Ch 8 Notesheet L Key V3 Note Sheets Chapter 8: rea In General ON LL PROBLEMS!!. State the relationship (or the formula).. Sustitute in known values. 3. Simplify or Solve the equation. Use the order of operations in the correct order. Simplify Lesson 8. reas of Rectangles and Parallelograms See Worksheet page. Order of Operations Parenthesis Exponents and Roots Multiplication and Division (or multiply y reciprocal) ddition and Sutraction (or add the opposite of) Solve rea The measure of the size of the interior of a figure, expressed in square units. Write the units as in, or sq. in, NOT in! ase (of a polygon) ny side of the polygon used for reference to determine an altitude or other features. altitude ny perpendicular segment from a ase to the opposite part (vertex or side) of the figure. n altitude for a parallelogram would go from the line containing one parallel ase to the line containing the other parallel ase. In a triangle, an altitude goes from a vertex perpendicular to the line containing the opposite side. height The length of an altitude. height can e inside or outside a parallelogram. Procedure for finding the height and ase: Choose the height first! Look for a right angle symol. Then find the side that the height is perpendicular to. Rectangle rea Conjecture The area of a rectangle is given y h Parallelogram rea Conjecture The area of a parallelogram is given y h where is the area is the length of the ase h is the height (or width) h where is the area is the length of the ase h is the height h S. Stirling Page of 0
Ch 8 Notesheet L Key V3 dditional Examples (Note: Drawing are not necessarily to scale!) represents area and P represents perimeter. Ex. B Find the height of a parallelogram that has an area 7.3 m and a ase length.3 m. Ex. =? 4 m h 7.3.3h 7.3 h 3..3 m 0 m ig = 46 44 sm = 05 50 44 50 94 m Or try sutraction! 6 m ig sm shape 4 64 45 76 64 76 94 Ex. Given the rectangle with P = 4 ft, =? 9 ft x Ex. 3 Given the parallelogram, find the shaded area =? 5 ft 3 ft P = 4 x9 x 9 4 x 8 4 x So x = = 9 08 ft 6 ft tri = ½ ase height tri = 6 3 9 ft Half the area of the parallelogram. Note: Choose the height st! Look for perpendiculars! Use algera!!! S. Stirling Page of 0
Ch 8 Notesheet L Key V3 Lesson 8. reas of Triangles, Trapezoids and Kites Triangle rea Conjecture The area of a triangle is given y h is the area is ase length h is the height h Trapezoid rea Conjecture The area of a trapezoid is given y h, is the area and are the ase lengths h is the height Examples (Note: drawing are not necessarily to scale!) h Kite rea Conjecture The area of a kite is given y dd, is the area d and d are the lengths of the diagonals Ex. Find h. h 6 cm Given = 40 cm, h =? h 40 (6) h 40 8h 40 h 5 cm 8 Ex. Given = 45.5 m, =? h 45.5 (9) 36 45.5 4.5 36 45.5 4.5 36 4.5 4.5 54.5 36 8.5 meters 9 m 36 m Ex. 3 Find the area of a kite with diagonals 7 feet and 6 feet. d d 7 6 8 7 56 feet Ex. 4 Find the area of the pentagon h 305 450 h 9 54 450 54 504 cm S. Stirling Page 3 of 0
Ch 8 Notesheet L Key V3 Lesson 8.3 rea Prolems Preparation for the rea Project. On all word prolems: You must show all work, as you have een doing, and also lael the sections of your work! For example: rea of walls and ceiling. If a drawing would help, make and lael the drawing. Make sure you are identifying the information you know and the question eing asked. Note: Be careful with unit conversions!!! If yard = 3 feet, then square yard = 9 square feet. Draw a square to prove this to yourself! Geometrically: 3 ft lgeraically: yard = 3 feet 3 ft yd yd If you square oth sides yard = 3 feet Example: You are painting your 5 ft y 3 ft room and your 0 ft y 7 ft family room. The ceilings in your house are 9.5 ft high. You want to do a good jo, so you will do one coat on the ceiling and two coats on the walls. (You may ignore the area taken up y windows and doors.) The product information