Geometry !"#$"%&'()#%"*( +"&,$"%"&(-./(0&"-( +-1"(2(#3(45( Perimeter and Area

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1 MAT 142 College Mathematics Geometry Terri L. Miller & Elizabeth E. K. Jones Module GM revised April 20, 2011 ""&)"* +"&,""&-./0&"- +-1"2345 Perimeter and Area "&)&*+**,+*,-./&0*.1*2*13-0)+)4/&,+5 We are going "14446**2)3-0)+)4-+*)7,)-*2/&,)-*2*13 to start our study of geometry with two-dimensional figures. We will look at the one-dimensional -0)+) :.*2/&,5 distance around the figure and the two-dimensional space covered by the figure. ;280*/+28+-/)-+*2-+*)7,)-*2+285<)7 The perimeter of a shape is defined as the distance around the shape. Since we usually 1,+,44.-+7,++*280*/84.&)+=74+-84)/&,+12+ discuss the perimeter of polygons closed plane figures whose sides are straight line segment), +-++*&2*4)+&0)*>?1:4*747,4*80*:.@,+* we are able--)&,8*24)&*2+/72/*2+-+5"2)1*46:,**2 to calculate perimeter by just adding up the lengths of each of the sides. When we talk about 80*/774?1744*:.*2+874)0/77,0/)75<)7 the perimeter of a circle, we call it by the special name of 77,0/)7 circumference Since we dont 1-)A*29+*&2*+-+*--,8/*277,0/)7=80*>/ have straight sides to add up for the circumference perimeter) of a circle, we have a formula 774?129/0,4/747,4*)&*2+5 for calculating this. Circumference Perimeter) of a Circle J7,0/)7=K0*>/J74 C = 2πr J = Lπ r = radius &9&-/,:*3;"<,&<=" >9;".:?"&;-,*-@@&A,-"/?BC7874DB π =the number that is approximated by BC084DE F)-*280*/*2/&,:41 Example 1. Find the perimeter of the figure below * <4,*)E G*+*08*)&*@,+*+**--)&/*2),0:+&9)*&*2? :,**2*144)*&9,+*280*5;2+)*2**144)* + *2* *2+ /&, 2+ <GH +-+ )- 1 )4. &9) /, ),0:+5 " 0,+* /+* -*0) *2 4)&*2+ / *2 *1 +-+ *2*)*4:4-:/17)/)-*280*5I*A+46 **2/&,&)*/)-*24)&*2+/*2*2+-+5 Solution:: It is tempting to just start adding of the numbers given together, but that will not give us the perimeter. The reason that it will not is that this figure has SIX sides and we are only given four numbers. We must first determine the lengths of the two sides that are not labeled before we can find the perimeter. Lets look at the figure again to find the lengths of the other sides. Since our figure has all right angles, we are able to determine the length of the sides whose length is not currently printed. Lets start with the vertical sides. Looking at the image below, we can see that the length indicated by the red bracket is the same as the length of

2 .0-)020,0,&;&)0-/.4&.89<+-00,&5-+&3&/212&- "&)*+)&,-.-//)+,0-+/&.12&-)&-3/&04&0&)5&0,&.&& /&+0, 0,-0 * 0,& 0,& /&+0,.4&.4-0&4 2,.& /&+0, 36 0,&. 0 )&4))&0/6 3)-<&0. 7)0&48 0,&.-5& 9&0:. -. 0,&/&+0,*0,&;&)0-/.4&2,.&/&+0,.=0.8>,.5&-..0-)020,0,&;&)0-/.4&.89<+-00,&5-+&3&/212&-.&& 0,-0 0,& /&+0, 4-0&4 36 0,& )&4 3)-<&0. 0,&.-5& ,-0 2& - -//-0& 0,& Geometry /&+0, * 0,& +)&&.&+5& )-0+ 0,&/&+0,*0,&;&)0-/.4&2,.&/&+0,.=0.8>,.5&-. *)5??8 0, ,-0 2& - -//-0& 0,& /&+0, * 0,& +)&&.&+5& )-0+ = *)5??8 >,. 5&-. 0,-0 0,& :5 77 the vertical side whose length is 4 units. This means that we can calculate the length of the green segment by subtracting 4 from 11. This means that the green segment is 7 units. 7798:5 6 6 A-.5/-)5-&)12&--//-0&0,&/&+0,*0,&0,&) &.+?= C = B 8>,.+;&..0,&/&+0,.*-//0,&.4&.-..,20,&*+)&3&/28 A-.5/-)5-&)12&--//-0&0,&/&+0,*0,&0,&) &.+?= C = B 8>,.+;&..0,&/&+0,.*-//0,&.4&.-..,20,&*+)&3&/28 78 In a similar manner, we can calculate the length of the other missing side using. This gives us the lengths of all the sides as shown in the figure below ; 8 77 ; 5 ""&)"* +"&,""&-./0&" " D20,-02&,-;&-//0,&/&+0,.*0,&.4&.12&-.57/6 6 "&)&*+**,+*,-./&0*.1*2*13-0)+)4/&,+5 -//-0&0,&7&)5&0&) ,&/&+0,.0+&0,&)0+&0 80* "14446**2)3-0)+)4-+*)7,)-*2/&,)-*2*13 = D20,-02&,-;&-//0,&/&+0,.*0,&.4&.12&-.57/6 +?= +?? + C + B = EF8 "&7&)5&0&).-)&G.00,&/&+0,. -0)+) :.*2/&,5 */&.10,&7&)5&0&)2/43&EF0.8 -//-0&0,&7&)5&0&) ,&/&+0,.0+&0,&)0+&0 = +?= +?? + C + B = EF8 "&7&)5&0&).-)&G.00,&/&+0,. ;280*/+28+-/)-+*2-+*)7,)-*2+285<)7 */&.10,&7&)5&0&)2/43&EF0.8 1,+,44.-+7,++*280*/84.&)+=74+-84)/&,+12+ -)&- +-++*&2*4)+&0)*>?1:4*747,4*80*:.