{VERSION 5 0 "IBM INTEL NT" "5.0" } {USTYLETAB {CSTYLE "Maple Input" -1 0 "Courier" 0 1 255 0 0 1 0 1 0 0 1 0 0 0 0 1 }{CSTYLE "2D Math" -1 2 "Times" 0 1 0 0 0 0 0 0 2 0 0 0 0 0 0 1 }{CSTYLE "2D Comment" 2 18 "" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 1 } {CSTYLE "" -1 256 "" 0 1 0 128 0 1 0 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 257 "" 0 1 0 128 0 1 0 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 258 "" 0 1 0 128 128 1 0 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 259 "" 0 1 128 128 128 1 0 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 260 "" 0 1 0 0 128 1 0 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 261 "" 0 1 0 0 128 1 0 0 0 0 0 0 0 0 0 0 } {CSTYLE "" -1 262 "" 0 1 0 0 128 1 0 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 263 "" 0 1 0 0 128 1 0 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 264 "" 0 1 0 0 128 1 0 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 265 "" 0 1 0 0 128 1 0 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 266 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 } {CSTYLE "" -1 267 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 268 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 269 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 270 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 271 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 } {CSTYLE "" -1 272 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 273 "" 0 1 0 0 128 1 0 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 274 "" 0 1 128 128 0 1 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 275 "" 0 1 128 128 0 1 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 276 "" 0 1 128 128 0 1 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 277 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 } {PSTYLE "Normal" -1 0 1 {CSTYLE "" -1 -1 "Times" 1 12 0 0 0 1 2 2 2 2 2 2 1 1 1 1 }1 1 0 0 0 0 1 0 1 0 2 2 0 1 }{PSTYLE "Heading 1" -1 3 1 {CSTYLE "" -1 -1 "Times" 1 18 0 0 0 1 2 1 2 2 2 2 1 1 1 1 }1 1 0 0 8 4 1 0 1 0 2 2 0 1 }{PSTYLE "Heading 2" -1 4 1 {CSTYLE "" -1 -1 "Times " 1 14 0 0 0 1 2 1 2 2 2 2 1 1 1 1 }1 1 0 0 8 2 1 0 1 0 2 2 0 1 } {PSTYLE "Normal" -1 256 1 {CSTYLE "" -1 -1 "Times" 1 12 0 0 0 1 1 2 2 2 2 2 1 1 1 1 }1 1 0 0 0 0 1 0 1 0 2 2 0 1 }} {SECT 0 {PARA 0 "" 0 "" {TEXT -1 0 "" }{TEXT 257 50 "High School Modul es > Geometry by Gregory A. Moore" }}{PARA 3 "" 0 "" {TEXT -1 4 " \+ " }{TEXT 256 36 "The Circumference & Area of a Circle" }}{PARA 0 "" 0 "" {TEXT -1 95 "\nAn exploration of the circumference, and development and application of the area of a circle.\n" }}{PARA 0 "" 0 "" {TEXT 258 153 "[Directions : Execute the Code Resource section first. Althou gh there will be no output immediately, these definitions are used lat er in this worksheet.]" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{SECT 0 {PARA 4 "" 0 "" {TEXT -1 1 " " }{TEXT 260 7 "0. Code" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 38 "restart; with(plots): with(geometry): " } }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 499 "sbl := 'axes = none, sc aling = constrained, \n thickness = 3, color = COLOR(RGB, .5,.4, .8)':\nsgy := 'axes = none, scaling = constrained, \n thicknes s = 3, color = COLOR(RGB, .5,.5,.6)':\nsbe := 'axes = none, scaling \+ = constrained, \n thickness = 3, color = COLOR(RGB, .8,.7,.5)': \nsor := 'axes = none, scaling = constrained, \n thickness = 3 , color = COLOR(RGB, .9,.7,.1)':\nsgn := 'axes = none, scaling = con strained, \n thickness = 3, color = COLOR(RGB, .4,.7,.4)':" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 233 "cbl := 'color = COLOR(RGB, \+ .5,.4,.8)':\ncgy := 'color = COLOR(RGB, .5,.5,.6)':\ncbe := 'color = C OLOR(RGB, .8,.7,.5)':\ncor := 'color = COLOR(RGB, .9,.7,.1)':\ncyl := \+ 'color = COLOR(RGB, .9,.8,.1)':\ncgn := 'color = COLOR(RGB, .5,.8,.5)' :" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 173 "scanft := 'scaling = \+ constrained, axes = none':\nscanft := 'filled = true, axes = none, sca ling = constrained':\nscant2 := 'thickness = 2, axes = none, scaling = constrained':" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 33 "point(o,0,0);\ncircle( UC ,[ o,1]):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 346 "RegPolyPlot := \+ proc( N )\n local k, n, ngon, c, UC;\n UC := plottools[circle]([0,0 ], 1, color=gray):\n ngon := n -> [seq([ cos(2*Pi*i/n), sin(2*Pi*i/n) ], i = 1..n)]: \n c := n -> COLOR(RGB, .2+n/(2*N), .2+n/(2*N), .5); \n display([ UC,seq( polygonplot(ngon(n), color=c(n)), n = 3..N)],\n \+ scaling = constrained, axes = none);\nend proc:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{SECT 0 {PARA 4 "" 0 "" {TEXT -1 1 " " }{TEXT 261 20 "1. The Circumference" }}{PARA 0 "" 0 "" {TEXT -1 1 "\n" } {XPPEDIT 18 0 "Pi" "6#%#PiG" }{TEXT -1 166 " (pronounced \"pie\") is d efined as the ratio of circumference to diameter. This definition lead s to simple formulas for circumference in terms of diameter and radius .\n" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 10 "C/d = Pi;" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 31 "DiamFormula := C = solve(%, \+ C);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 32 "RadFormula := subs( \+ d = 2*r, %);" }}}{PARA 0 "" 0 "" {TEXT 267 16 "\n\n Example : " } {TEXT -1 70 " If circle has diameter of 10 feet. How long is the the c ircumference?" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 27 "subs( d = 1 0, DiamFormula);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 9 "evalf(%) ;" }}}{PARA 0 "" 0 "" {TEXT -1 1 "\n" }}{PARA 0 "" 0 "" {TEXT -1 1 "\n " }{TEXT 268 13 " Example : " }{TEXT -1 66 " If circlular pond has a radius of 43 feet. How long is the shore?" }}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 26 "subs( r = 43, RadFormula);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 9 "evalf(%);" }}}{PARA 0 "" 0 "" {TEXT -1 62 "\nWe c an turn it around and solve in the other direction too.\n\n" }{TEXT 269 13 " Example : " }{TEXT -1 112 " If the diameter of a circular l og of a giant Sequoia tree is 110 feet\n around, what is the radius of the tree?\n" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 27 "subs( C = 11 0, RadFormula);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 11 "solve(%, r);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 13 "r = evalf(%);" }}} {PARA 0 "" 0 "" {TEXT -1 1 "\n" }{TEXT 266 11 " Example" }{TEXT -1 258 " : Suppose the earth were a sphere with a radius of 6,000 miles ( one mile = 5,280 feet), and a tight metal band went around the equator . If 6 inches of slack were put into the band, how tall of an object ( in feet) could you pass under the now loosened band?\n" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 37 "New_R := 'New_R';\nOld_R := 6000*52 80;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 68 "Band_C := 2*Pi*Old_R ; \n`Original circumference of the band in feet`;" }}}{EXCHG {PARA 0 " > " 0 "" {MPLTEXT 1 0 22 "New_C := Band_C + 1/2;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 19 "New_C = 2*Pi*New_R;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 17 "solve( %, New_R);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 18 "New_R := evalf(%);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 89 "New_R - Old_R;\n`The height of the tallest object you could pass under the band, in feet`;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 26 "%%*12;\n`...and in inches`;" }}}{PARA 0 "" 0 "" {TEXT -1 63 "\nAbout one inch .... the band is about 1 inch off the gr ound.