you got from the wesite. How much is it going to cost to paint your room? So Likewise: 3 3 yard = 7 feet yard = 9 feet rea ceilings: 5 3 95 ft ed 0 7 340 ft family total 95 340 535 ft gallons 535 300.783 So need gallons ceiling paint. rea walls: 5 3 9.5 53 ft ed family 0 7 9.5 703 ft 53 703 35 ft total gallons 35 400 3.0875 So need 4 gallons ceiling paint, proaly. BEHR Premium Plus -Gal. Ceiling Paint Premium Plus Interior $3.98 container covers up to 300 sq. ft. BEHR Premium Plus -Gal. Swiss Coffee Semi-Gloss Enamel Zero VOC Interior Paint $8.96 container covers up to 400 sq. ft. Cost: Cost 3.98 4 8.96 63.80 $63.80 to paint oth rooms. Caution: Say you are painting something ojects measured in inches: 3 in y 45 in, 57 in y 9 in and 66 in y 98 in. To figure out how much paint, you will need square feet! The easiest way is to find the square inches and convert to square feet. There are 44 square inches in a square foot! 3 45 57 9 66 98 8706 in now 8706 in ft 44 in 60.49 ft S. Stirling Page 4 of 0
Ch 8 Notesheet L Key V3 Lesson 8.4 reas of Regular Polygons center (of a regular polygon) The point that is the center of the circle that is circumscried aout the polygon. radius (of a regular polygon) segment from the center to a vertex of the polygon. lso, it can mean the length of that segment. When you draw all radii into an n-gon, you reak the regular n-gon into n congruent isosceles triangles. apothem (of a regular polygon) perpendicular segment from the center of the polygon s circumscried circle to a side of the polygon. It is also the height of each of the n congruent isosceles triangles. Deriving the Regular Polygon rea Conjecture Try to find a formula for a regular polygon with n-sides. Hint: make congruent triangles! poly = 3 as poly = 4 as poly = 5 as n = # sides, # angles, or # congruent triangles r a s r a s r a n as s Regular Polygon rea Conjecture The area of a regular polygon is given y OR ans ap Ex Find the area of a regular hexagon if the apothem measures 6.93 m and a side measures 8 m. ans (6.93)(6)(8) 66.3 m is the area, p is the perimeter, a is the apothem, s is the length of each side, and n is the numer of sides. Ex Find the perimeter of a regular octagon if a =.07 inches and = 48.8 square inches. ap 48.8 (.07) p 48.8 6.035 p 48.8 p 80 in. 6.035 P 443 #8 Find the approximate length of each side of a regular n-gon if a = 80 feet, n = 0, and 0,000 square feet. ans 0000 (80)(0) s 0000 800 s 00 s 5 ft 8 S. Stirling Page 5 of 0
Ch 8 Notesheet L Key V3 Lesson 8.5 reas of Circles WRNING! It is very easy to confuse the formulas for area and circumference. Just rememer that area is measured in square units so the formula for area contains squaring. Circle rea Conjecture The area of a circle is given y r is the area r is the radius Circle Circumference Conjecture The circumference of a circle is given y C r d C is the circumference r is the radius & d is the diameter s efore, when we studied circumference, if the prolem asks for an exact answer, DO NOT sustitute in for π. Only use the π key or 3.4 or /7 if they want an approximate answer. Look for, which means approximate. P 450 Ex. The small apple pie has a diameter of 8 inches, and the large cherry pie has a radius of 5 inches. How much (what percent) larger is the large pie? sm (4) 6 50. lg (5) 5 78.5 5.56 or 00% + 56% so large is 56% larger 6 than the small. P 450 Ex. B The area of a circle is 56π m, what is the circumference of the circle? Start with area: r 56 r r 56 so r = 6. On calculator: Use the key. Type 56. Now: C r C (6) C 3 meters key. ove the x Ex. Find the area of the shaded region. The rectangle is y 6 inches and the radius of the circle is 0 inches. Ex. Find the area of the shaded region. ll measures in cm. rea Shaded = rea Square rea 4 3 6 4 (3) = = 44 36 cm. = 30.90 in rea Shaded = rea Circle rea of rectangle = (0) ()(6) 00 9 or approx..59 in S. Stirling Page 6 of 0