@,+* -)&- --)&,8*24)&*2+/72/*2+-+5"2)1*46:,**2 80*/774?1744*:.*2+874)0/77,0/)75<)7 77,0/)7 1-)A*29+*&2*+-+*--,8/*277,0/)7=80*>/ 774?129/0,4/747,4*)&*2+5 J7,0/)7=K0*>/J74 Area of a Rectangle J A = L= π bh b = base of the rectangle &9&-/,:*3;"<,&<=" >9;".:?"&;-,*-@@&A,-"/?BC7874DB h = the height of the rectangle BC084DE F)-*280*/*2/&,:41 Now that we have all the lengths of the sides, we can simply calculate the perimeter by adding the lengths together to get Since perimeters are just the lengths of lines, the perimeter would be 50 units. The area of a shape is defined as the number of square units that cover a closed figure. For most of the shape that we will be dealing with there is a formula for calculating the area. In some cases, our shapes will be made up of more than a single shape. In calculating the area of such shapes, we can just add the area of each of the single shapes together. We will start with the formula for the area of a rectangle. Recall that a rectangle is a quadrilateral with opposite sides parallel and right interior angles. Example 2. Find the area of the figure below <4,*)E G*+*08*)&*@,+*+**--)&/*2),0:+&9)*&*2? :,**2*144)*&9,+*280*5;2+)*2**144)* + *2* *2+ /&, 2+ <GH +-+ )- 1 )4. &9) /, ),0:+5 " 0,+* /+* -*0) *2 4)&*2+ / *2 *1 +-+ *2*)*4:4-:/17)/)-*280*5I*A+46

3 MAT Module 3 Solution: This figure is not a single rectangle. It can, however, be broken up into two rectangles. We then will need to find the area of each of the rectangles and add them together to calculate the area of the""&)"* whole figure. +"&,""&-./0&"- +-1"2324 ""&)"* There is more than +"&,""&-./0&"- one way to break ""&)"* this figure into rectangles. +-1"2324 We will +-1"2324 only illustrate one below. +"&,""&-./0&" " "&)*+&,,)-*+./,0102*, ,* *57,8*34.*.-,* *.24,89" "&)*+&,,)-*+./,0102*,.3 "&)*+&,,)-*+./,0102*,.3 &,5*4,7+,237,.3.-,*459:,3*,.-,* *57,8*34.*.-,* *.24,89".-,* *57,8*34.*.-,* *.24,89" ;,,.)2-*)-55,+*3)2-)-55,24,9< &,5*4,7+,237,.3.-,*459:,3*, &,5*4,7+,237,.3.-,*459:,3*,.7,.3.-,*452 ;,,.)2-*)-55,+*3)2-)-55,24,9< A = + = => ;,,.)2-*)-55,+*3)2-)-55,24,9<.7,.3.-,*452 A = + = A +?@ = =>?@ = =>.7,.3.-,*452 "&,352,,5;.)./,72*3,.7,4.* A = + = =>.-,*459"/*),,,5*4,7*7,232B1*2,9" "&,352,,5;.)./,72*3,.7,4.* "&,352,,5;.)./,72*3,.7,4.* 3,72*3,5*4,7,,.237,4.*.-,*45)*.-,*459"/*),,,5*4,7*7,232B1*2,9".-,*459"/*),,,5*4,7*7,232B1*2,9" )-5-15,3,0.2;,.2*CD;05?+&9:,5*4,)E 3,72*3,5*4,7,,.237,4.*.-,*45)* 3,72*3,5*4,7,,.237,4.*.-,*45)* 1*2,9 )-5-15,3,0.2;,.2*CD;05?+&9:,5*4,)E 1*2,9 )-5-15,3,0.2;,.2*CD;05?+&9:,5*4,)E 1*2,9 67 1*2, We have shown above that we can break the shape up into a red rectangle figure on left) and a green rectangle figure on right). We have the lengths of both sides of the red rectangle. It does not matter which one we call the base and which we call the height. The area of the red rectangle is A = bh = 4 14 = 56. We have to do a little more work to find the area of the green rectangle. We know that the length of one of the sides is 8 units. We had to find the length of the other side of the green rectangle when we calculated the perimeter in Example 1 above. Its length was 7 units <1,.7,4.*.-,*452 5 A = 5 + = B E = => 9<1, 5.7,) <1,.7,4.*.-,*452 A = + = A = B + E = B => 9<1, E = => 9<1, <1,.7,4.*.-,*452.7,) ,*45 +.-,*45 A = + = B => E+ = => => = 9<1,??F9:*..7,) ,*45.3.-,* * 7.-,*45 4.*.-,*45 = => + => = => =??F + => 9:* =??F9:* ,* *.-,*45 = => + => =??F9:* Thus the area of the green rectangle is A = bh = 8 7 = 56. Thus the area of the whole figure is area of red rectangle + area of green rectangle = = 112. In the process of calculating the area, we multiplied units times units. This will produce a final reading of square units or units squared). Thus the area of the figure is 112 square units. This fits well with the definition of area which is the number of square units that will cover a closed figure. Our next formula will be for the area of a parallelogram. A parallelogram is a quadrilateral with opposite sides parallel. 5 5

