Read pages in the book, up to the investigation. Pay close attention to Example A and how to identify the height.

Transcription

1 C 8 Noteseet L Key In General ON LL PROBLEMS!!. State te relationsip (or te formula).. Sustitute in known values. 3. Simplify or Solve te equation. Use te order of operations in te correct order. Order of Operations Parentesis Exponents and Roots Multiplication and Division (or multiply y reciprocal) ddition and Sutraction (or add te opposite of) Lesson 8. reas of Rectangles and Parallelograms Read pages 4 44 in te ook, up to te investigation. Pay close attention to Example and ow to identify te eigt. rea Te measure of te size of te interior of a figure, expressed in square units. Write te units as in, or sq. in, NOT in! ase (of a polygon) ny side of te polygon used for reference to determine an altitude or oter features. altitude ny perpendicular segment from a ase to te opposite part of te figure. n altitude for a parallelogram would go from te line containing one parallel ase to te line containing te oter parallel ase. In a triangle, an altitude goes from a vertex perpendicular to te line containing te opposite side. eigt Te lengt of an altitude. Rectangle rea Conjecture Te area of a rectangle is given y = Parallelogram rea Conjecture Te area of a parallelogram is given y = were is te area is te lengt of te ase is te eigt (or widt) were is te area is te lengt of te ase is te eigt Page 44 Investigation: Deriving te rea Formula for Parallelograms Follow Steps and of te investigation in your ook. In Step, eac new sape you form as te same area as te original parallelogram ecause you ave simply rearranged pieces, witout adding or removing any cardoard. Form a rectangle wit te two pieces. Notice tat te ase and eigt of te rectangle are te same as te ase and eigt of te original parallelogram. Because te area of te rectangle and te parallelogram are te same, te area of te parallelogram is. Te side lengt, s, as noting to do wit finding te area. Page 45. Pay close attention to ow to sow your work, Example B. S. Stirling Page of 0

2 C 8 Noteseet L Key dditional Examples (Note: drawing are not necessarily to scale!) represents area and P represents perimeter. Ex. =? 4 m Ex. P = 4 ft, =? x Ex. 3 Saded area =? m 6 m 9 ft 5 ft 3 ft 0 m ig = 4 6 = 44 sm = 0 5 = 50 P = 4 = x x = x = x So x = tri = 6 ft ft = = 94 m Or try sutraction! = 9 = 08 ft Use algera!!! Half te area of te parallelogram. Note: Coose te eigt st! Look for perpendiculars! Lesson 8. reas of Triangles, Trapezoids and Kites Triangle rea Conjecture Te area of a triangle is given y = is te area is ase lengt is te eigt Trapezoid rea Conjecture Te area of a trapezoid is given y ( ) = +, is te area and are te ase lengts is te eigt Kite rea Conjecture Te area of a kite is given y = dd, is te area d and d are te lengts of te diagonals Deriving te Triangle rea Conjecture para = tri = Note: ase must e a side of te triangle! Te eigt does not. Deriving te Trapezoid rea Conjecture Put two trapezoids togeter, forms a parallelogram para = ( + ) Trapezoid is alf trap = ( + ) Deriving te Kite rea Conjecture Kite makes two congruent triangles, so tri kite = 4 dd = dd 4 = dd = dd S. Stirling Page of 0

3 C 8 Noteseet L Key Examples (Note: drawing are not necessarily to scale!): Ex. Find. 6 cm Given = 40 cm, =? = 40 = (6) 40 = 8 Ex. Given = 45.5 m, =? = ( + ) 45.5 = (9)( + 36) 45.5 = = = ( ) 9 m 36 m Ex. 3 Find te area of a kite wit diagonals 7 feet and 6 feet. = d d = ( 7)( 6 ) 87 ( ) 56 = = feet 40 = = 5 cm 8 8. Page 433 Exercise #30 S. Stirling Page 3 of 0