Ch 8 Notesheet L Key V3 Lesson 8.6 ny Way You Slice It sector of a circle The region etween two radii and an arc of the circle. (aka a slice of pizza) segment of a circle The region etween a chord and an arc of the circle. (aka just keep a slice of the crust) annulus The region etween two concentric circles of unequal radius. (aka just the crust, cut out the center) Sector of a Circle Segment of a Circle nnulus or Washer fraction of the area of a circle = a piece of pizza. rea of sector minus area of triangle = area of the crust. rea of the whole ig circle minus the area if the inner small circle. Just like arc length, ut its part of the area. CUTION: The choice for the ase and the height of the triangle will depend on the type of triangle. If it is a right triangle, the radii will e the ase and the height. Ex. Find the area of the shaded sector. Ex. (P 454 Example ) Find the area of the shaded segment. sector fract circle 70 (8) 360 3 (64) 48 cm 4 50.796 cm Think: How much do you want to keep? sector fract circle 45 (0) 360 (400) 8 50 57.080 cm S. Stirling Page 7 of 0
Ch 8 Notesheet L Key V3 Ex. 3 (P 454 Example B) Find the area of the shaded segment. shaded sector triangle 9 8 0.7 ft 90 360 triangle (6)(6) 8 sec tor (6) 9 Ex. 4 (P 454 Example C) The area of the shaded region is 4π and the radius is 6 cm. Find x. sector fract x 4 (6) 360 x 36 4 360 0 4 x 0 0 40 x circle Ex. 5 Find the area of the shaded annulus. Need radii: R = 4 + 3 = 7 and r = 4 in. shaded ig small 7 4 49 6 33 03.673 in 4 in 3 in Ex 6. Find x. The area of the shaded region is 3π cm. The large circle has a radius 0 cm and the inner circle has a radius 8 cm. shaded fract washer x 3 36 360 x 36 3 360 0 3 0 x x = 30 degrees 0 washer outside inside (0) (8) 36 S. Stirling Page 8 of 0
Ch 8 Notesheet L Key V3 Lesson 8.7 Surface rea surface area The sum of the areas of all the surfaces, faces, of a solid. ase (of a solid) polygon or circle used for reference to determine an altitude or other feature of the solid, or used to classify the solid. lateral face face of a solid other than a ase. Does not have to e vertical! slant height The height of each triangular lateral face of a pyramid. They usually use l for slant height. Prism Pyramid height of prism pentagonal ase Surface rea = (ase area) + (lateral surface area) = (ase area) + (perimeter ase * height) Use appropriate area formula. rectangular lateral face Is the large rectangle that wraps around the ase. height of prism Note: The ase is a regular polygon. Surface rea = (ase area) + (lateral surface area) = (ase area) + (area one face * n) = (ase area) + (½ l * n) = (ase area) + (½ pl ) n = numer of sides of regular polygon ase = length of ase edge l = slant height p = perimeter of the ase l USE THE FORMULS BOVE TO GET THE FORMULS FOR CYLINDER ND CONE!! Cylinder Same as a prism with circular ases. Cone Same as a pyramid with circular ase. circular ase l l h = height rectangular lateral face the can s lael Circumference circular ase triangular-ish lateral face Surface rea = (ase area) + (perimeter ase * height) = (π r ) + ( π r * h) Surface rea = (ase area) + (lateral surface area) = (ase area) + (½ pl ) = (π r ) + (½ * π r * l ) S. Stirling Page 9 of 0
Ch 8 Notesheet L Key V3 Ex. Find the surface area of a rectangular prism with dimensions 3 m, 6 m and 8 m. h = 3 p 6 8 6 8 8 ase ase B 6 8 48 S B ph S 48 83 96 84 80 m 8 6 Ex. B Find the surface area of a cylinder with ase diameter 0 in. and height in. p C 5 0 ase ase ase B 5 5 S B ph S 5 0 50 0 70 S 534.07 0 h = Ex. C Find the surface area of a cone with ase radius 5 cm. and slant height 0 cm. Ex. D The surface area of the right square pyramid is 95 in with ase edge of 5 in. What is the measure of the slant height? p C 5 0 ase ase ase B 5 5 0 5 p 4 5 0 ase ase B 5 5 5 l S B pl S 5 0 0 5 50 75 S 35.6 S B pl 95 5 0l 70 0 l l 7 in 5 5 S. Stirling Page 0 of 0