4 6"-)"+&+*"*.0-&.)::;)7-+&-/.)56".)*.)2,,2."" 4*./../*+&+2".".)"/-3<&*)7-+&-/.)"+2.,,=&+,)4*.0-&5 4 Geometry >-&/?*&3-,+2.,,<*&"+&+*++&+,,,0&+35@+&+,,,0&+3.) +&+,,,0& &.,+&+,2.").).4)+&+,,,5 Area of a A = = <" b = base of the parallelogram 789"7-*"39":-&-;;";1&- 989"9",1939":-&-;;";1&- h = the height of the parallelogram A-2.,,/."+".).)")+3+)"*&3-,+*&"+&+*+ &+/0,5@&+/0,.)B-)+).+,C*+&+,,,0&+356"".0"* ++&+,,,0&+3.)+)03/"+//)"*"+&+,,,0&+3+/4 "<+)*"+&+,,,0&+3+/4.)&/4.-,+&<""+/4" <+)5D/"+)*+&+/0,1".).)")+3+)/*").4)*" of the rectangle &+/0,"+.)&/4.-,+&"<+)5 that is perpendicular to the base. Example 3. E?+3,FG Find the area of the figure below. H./4"+&+*"*.0-&<,2 You will notice that this is the same as the formula for the area of a rectangle. A rectangle is just a special type of parallelogram. The height of a parallelogram is a segment that connects the top of the parallelogram and the base of the parallelogram and is perpendicular to both the top and the base. In the case of a rectangle, this is the same as one of the sides I,-./G D/".)*.0-&1"<+)*"+&+,,,0&+3.):J-/.)+/4" ".0".)K-/.)56".)3+/"+2/,C/43-,.,C*./4 "+&+*@ = <" = :J K = LM )7-+&-/.)5 A-)"-,4/."+2+//*./4"&.3&*".)*.0-& )./24/"+=",/0")*+,,*").4)1+/42"+=/ Solution: In this figure, the base of the parallelogram is 15 units and the height is 6 units. This mean that we only need to multiply to find the area of A = bh = 15 6 = 90 square units. You should notice that we cannot find the perimeter of this figure since we do not have the lengths of all of the sides, and we have no way to figure out the lengths of the other two sides that are not given. Our next formula will be for the area of a trapezoid. Area of a Trapezoid A = 1 2 b 1 + b 2 )h b 1 = one base of the trapezoid b 2 = the other base of the trapezoid h = the height of the trapezoid A trapezoid is a quadrilateral that has one pair of sides which are parallel. These two sides are called the bases of the trapezoid. The height of a trapezoid is a segment that connects the one base of the trapezoid and the other base of the trapezoid and is perpendicular to both of the bases. Example 4. Find the area of the figure below

5 +"8,9&10"0,)6,/-"<&//,<0-,&/,7"0,&-,+"8,9&1"/1-, &-,+7"0,&-,+"8,9&1"/108,+8,/1<*."+&7&-&-,7"0,03 MAT Module 5 A/1-,"+,"&-,)*+, Solution: For this trapezoid, the bases are shown as the top and the bottom of the figure. The lengths of these sides are 45 and 121 units. It does not matter which of these we say is b 1 and which is b 2. The height of the trapezoid is 20 units. When we plug all this into the formula, we get A = 1b b 2 )h = 1 2 B&.*&/@ A&+-0+"8,9&1C-,7"0,0"+,0-&/"0-,&8"/1-,7&&6 &-,)*+,3=-,.,/)-0&-,0,01,0"+,?D"/1EFE*/03G 1&,0/&6",+-<-&-,0,,0" )20 = 1660 E square "/1-<-07 units. F 3=-, -,)-&-,+"8,9&10FH*/03I-,/,8.*)"..-0/&-, E E &+6*."C,),: = 7E + 7F )- = EFE +?D) FH = EJJH 0;*"+, F F */03 Our next formulas will be for finding the area of a triangle a three-sided polygon). We will have more than one formula for this since there are different situations that can come up which will require different formulas Triangle Area Formulas Area of a Triangle for a triangle with a base and a height A = 1bh 2 b = the base of the triangle h = the height of the triangle Heron s Formula for a triangle with only sides A = ss a)s b)s c) a = one side of the triangle b = another side of the triangle c = the third side of the triangle s = 1 a + b + c) 2 The height of a triangle is the perpendicular distance from any vertex of a triangle to the side opposite that vertex. In other words the height of triangle is a segment that goes from the vertex of the triangle opposite the base to the base or an extension of the base) that is perpendicular to the base or an extension of the base). Notice that in this description of the height of a triangle, we had to include the words or an extension of the base. This is required because the height of a triangle does not always fall within the sides of the triangle. Another thing to note is that any side of the triangle can be a base. You want to pick the base so that you will have the length of the base and also the length of the height to that base. The base does not need to be the bottom of the triangle.

