangles), which have no interior points in common. This principle is an example of the problem solving strategy of using sub problems. To use sub problems, find simpler problems that you know how to solve and use then them to help you solve the larger problem. Example Find the area of the figure below. First fill in the missing lengths. The right vertical side of the figure is a total of 8 units tall. Since the left side is also 8 units tall, the missing vertical length in the left middle cutout is 8 = units. Horizontally, the bottom length is 8 units ( + ). Therefore, the top is also 8 units. The missing horizontal length at the top center is 8 = units. Enclose the entire figure in an 8 by 8 square and then remove the area of the three rectangles that are not part of the figure. The area is: Enclosing square: 8! 8 = u Top cut out:! = u Left cut out:! = u Lower right cut out:! = u Area of figure:!!! = 8u Example Find the area of the figure below. This figure consists of five familiar figures: a central square, units by units; three triangles, one on top with b = and h =, one on the right with b = and, and h = one on the bottom with b = and h = ; and a trapezoid with an upper base of, a lower base of and a height of. The area is: square! = u top triangle bottom triangle right triangle!! = 7.u!! = u!! = u ( +)i trapezoid = u total area = 9. u

2 Example Find the perimeter of this figure. h x The first sub problem is to find the hypotenuse of the smaller triangle: h = + = so h =. Then find the hypotenuse of the larger triangle: x = + = 9 so x =. The perimeter is = units. Problems Find the area of each of the following figures. Assume that anything that looks like a right angle is a right angle and : Find the surface area of each polyhedra. 0 0

3 Find the area and perimeter of each of the following figures AN APPLIATION: TILING THE KITHEN FLOOR The kitchen in Susan s apartment has an ugly floor covering and she wants to replace it with various sized ceramic tiles. There will be a rectangular design made of " square teal tiles (Susan s signature color!) set in one foot from the edge of the walls as shown with dashed lines on the diagram below. Your task is to find the number of tiles needed for this rectangle, as well as the square footage of the kitchen. a. Use the floor plan of the kitchen to determine the area in square feet of her kitchen floor. Show your work for the sub problems you use. b. The tiles are currently selling for $. per square foot. How much will tile for the whole floor cost? Be sure to include 7% sales tax! c. The teal tiles are by and are positioned in a rectangle one foot away from the walls. How many teal tiles will Susan need to purchase? Remember that the dashed lines indicate the position of this rectangle. 8' 7' ' '

4 Answers. u. u. 8 u. u. u. u 7. u u 9. u 0. 8 u. SA = 9 u. u. A = u P = + +! 8.u. A = u P = +!.8u. A =.u P = !.89u. A = +!.9u P = +! 8.899u 7. a. 7 ft b. $7.0 without tax, $88.7 with tax c. 8 tiles

5 AREA AND PERIMETER OF SETORS SETORS of a circle are formed by the two radii of a central angle and the arc between their endpoints on the circle. For example, a 0 sector looks like a slice of pizza. The area of a sector is found by multiplying the fraction of the circle that is in the sector by the area of the whole circle. The perimeter of the sector is found by multiplying the fraction of the circle that is in the sector by the circumference of the whole circle and adding the length of two radii. 0 Example Find the area of the sector. A ' Example Find the area and perimeter of the 0 sector. A 8' 0 B B First find the fractional part of the circle involved. mzy! 0 Next find the area of the circle: A =! =! ft Finally, multiply the two results to find the area of the sector.!" = " #.8 ft 8 = = 0 8 Next find the perimeter of the sector. The circumference of the whole circle is: =! " = 8! ft The length of the arc is 8! 8" = " ft so the perimeter is: + +! ". ft Fractional part of circle is: mab! 0 = 0 = 0 Area of circle is: A =!r =!8 =! ft Area of the sector is: 0"! " = = 80" # 8.7 ft ircumference of circle is: =!r =! " 8 =! ft 0" Perimeter of sector:!" = + #.9 ft

6 Problems. Find the area of the shaded sector in each circle below. Points A, B and are the centers. a) A b) 0 7 B c) alculate the perimeter of the following shaded sectors. Point O is the center of each circle.. O. 0 O. 7 O 70. O 0 8. The shaded region in the figure is called a segment of the circle. It can be found by subtracting the area of MIL from the area of sector MIL. Find the area of the segment of the circle. M I L Find the area of the shaded regions. 7. I N H YARD is a square; A and D are the centers of the arcs. Y A D R 0. Find the area of a circular garden if the diameter of the garden is 0 feet.. Find the area of a circle inscribed in a square whose diagonal is 8 feet long.. The area of a 0 sector of a circle is 0π m. Find the radius of the circle.. The area of a sector of a circle with a radius of mm is 0π mm. Find the measure of its central angle.

7 Find the area of each shaded region.. 0 m. ft. 7. in 0 m 0 0 ft r = Find the radius. The shaded area is π cm. 0 r sm = 0 r r lg = 8 Find the perimeter of these shaded figures. figure 7. figure 9. figure. figure 7. AN APPLIATION: THE BROKEN WINDOW The window in Susan's living room needs new glass. The glass shop prices depend on the area of each pane of glass, and their price chart is shown below: Area Price Area Price 0-99 cm $ cm $ cm $ cm $ cm $ cm $ cm $ cm and up $0.0 per cm a. Use the design of her window to determine the cost of replacing all panes of glass. Assume that the four panes of glass making up the outer circular region are equal in area. 0 cm b. How much money would Susan save if she replaced the entire window with one solid piece of glass? 0 cm 0 cm 0 cm

8 Answers. a) π u b) 9! c) u! u.! + u. π + u. 9! + u. π + 0 u 8. π u 7. 00! " u 8. 0π 0 u 9. 8π u 0. π ft. 8π ft. m. degrees.00! " m. 9 9π ft. 0π in 7. 8π + u 8.! u 9..8 u 0. u. 0! + 0 u.! u.! u.! + 8 u. $7, $79