4 C 8 Noteseet L Key Lesson 8.3 rea Prolems On all word prolems: You must sow all work, as you ave een doing, and also lael te sections of your work! For example: rea of walls and ceiling. Note: Be careful wit unit conversions!!! If yard = 3 feet, ten square yard = 9 square feet. Draw a square to prove tis to yourself! Geometrically: 3 ft lgeraically: yard = 3 feet 3 ft If you square ot sides yard = 3 feet yd yd So yard = 9 feet Likewise: 3 3 yard = 7 feet Lesson 8.4 reas of Regular Polygons center (of a regular polygon) Te point tat is te center of te circle tat is circumscried aout te polygon. radius (of a regular polygon) segment from te center to a vertex of te polygon. lso, te lengt of tat segment. apotem (of a regular polygon) perpendicular segment from te center of te polygon s circumscried circle to a side of te polygon. lso, te lengt of tat segment. Deriving te Regular Polygon rea Conjecture Try to find a formula for a regular polygon wit n-sides. Hint: make congruent triangles! poly = 3 as poly = 4 as poly = 5 as s r a r a s r a s Regular Polygon rea Conjecture Te area of a regular polygon is given y = as n i = ap OR is te area, p is te perimeter, a is te apotem, s is te lengt of eac side, and n is te numer of sides. P 443 #8 Find te approximate lengt of eac side of a regular n- gon if a = 80 feet, n = 0, and 0,000 square feet. = asn 0000 = (80)(0) s 0000 = 800i s 00 s = 5 ft 8 S. Stirling Page 4 of 0

5 C 8 Noteseet L Key Lesson 8.5 reas of Circles Read page 449 in te ook, including te investigation. Ten Deriving te Circle rea Conjecture Cut up te circle into 6 congruent wedges. Dimensions of parallelogram? ase = ½ Circ. eigt = radius of circle Using te wedges as triangles: wedge = = πr r 6 πr sape = 6 r 6 = π r Using te parallelogram: para = ase * eigt = ½ Circ * eigt = iπri r = π r WRNING! It is very easy to confuse te formulas for area and circumference. Just rememer tat area is measured in square units so te formula for area contains squaring. Circle rea Conjecture Te area of a circle is given y = πr, is te area r is te radius Circle Circumference Conjecture Te circumference of a circle is given y C= π r = dπ, C is te circumference r is te radius & d is te diameter s efore, wen we studied circumference, if te prolem asks for an exact answer, DO NOT sustitute in for π. Only use te π key or 3.4 or /7 if tey want an approximate answer. Look for, wic means approximate. P 450 EXMPLE Te small apple pie as a diameter of 8 inces, and te large cerry pie as a radius of 5 inces. How muc (wat percent) larger is te large pie? sm π (4) 6π = = 50. lg = π (5) = 5π 78.5 difference 5π 6π = 9π sq in larger (or 8.3) 9π i 8π so more tan 50% larger P 450 EXMPLE B Te area of a circle is 56π m, wat is te circumference of te circle? Start wit area: = πr 56π = πr r = 56 so r = 6. Now: C = π r C = π (6) C = 3π meters OR 5 π =.56 so large is 56% larger. 6π S. Stirling Page 5 of 0

6 C 8 Noteseet L Key Ex. Find te area of te saded region. ll measures in inces. (0, 0) (6, 8) (0, 0) Use location (x, y), to find distances. radius = 0 eigt = 6 widt = rea Saded = rea Circle rea of rectangle = π (0) ()(6) Ex. Find te area of te saded region. ll measures in cm. rea Saded = rea Square rea 4 4 π (3) = 44 36π cm. = in = ( ) 00π 9 or approx..59 in Lesson 8.6 ny Way You Slice It sector of a circle Te region etween two radii and an arc of te circle. (aka a slice of pizza) segment of a circle Te region etween a cord and an arc of te circle. (aka just keep a slice of te crust) annulus Te region etween two concentric circles of unequal radius. (aka just te crust, cut out te center) Sector of a Circle Segment of a Circle nnulus or Waser fraction of te area of a circle = a piece of pizza. rea of sector minus area of triangle = area of te crust. rea of te wole ig circle minus te area if te inner small circle. Note: Te coice for te ase and te eigt of te triangle will depend on te type of triangle. If it is a rigt triangle, te radii will e te ase and te eigt. S. Stirling Page 6 of 0