6 6 Geometry You will notice that we can still find""&)"* the area of a triangle if we dont have its height. This can be done +"&,""&-./0&"- in the case where we have the lengths of all the sides+-1"2345 of the triangle. In this case, we would use Herons formula. ""&)"* "&)* +"&,""&-./0&"- +-1"2345 +,-./0123/03,4516 "&)* +,-./0123/03,4516 Example 5. Find the area of the figure below 9 :74 9 : &5/,2-* 82/,9/0/,-/0,:3,4510:.:0.&,-/0/,::02;-/2< 72&5/,2-* 1-.,95&1/2/0:,./0/,:=6>5-,/:,-&-4/06?0,:,:02;;,-.,9//00,40/23/0/1,-4&@/0.:0.&,-A-./0<: 82/,9/0/,-/0,:3,4510:.:0.&,-/0/,::02;-/2< 23/0/1,-4&@/0:,./0//0.:0.&,-,:1-.,95&1/2A6 1-.,95&1/2/0:,./0/,:=6>5-,/:,-&-4/06?0,:,:02;;?0/-:;0B<2/0/00,40/-./0<:23/0,:/1,-4&C,-.,9//00,40/23/0/1,-4&@/0.:0.&,-A-./0<: :2;9-D5:/&54/0:-5<1:,-/2/03215&/24/ 23/0/1,-4&@/0:,./0//0.:0.&,-,:1-.,95&1/2A6?0/-:;0B<2/0/00,40/-./0<:23/0,:/1,-4&C F F G = <0 = F=6E) :H515-,/:6 :2;9-D5:/&54/0:-5<1:,-/2/03215&/24/ > F F G = A = 1 > <0 = F=6E) :H515-,/:6 > 2 bh = 1 8.2)4.5) = 18.45square units. 82/,9/0//0-5<1I,:4,B-:/0&-4/ /0:,.: > 2 23/0/1,-4&6?0,::,.,:-2/0,40/23/0/1,-4&:,-9,/,: -2/1-.,95&1/2-2/01:,.23/0/1,-4&6J/,:&:2-2/ 82/,9/0//0-5<1I,:4,B-:/0&-4/ /0:,.: <:23/0/1,-4&C:,-9/01,:-2,-.,9/,2-23/0 23/0/1,-4&6?0,::,.,:-2/0,40/23/0/1,-4&:,-9,/,: 1-.,95&1.,:/-9</;-/0/:,.-./022:,/B1/"6-2/1-.,95&1/2-2/01:,.23/0/1,-4&6J/,:&:2-2/?0,:-:/0/,/,:-2/5:.,-/09&95&/,2-23/0123/0 <:23/0/1,-4&C:,-9/01,:-2,-.,9/,2-23/0 /1,-4&6 1-.,95&1.,:/-9</;-/0/:,.-./022:,/B1/"6?0,:-:/0/,/,:-2/5:.,-/09&95&/,2-23/0123/0 /1,-4&6 "&I* +,-./0123/03,4516 "&I* +,-./0123/03,4516 Solution: Notice that in this figure has a dashed line that is shown to be perpendicular to the side that is 8.2 units in length. This is how we indicate the height of the triangle the dashed line) and the base of the triangle the side that the dashed line is perpendicular to). That means we have both the height and the base of this triangle, so we can just plug these numbers into the formula to get Notice that the number 6 is given as the length of one of the sides of the triangle. This side is not a height of the triangle since it is not perpendicular to another side of the triangle. It is also not a base of the triangle, since there is no indication of the perpendicular distance between that side and the opposite vertex. This means that it is not used in the calculation of the area of the triangle. Example 6. Find the area of the figure below ;75 ;75 : : Solution: In this figure there are two dashed lines. One of them is extended from the side of the triangle that has a length of 1.7. That dashed line is show to be perpendicular to the dashed line that has a length of 2.6. This is how we indicate that the dashed line that has a length of 2.6 is the height of the triangle and the base of the triangle is the side of length 1.7. The height here is outside of the triangle. Also, the dashed line that is extended from the base is not part of the triangle and its length is not relevant to finding the area of the triagnle. Since we have both the height and the base of this triangle, we can just plug these numbers into the formula to get A = 1bh = 1 1.7)2.6) = square units. 2 2 Example 7. Find the area of the figure below

7 = ;+ = E849DE>4?D = > 4 >8) > >,F1/0)&,4) ) G716<0)9) MAT Module 7 H&3)+0)1/01)"-)+0)-&./04) ) ) "&") I"),+"3)"&=0)+1)20)3")")+1B0)1)+0&.+)-"/)+&,)/&1.04)) :+&,)601,)+1)20)=1"),0)+0)-"/61)+1)20)+1B0) K0/"L,)H"/61) ;00),&.)")-&3)+0)1/01)"-)+&,)/&1.04))J0)3")+1B0) ""&)"* =, E, 1+"&,""&-./0&"- DE, ; DE, = D +0)0.+)"-)1)+/00),&30,)"-)+0)/&1.04)):+&,)601,) +-1"22345 > +1)20)=1),0)K0/"L,)H"/61)")-&3)+0)1/01)"-)+&,) -9."*,/"4:"&,-.1;" " = /&1.04) <9-.:"&*,/"4:"&,-.1;" + & + ) = ) = *+, +-.// :;<=37=3;. ) =9:":,&/*,/"4:" >09.3?"@.9A652;.@=3B;<02902.@;<=";9=23750;.C0 H"/)+&,)-"/61A)&)3"0,)")610/)2+&=+),&30)20)1;0)1A) &,-.1;" ;A)"/)=4))H"/)"/)</<",0,A)20)2&)0)1);0)?A););0)9A)13)=);0) *9>?- O);)O)=D D = "" 2)" C)" = *+,*+, )*+, &)*+, ) = *+, M4))N"2)+1)20)+1B0)1A);A)13)=A)20)003)")=1=10),),") "E629063=;"+ +1)20)=1)<.)1A);A)=)A)13),)&")+0)-"/614))J0).0) D3.;<09@.9A652;<2;/0290=3;090";0B=3=";<0F:;< G<0.90A+ G<="2445=0";..35:9=7<;;9=23750"+G<0F:;< G<0.90A9052;0";<0 5037;<".