\n " }}}{SECT 0 {PARA 4 "" 0 "" {TEXT -1 1 " " }{TEXT 273 43 "2. \+ Examining Area and Perimeter of Polygons" }}{PARA 0 "" 0 "" {TEXT -1 129 "\nIts very clear where the formula for the circumference of a cir cle comes from, because it flows directly from the definition of " } {XPPEDIT 18 0 "Pi" "6#%#PiG" }{TEXT -1 159 ". However, it is not clear how we get from there to finding a formula for the area of a circle. \+ In fact, the derivation for this rarely covered at this level.\n" }} {PARA 0 "" 0 "" {TEXT -1 390 "Its easy enough to compute the area of t riangles and rectangles. In fact, the area of any region bounded by st raight lines can be computed by cutting it into triangles or other sim ple shapes. However, a curved shape like a circle defies this method, \+ because it can not be cut into a finite number of polygons. \n\nLets b egin by examining regular polygons as approximations to circles. \n\n \n " }{TEXT 274 21 "Geometry View of Area" }{TEXT -1 196 "\n\nA cru de estimate for the area of circle would be to find the area of an ins cribed regular polygon - that is, a polygon which has all of its verti ces on the circle, and each edge is same length. " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "RegPolyPlot( 3);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "RegPolyPlot(6);" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 16 "RegPolyPlot(10);" }}}{PARA 0 "" 0 "" {TEXT -1 2 "\n\n" }{TEXT 275 36 " Numeric View of Area of Polygons" }{TEXT -1 184 "\n\nLets look at these areas numerically. He re are some tables which show the area of various regular polygons. No tice that as n increases, the area gets closer and closer to 3.14159.. .\n" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 141 "A := array( [seq( [ \+ n, evalf(n*sin(Pi/n)*cos(Pi/n) )], n = 2..10) ]):\nA[1,1] := `Number of sides`: A[1,2] := `Area of Polygon`:\nprint(A);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 144 "A := array( [seq( [ n, evalf(n*sin (Pi/n)*cos(Pi/n) )], n = 11..20) ]):\nA[1,1] := `Number of sides`: \+ A[1,2] := `Area of Polygon`:\nprint(A);" }}}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 171 "A := array( [seq( [ 10*n, evalf(10*n*sin(Pi/(10*n) )*cos(Pi/(10*n)) )], \n n = 1..10) ]):\nA[1,1] := `Number \+ of sides`: A[1,2] := `Area of Polygon`:\nprint(A);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 171 "A := array([seq([100*n,evalf(100*n *sin(Pi/(100*n))*cos(Pi/(100*n)) )], \n n = 1..10) ]):\nA[ 1,1] := `Number of sides`: A[1,2] := `Area of Polygon`:\nprint(A); " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 188 "A := array([seq([1000* n,\n evalf(1000*n*sin(Pi/(1000*n))*cos(Pi/(1000*n)) )], \n n = 1..10) ]):\nA[1,1] := `Number of sides`: A[1,2] := `Area of Polygon`:\nprint(A);" }}}{PARA 0 "" 0 "" {TEXT -1 113 "\n\nA s you can see, when we have a regular polygon with 10,000 sides, its a rea is very, very close to the value of " }{XPPEDIT 18 0 "Pi" "6#%#PiG " }{TEXT -1 12 ". How close?" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 12 "evalf(2*Pi);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 29 "deviati on := % - 3.141592447;" }}}{PARA 0 "" 0 "" {TEXT -1 3 "\n\n\n" }{TEXT 276 42 " Numeric View of Perimeter of Polygons" }{TEXT -1 87 "\n\n We can also look at the perimeter of regular polygons. Just as the are a tends toward " }{XPPEDIT 18 0 "Pi" "6#%#PiG" }{TEXT -1 48 ", the per imeters seem to approach the value of 2" }{XPPEDIT 18 0 "Pi" "6#%#PiG " }{TEXT -1 18 " = 6.283185308...\n" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 249 "A := array( [seq( [ 10*n, evalf(10*n*sin(Pi/(10*n))* cos(Pi/(10*n)) ),\n evalf(2* 10*n*sin(Pi/(10*n)))] , \n n = 1..10) ]):\nA[1,1]:=`Number of sides`: A[1,2]:=`A rea of Polygon`:\nA[1,3] := `Perimeter of Polygon`:\nprint(A);" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 255 "A := array( [seq( [ 100*n, \+ evalf(100*n*sin(Pi/(100*n))*cos(Pi/(100*n)) ),\n e valf(2* 100*n*sin(Pi/(100*n)))], \n n = 1..10) ]):\nA[1,1] :=`Number of sides`: A[1,2]:=`Area of Polygon`:\nA[1,3] := `Perimeter \+ of Polygon`:\nprint(A);" }}}{PARA 0 "" 0 "" {TEXT -1 1 "\n" }}{PARA 0 "" 0 "" {TEXT -1 79 "Maybe we can use this observation to find a formu la for the area of a circle.