7 C 8 Noteseet L Key Ex. Find te area of te saded sector. Ex. (P 454 Example B) Find te area of te saded segment. 6 ft 8 cm sector = fract i circle 70 (8) = π (8 8) 9 cm = π i = π 4 = saded sector triangle 90 = π 360 = 9π ft (6) (6)(6) Ex 3. Find x. Te area of te saded region is 0π cm. Te large circle as a radius 0 cm and te inner circle as a radius 8 cm. saded = fract ( waser ) ( ) = fract x 0 π = π(0) π(8) 360 x 0π = 36π and 0π = 360 saded outside inside π x 0 x x = 00 degrees lso examine Example and Example C on page 454 of te ook. 8.6 Page 456 Exercise #5 S. Stirling Page 7 of 0

8 C 8 Noteseet L Key Lesson 8.7 Surface rea surface area Te sum of te areas of all te surfaces of a solid. ase (of a solid) polygon or circle used for reference to determine an altitude or oter feature of te solid, or used to classify te solid. lateral face face of a solid oter tan a ase. Does not ave to e vertical! slant eigt Te eigt of eac triangular lateral face of a pyramid. Tey usually use l for slant eigt. Prism Pyramid eigt of prism pentagonal ase rectangular lateral face eigt of prism l Surface rea = (ase area) + (lateral surface area) = (ase area) + (perimeter ase * eigt) 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 = lengt of ase edge l = slant eigt p = perimeter of te ase USE THESE FORMULS TO GET THE FORMULS FOR CYLINDER ND CONE!! Cylinder Same as a prism wit circular ases. Cone Same as a pyramid wit circular ases. circular ase l l rectangular lateral face te can s lael circular ase triangular-is lateral face Surface rea = (ase area) + (perimeter ase * eigt) = (π r ) + ( π r * ) Surface rea = (ase area) + (lateral surface area) = (ase area) + (½ pl ) = (π r ) + (½ * π r * l ) = π r + π r l S. Stirling Page 8 of 0

9 C 8 Noteseet L Key EXMPLE Find te surface area of a rectangular prism wit dimensions 3 m, 6 m and 8 m. EXMPLE B Find te surface area of a cylinder wit ase diameter 0 in. and eigt in. S = B + lateral face area S = (6i 8) + 3( ) = (48) + 3(8) = = 80 m S = B + lateral face area S = π5 + (iπ i 5) ( ) S = 50π + 0π = 70π in EXMPLE C Find te surface area of a cone wit ase radius 5 cm. and slant eigt 0 cm. EXMPLE D Te surface area of te rigt square pyramid is 95 in wit ase edge of 5 in. Wat is te measure of te slant eigt? S = B + lateral face area S = π5 + πii 50 S = 5π + 50π = 75π 35.6 cm S = B + lateral face area 95 = 5 + 4i i5il 95 = 5 + 0il 70 = 0il l = 7 7 in S. Stirling Page 9 of 0

10 C 8 Noteseet L Key 8.7 Page 467 Exercise # Paint rea: tri = i.5 i = 5 trap = ()( + 30) = 5 = 30i 4 = 70 rect One face 987 ft T = (987) + (4 40) T = 3894 ft Paint (cont.): = gallon cost (6)($5) = $400 Roof rea: next = 40(5) = 600 = 40(6.5) = 60 top rect ( ) = 70 ft = 7. undles cost (8)($65) = $70 Total Cost = = $ S. Stirling Page 0 of 0