@;<0"=B0".@29=7<;;9= Pythagorean Theorem F:;< G<0.90A a 2 + b 2 = c C = 1 a = leg one side of the triangle that makes up the right angle) -9:"1;."*,/"3<"&,-.1:"<--="*>?<"&,1<-.1:"@ b = leg another A9:"1;-.<"&*,/"3<"&,-.1:"<--="*>?<"&,1< side of the triangle that makes up the right angle -.1:"@ c B9<?".>*";*,/"??*,"<"&,1<-.1:"@ = hypotenuse side opposite the right angle) H<036"=37;<0F:;< G<0.90AI=;="=A4.9;23;;.A2J0"690;<2;/0 25/2:"6"0;<0507".@;<0;9=23750@.9223BC23B;<0<:4.;036"0@.91+ KL2A450M N=3B;<05037;<.@;<0;<=9B"=B0.@;<0;9=23750C05./+ Solution: You should notice that we do not have a height for this triangle. This means that we cannot use the formula that we have been using to find the area of this triangle. We do have the length of all three sides of the triangle. This means that we can use Herons Formula to find the area of this triangle. For this formula, it does not matter which side we label a, b, or c. For our purposes, we will let a be 6, b be 7, and c be 8. Now that we have a, b, and c, we need to calculate s so that we can plug a, b, c, and s into the formula. We get s = ) = 1 21) = Now 2 2 we can plug everything in to Herons formula to find the area of this triangle to be A = ss a)s b)s c) = )10.5 7)10.5 8) square units. Another formula that we are interested in is the Pythagorean Theorem. This applies to only right triangles. The Pythagorean Theorem relates the lengths of the sides of a right triangle. When using the Pythagorean Theorem, it is important to make sure that we always use the legs of the triangle for a and b and the hypotenuse for c. Example 8. Find the length of the third side of the triangle below O.56;=.3M G<0@=7690="29=7<;;9=237502"=3B=12;0BC:;<0C.L=3.30.@;< ".@;<0;9=23750)+H0300B;.B01=B0/<2;;<0"=B0/ J=37@.9="=3;09A".@2507.9;<0<:4.;036"0.@;<0;9= G<0<:4.;036"0=";<0"=B0.@;<0;9= "=;0;<09=7<; G<2;/.65BC0;<0"=B0;<2;<2"5037;<=3.694=1;690+G<6"1 Solution: The figure is a right triangle as indicated by the box in one of the angles of the triangle). We need to decide what the side we are looking for is in terms of a leg or the hypotenuse of the triangle. The hypotenuse is the side of the triangle opposite the right angle. That would be the side that has length 26 in our picture. Thus c will be 26 in our formula. This means that the other two sides of the triangle are legs a and b. We will let a be 10, and we will thus be looking for b. When we plug all this into the formula, we get

8 8 Geometry a 2 + b 2 = c b 2 = b 2 = 676 We now need to solve this equation for b b 2 = 676 b 2 = 576 b = 576 = 24 Since we are asked for a length, we have that the third side is 24 units. Notice that this is not square units since we are not finding an area). We are now going to move on to circles. We already mentioned the perimeter circumference) of a circle in the perimeter sections. We also need a formula for finding the area of a circle. Area of a Circle A = πr 2 r = radius of the circle π = the number that is approximated by For circle problems we need to remember that the circumference perimeter) of a circle is only the curved part of the circle. It does not include either the radius or diameter of the circle. In order to find the perimeter and area, we will need the radius of the circle. Recall that the diameter of a circle is a segment from on point on the circle to another point on the circle that passes through the ""&)"* center of the circle. A radius of a circle is half of a +"&,""&-./0&"- +-1"23456 diameter i.e. a segment that from one point on the circle to the center of the circle). Example &51*67* Find the perimeter and area of the circle below. 8")9*-:*5"&-*0)9*00*/*-:*"1*;1/<=* 22 * * >/1-"/)7*?:*)&;*@@*")*-:*"A*0;/B*"C*-:*1)A-:*/*-:*9"0&-*/* "&)* -:"C*"1=**D*)9*-:*09"C*-/*;*0;1*-/*C*-:*/&10C=**?:* 11) = ,"&-.*/*0* "1* 09"C*"C*:01*/*-:*9"0&-=**?:C* ** 2 +9"0&-. +@@. = E=E=** 2 = 2π D*0*)/<*09F*-/*")9*;/-:*-:*5"&-*+"&).*0)9* C = 2πr = 2π)5.5) = 11π *;F*51AA")A*E=E*")-/*0:*/&10*/*=*,"&-7*** Solution: The number 11 in the figure above is the length of the diameter of this circle. We need the radius to be able to use the formulas. The radius is half of the diameter. Thus r = 1diameter) = 1 We are now ready to find both the perimeter 2 2 circumference) and area by plugging 5.5 into each formula for r. Perimeter: &7&-/,8*49" :,&:;" <79".8="&9-,*->>&?,-"/= 3@2A25B3 ** = 2π = 2+π.+E=E. GH=EEIE@6***?:*0)C<*@@π*)"-C*"C*0)*40-*0)C<*/*-:*5"&-=**?:* 0)C<*GH=EEIE@6*)"-C*"C*0)*055/4"&0-"/)=* *