\n " }}}{SECT 0 {PARA 4 "" 0 "" {TEXT -1 1 " " }{TEXT 262 42 "3. Finding the Area Formula Using Polygons" }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 75 "Lets look at regular polygons again, and find an expression for their area." }} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 305 "RegularPolygon( RegPoly ,7, o,1):\npoint(A,[cos(2*Pi/7), sin(2*Pi/7)]):\npoint(B,[cos(4*Pi/7), sin (4*Pi/7)]):\npoint(C,[cos(3*Pi/7)*cos(Pi/7), sin(3*Pi/7)*cos(Pi/7)]): \nsegment( rad1, [o, A]): \nsegment( rad2, [o, B]): \nsegment( h, [o, \+ C]):\nsegment( side, [A,B]): \n`(no output ... proceed to next command )`;\n \n\n" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 311 "display( dra w( \n [ UC(thickness=1,cbe), RegPoly(thickness=2,cgn),\n \+ rad1(color=red,thickness = 2), \n rad2(color=gold, l inestyle = 3),\n h(thickness = 2, color = coral), \n \+ side(thickness = 3, color = green)\n ]), axes=none, scal ing=constrained );" }}}{PARA 0 "" 0 "" {TEXT -1 402 "\n\nLets s ay that this polygon has n sides, each length s, and the polygon is in scribed in a circle of radius r. (In the diagram above, r is red, and \+ s is the thick green segment). And lets let h be the height of the tri angle (the gold segment perpindicular to the green triangle base).\n\n The area of this kind of regular polygon in general, is n times the ar ea of each of the n triangles that make it up." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 78 "n:='n': s:=' s': h:='h': C:='C': P:='P':\n`Area of one triangle` = (1/2)*s*h;" } }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 90 "`Area of Polygon = Total A rea of n Triangles` = n*(1/2)*s*h;\nAreaPolygon := n*(1/2)*s*h;" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 57 "Perimeter := n*s;\n`Area of \+ Polygon` = Perimeter*(1/2)*h;\n" }}}{PARA 0 "" 0 "" {TEXT -1 107 "\nNo w here is where the magic occurs. As we saw in the previous section (a bove), the perimeter tends toward " }{XPPEDIT 18 0 "2*Pi" "6#*&\"\"#\" \"\"%#PiGF%" }{TEXT -1 24 ". Also h tends toward r." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 53 "Perimeter*(1/2)*h, `tends toward `, (2*Pi*r )*(1/2)*r;" }}}{PARA 0 "" 0 "" {TEXT -1 36 "\nIn other words, the area tends to \n" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 29 "`Area of a C ircle ` = Pi*r^2;" }}}}{SECT 0 {PARA 4 "" 0 "" {TEXT -1 1 " " }{TEXT 265 18 "4. Area of Circles" }}{PARA 0 "" 0 "" {TEXT -1 71 "\nNow we ca n use the fruits of our efforts in the form of this formula.\n" }} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 19 "Area := A = Pi*r^2;" }}} {PARA 0 "" 0 "" {TEXT -1 2 "\n " }{TEXT 270 15 "\n Example : " } {TEXT -1 124 " If a goat is chained to a 20 foot rope, and eats all of the grass he is able to eat. How big is the patch of eaten grass? \n " }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 31 "subs( r = 20, Area);\nev alf(%); " }}}{PARA 0 "" 0 "" {TEXT -1 2 "\n\n" }{TEXT 271 14 " Exam ple : " }{TEXT -1 42 " A plate is 30 cm wide. What is the area?\n" }} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 40 "r = 30/2;\nsubs( r = 15, Are a);\nevalf(%);" }}}{PARA 0 "" 0 "" {TEXT -1 1 "\n" }}{PARA 0 "" 0 "" {TEXT 272 16 " Example : " }{TEXT -1 192 "Pizza dough is spread o ut to cover a baking pan 12 inches by 21 inches. If the baker finds a \+ round pan, and re-rolls the dough into the shape of a circle, what is \+ the diameter of the circle? \n" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 18 "PizzArea := 12*21;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 26 "subs( A = PizzArea, Area);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "r^2 = solve(%, r^2);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 28 "r = sqrt(252/Pi); \nevalf(%);" }}}}{SECT 0 {PARA 4 "" 0 "" {TEXT -1 1 " " }{TEXT 263 17 "5. Area of Annuli" }}{PARA 0 "" 0 "" {TEXT -1 61 "\nAn annulus is a ring shape - the difference of two circles.