9 ** 2 2 D*0*)/<*09F*-/*")9*;/-:*-:*5"&-*+"&).*0)9* 00*;F*51AA")A*E=E*")-/*0:*/&10*/*=*,"&-7*** MAT Module 9 ** = 2π = 2+π.+E=E. GH=EEIE@6***?:*0)C<*@@π*)"-C*"C*0)*40-*0)C<*/*-:*5"&-=**?:* 0)C<*GH=EEIE@6*)"-C*"C*0)*055/4"&0-"/)=* * J07* 2 A = πr 2 2 J = π = π +E=E. = π5.5) = GK=2E 2 = π 30.25π 6E=KGG@IIL **?:*0)C<*GK=2Eπ* The <79".8="&9- answer 11π units is an exact answer for the perimeter. The answer units is,*->>&?,-"/= an approximation. Area: = 2π &7&-/,8*49" :,&:;" 3@2A25B3 The answer 30.25π square CM0*)"-C*"C*0)*40-*0)C<*/*-:*00=**?:*0)C<* units is an exact answer for the area. The answer E=KGG@IIL*CM0*)"-C*"C*0)*055/4"&0-"/)=* square units is an approximation. ""&)"* * +"&,""&-./0&"- +-1"23456 Example &51*@K7* Find the perimeter and area of the semicircle below. 8")9*-:*5"&-*0)9*00*/*-:*C&""1*;1/<=* "&") *)+&,)-."/012)30)+450)4),01&6&.60)74)+48)"8)4)6&.609:));+0) <&4100.)"8)+&,)+48)6&.60)&,)=>?:));+,)+0).4<&,)&,) * =. = 7=>?9 :) > ) 2CD Solution: In this A0.&100.) problem, we have a semicircle a half of a circle). The diameter of this half circle is 120. Thus B0)"3)64)8&<)+0)-0.&100.:))B0)3&)8&<)+0)6& ) the radius is r == 1 120) = "8)+0)3+"0)6&.60)4<)+0)<&5&<0)&)/C)>),&60)30)"C)+450) Perimeter: We now +48)"8)4)6&.60:));+&,)3&)D&50),) can find the perimeter. We will E = > find π 7@?9the = =>? circumference π )4,)+0) of the whole circle and then divide it by 6& )8".)+0)3+"0)6&.60)4<)@?F)&,)4,)+0) 2 since we only have half of a circle. This will give us C = 2πr = 120π as the circumference6& )"8)+48)"8)+0)6&.60:))) for the whole circle and 60π units as the circumference of half of the circle. ) G8".40C)+&,)&,)")+0)4,30.)8".)+0)-0.&100.)"8)+0) Unfortunately this is 8&D.0)D&50:));+0)6& )"8)4)6&.60)&,)+0)6.50<)-4.:)) not the answer for the perimeter of the figure given. The circumference of a circle is the curved ;+0),.4&D+)&0),0D10)&)".)8&D.0)3"<)")+450)/00) part. The straight line segment in our figure would not have been part of the circumference -4.)"8)+0)6& )"8)+0)3+"0)6&.602)4<)+,)&)&,)") of the whole circle, and thus it is not included as part of half of the circumference. We &6<0<)4,)-4.)"8)+48)"8)+0)6& :))B0)"C)+450) only have the red part shown in the picture below. +0).0<)-4.),+"3)&)+0)-&6.0)/0"3:) * 278 B0),&)00<)")&6<0)+0),.4&D+),0D10)+4)&,)=>?)&,) "D:));+,)+0)3+"0)-0.&100.)"8)+0)8&D.0)&,) 6.50< -4. +,.4&D+ -4. π + =>? K?L:HIJJJI &,:) We still need to include the straight segment that is 120 units long. Thus the whole perimeter of the figure is ) M.04) B0)64)4,")8&<)+0)4.04:))N0.0)30)3&)-D)@?)&)8".).)&)+0) 4.04)8".14)8".)4)3+"0)6&.60)4<)+0)<&5&<0)/C)>)8".)+0)+48) 6&.60:));+&,)3&)D&50),) M = π 7@?9 > = K@?? π ),O4.0)&,)8".) curved part + straight part = 60π units. Area: We can also find the area. Here we will plug 60 in for r in the area formula for a whole circle and then divide by 2 for the half circle. This will give us A = π60) 2 = 3600π square units for the area of the whole circle and 1800π square units for the area of the half +0)4.04)"8)+0)3+"0)6&.60)4<)=L??F),O4.0)&,)8".)+0) circle. Now are we 4.04)"8)+0)+48)6&.60:))P"3)4.0)30)<"0)3&+)8&<&D)+0)4.04) done with finding the area or is there more that we need to do like we did in finding the perimeter? ".)&,)+0.0)1".0)+4)30)00<)")<")&Q0)30)<&<)&)8&<&D)+0) If we think back to the definition of the area, it is the number of square units needed -0.&100.R))*8)30)+&Q)/46Q)")+0)<08&&&")"8)+0)4.042)7&)&,) to cover the figure) we should see that there is nothing further to do. The inside of the figure +0)1/0.)"8),O4.0)&,)00<0<)")6"50.)+0)8&D.09)30) was covered already. We just have half as much inside as we did before.,+"<),00)+4)+0.0)&,)"+&d)8.+0.)")<":));+0)&,&<0)"8) Example 11. Walk 10 yards south, then 12 west and then 3 yards south. How far from the original starting point are you? Solution: To solve this problem, it would help to have a picture.