\n" }} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 226 "display([\n polarp lot( 5, theta = 0..2*Pi, \n filled = true, color = white),\n \+ polarplot( 7, theta = 0..2*Pi, \n filled = true, col or = COLOR(RGB,.4,.35,.25)) \n ],scaling = constrained );" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 225 "display([\n polarp lot( 2, theta = 0..2*Pi, \n filled = true, color = white),\n \+ polarplot( 9, theta = 0..2*Pi, \n filled = true, col or = COLOR(RGB,.8,.0,.25)) \n ],scaling = constrained );" }}} {PARA 0 "" 0 "" {TEXT -1 114 "\nSince the shape is the difference of t wo circles - a big circle minus a little one - the formula comes direc tly.\n" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 64 "r := 'r': R := 'R ':\nA = Pi*R^2 - Pi*r^2;\nA = Pi*(R^2 - r^2);\n" }}}{PARA 0 "" 0 "" {TEXT -1 36 "\nThe area of the shapes above are :\n" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 48 "`Area of 1st Annulus` = (7^2 -5^2)*Pi;\neva lf(%);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 48 "`Area of 1st Annu lus` = (9^2 -2^2)*Pi;\nevalf(%);" }}}{PARA 0 "" 0 "" {TEXT -1 78 "\nCa n you find the area of the figure below?... sort of a \"double annulus \"....." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 407 "display([\n \+ polarplot( 2, theta = 0..2*Pi, \n filled = true, color = white),\n polarplot( 5, theta = 0..2*Pi, \n filled \+ = true, color = COLOR(RGB,.4,.8,.25)), \n polarplot( 8, theta = 0..2*Pi, \n filled = true, color = white),\n pola rplot( 9, theta = 0..2*Pi, \n filled = true, color = COLOR(RG B,.4,.8,.25)) \n ],scaling = constrained );" }}}{PARA 0 "" 0 " " {TEXT -1 2 "\n " }}}{SECT 0 {PARA 4 "" 0 "" {TEXT -1 1 " " }{TEXT 264 19 "6. Area of Ellipses" }}{PARA 0 "" 0 "" {TEXT -1 36 "\nAn ellip se is a round figure also.\n" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 139 "plot( [5*cos(theta),2*sin(theta), theta = 0..2*Pi], filled = true ,\n scaling = constrained, thickness = 3, color = COLOR(RGB,.3,.6,. 3)); " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 320 "display( [plot( \+ [cos(theta),sin(theta), theta = 0..2*Pi], thickness=2,\n scaling = constrained, axes = none, color = COLOR(RGB,.7,.5,.2)),\n seq ( \n plot( [k*cos(theta)/6,sin(theta), theta = 0..2*Pi], \n \+ scaling = constrained, axes = none, color = COLOR(RGB,.2,.1,.4)),\n k = 1..18)\n ]);" }}}{PARA 0 "" 0 "" {TEXT -1 231 "\n\nIn a certain sense, an ellipse is a circle with two different radii - a \+ horizontal radius, a, and a vertical radius, b. If you think of an ell ipse in this way, its easy to remember the formula for the area of an \+ ellipse.\n " }}{PARA 0 "" 0 "" {TEXT 277 15 " " } {XPPEDIT 18 0 "A = a*b*Pi" "6#/%\"AG*(%\"aG\"\"\"%\"bGF'%#PiGF'" }} {PARA 0 "" 0 "" {TEXT -1 1 "\n" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 151 "plot( [5*cos(theta),2*sin(theta), theta = 0..2*Pi], filled = tr ue,\n scaling = constrained, thickness = 3, color = COLOR(RGB,.2,.1 ,.4));\nA = 5*2*Pi; " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }} }{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 150 "plot( [3*cos(theta),7*sin( theta), theta = 0..2*Pi], filled = true,\n scaling = constrained, t hickness = 3, color = COLOR(RGB,.7,.7,.5));\nA = 3*7*Pi;" }}}}{PARA 0 "" 0 "" {TEXT 259 36 "\n \251 2002 Waterloo Maple Inc " }}} {MARK "0 1" 9 }{VIEWOPTS 1 1 0 1 1 1803 1 1 1 1 }{PAGENUMBERS 0 1 2 33 1 1 }