10 ")&0+,+2""+,;0+"&30-D ")&*+,-&./01&&2&345&6-")1&)2*7-/) ++&.&+8&*0&4 10 E02-"+0= Geometry 9:7;2&<<= "" ")&0+,+2""+,;0+"&30-D E02-"+0= 26-&/* "" 7-&/* & 28-&/* 26-&/* G801&.&/&&&/+;"+00*)0..&.2>&+")&;082&74 28-&/* 5)".&&>&"0*++)0.*+"+*07.)&&.&""& 7-&/* "0.)&&.&&&4F)"+"/&.0-28&&;&&"&83 "+,)"2+&*07")&"""0")&*++)4F)+.+228&)0.+")& & ;+/"-&8& />)&2+&4 "" G801&.&/&&&/+;"+00*)0..&.2>&+")&;082&74 5)".&&>&"0*++)0.*+"+*07.)&&.&""& "0.)&&.&&&4F)"+"/&.0-28&&;&&"&83 "+,)"2+&*07")&"""0")&*++)4F)+.+228&)0.+")& ;+/"-&8& />)&2+&4 26-&/* "" 28-&/* 7-&/* 26-&/* & 28-&/* 7-&/* Above we can see a description of how we walked in the problem. What we are asked to find is how far it is from where we started to where we ended. That distance would be represented by a straight line from the start to the finish. This will be shown in the picture below by a black dashed line. & We need to figure out how long this line is. We might be tempted to say that we have two right triangles and that the line we are looking for is just the sum of each triangles hypotenuse. The only problem with that is that we do not know the lengths of the legs of each of the right triangles. Thus we are not able to apply the Pythagorean Theorem in this way. We are on the right track though. This is a Pythagorean Theorem problem. If we add a couple of line segments, we can create a right triangle whose hypotenuse is the line we are looking for.

11 =6"/-*& </&*&3 H + 4 H = : H -:;"1 <:;"1 =:>?".@*" )&&*"/&,-/"":8"/*.-/2</,,=6"/-*&</&*&3 5*41&32>+0&:*.51&*+1,&&-3&"?0&::&"&,-/"",-1&0/*&/65*"&.&,"/&1,&0&&1**8,-+*2 MAT Module 11 "" 27-&/* 29-&/* 8-&/* &-3&""/"0AB6"/&/&&-3&"38&.5,-/"",-1&0/*&/65*"&.&,"/&/&&-3&"2)&: +,-.&*."/*01*-&:/*+"/&1&-*+"/,",-1&&+,16 &,162</&*-&&-3&""/",&C"&,**+"/&& &-3&",*"/&D62</."/&7&",:11&-*+"/,",-1&,ABEDFAD61*-2</&*-&&-3&"0/,:/,/*,G*"1, "/&3&1&-"/"/&-&&&-3&"2</,3&"/""/& /*,G*"11&-*+"/,",-1&,AH61*-2)&:*0.&"/& =6"/-*&</&*&3"*:1:.1"&/*0+06+*3"/&*,-,1 "",-5*,"0&&2 As you can see above, the orange segments along with the red segment that was 10 yards and the dashed segment make up a right triangle whose hypotenuse is the dashed segment. We can figure out how long each of the legs of this triangle are fairly easily. The orange segment that is an extension of the red segment is another 3 yards. Thus the vertical leg of this triangle is = 13 yards long. The orange segment which is horizontal is the same length as the green segment. This means that the horizontal leg of this triangle is 12 yards long. We can now use the Pythagorean Theorem to calculate how far away from the original starting point we are = c = c = c = c Here an exact answer does not really make sense. It is enough to approximate that we end up yards from where we started. Volume and Surface Area We will continue our study of geometry by studying three-dimensional figures. We will look at the two-dimensional aspect of the outside covering of the figure and also look at the three-dimensional space that the figure encompasses. The surface area of a figure is defined as the sum of the areas of the exposed sides of an object. A good way to think about this would be as the area of the paper that it would take to cover the outside of an object without any overlap. In most of our examples, the exposed sides of our objects will polygons whose areas we learned how to find in the previous section. When we talk about the surface area of a sphere, we will need a completely new formula. The volume of an object is the amount of three-dimensional space an object takes up. It can be thought of as the number of cubes that are one unit by one unit by one unit that it takes to fill up an object. Hopefully this idea of cubes will help you remember that the units for volume are cubic units.

12 12 Geometry Surface Area of a Rectangular Solid SA = 2lw + lh + wh) l = length of the base of the solid w = width of the base of the solid h = height of the solid Volume of a Solid with a Matching Base and Top V = Ah A = area of the base of the solid h = height of the solid Volume of a Rectangular Solid specific type of solid with matching base and top) V = lwh l = length of the base of the solid w = width of the base of the solid ""&)"* h = height of the solid +,-"./01-&2.3"4&". 5.6"7829: "&)* +,-./012&3-./ /06,8359&2: Example 12. Find the volume and the surface area of the figure below =<@ ;<= ><? A2&3/,2-* B0,46,835,492"C266,7,&&D7&&.57/-83&55,4E<F5 8,1-/0&-8/ /0&-8/0G:,./0G-.0,80/26/0 92"G/034:2-&D-./2&38,-/2/06253&<H4.2-/0:D 23592",44,//,-8G:7-4D/0//0&-8/026/094,4;<=I /0:,./026/094,4><?I-./00,80/26/042&,.,4=<@< B034:7-J3,7K&D6,-./012&326/092"/29 N = &:0 = C ;<=EC><?EC=<@E = ;><LM=739,7/54< Solution: This figure is a box officially called a rectangular prism). We are given the lengths of each of the length, width, and height of the box, thus we only need to plug into the formula. Based on the way our box is sitting, we can say that the length of the base is 4.2 m; the width of the base is 3.8 m; and the height of the solid is 2.7 m. Thus we can quickly find the volume of the box to be V = lwh = 4.2)3.8)2.7) = cubic meters. Although there is a formula that we can use to find the surface area of this box, you should notice that each of the six faces outside surfaces) of the box is a rectangle. Thus, the surface area is the sum of the areas of each of these surfaces, and each of these areas is fairly straight-forward to calculate. We will use the formula in the problem. It will give us SA = 2lw + lh + wh) = ) ) )) = square meters. O&/02380/05,46253&/0/:7-34/26,-./ /0,492"GD234023&.-2/,7/0/7026/04,"674 C23/4, E26/092",457/-8&<B034G/ ,4/04326/ / G /0454,46,5&D4/5,80/P625:5./27&73&/<F:,&&34 /06253&,-/0529&<Q/:,&&8,134 AO = = C&: + &0 + :0E = =C;<= S ><? + ;<= S =<@ + ><? S =<@E A cylinder is an object with straight sides and circular ends of the same size. The volume of a cylinder can be found in the same way you find the volume of a solid with a matching 4J35/54< O7D&,-.5,4-29T7/:,/04/5,80/4,.4-.7,573&5-.426/04 4,U<B012&3267D&, ,-/04:DD236,-./0 7D&,-.5

13 MAT Module 13 base and top. The surface area of a cylinder can be easily found when you realize that you have to find the area of the circular base and top and add that to the area of the sides. If you slice the side of the cylinder in a straight line from top to bottom and open it up, you will see that it makes a rectangle. The base of the rectangle is the circumference of the circular base, and the height of the rectangle is the height of the cylinder. ""&)"* +,-"./01-&2.3"4&". 5.6"7829: "&)* Volume of a Cylinder V = Ah L2326M?,-.5 A = area of the base of the cylinder L = E0 h = height of the cylinder 4;<".&".2<"=.*"2<"3,>/0"& <;<"<">6<2<"3,>/0"& Surface Area of a Cylinder =3567E526M?,-.5 SA = 2πr 2 ) + 2πrh ) =E = )Cπ 5 B + ) π50 r = the radius of the circular base of the cylinder &;<"&.0>-*2<"3>&3-,.&=.*"2<"3,>/0"& <;<"<">6<2<"3,>/0"& h = height of the cylinder?;<"/-="&<.>*.@@&a>."0=bc8d89eb π = the number approximated by Example 13. Find the volume and surface area of the figure below +,-./ /06,83592: ;<,- ;),- =23/,2-* >0,46,835,47?,-.5@>0.,/526,/47,573594,4;),-704@>0,4-4/0//05.,3426/07,573594,4 ; ; 5 =. = C;)B = A,-704@>00,80/26/07?,-.5,4;<,-704@ ) ) >2773// D:4,&?-./2&38,-/2 /062534@ =3567E5* ) ) =E = )Cπ 5 B + ) π50 = )Cπ A B + ) πcabc;<b = G) π + ;)< π = ;F) π 4H35 Solution: This figure is a cylinder. The diameter of its circular base is 12 inches. This means that the radius of the circular base is r = 1d = 1 12) = 6 inches. The height of the cylinderi s 10 inches. To calculate the volume and 2 surface 2 area, we simply need to plug into the formulas. Surface Area: SA = 2πr 2 ) + 2πrh = 2π 6 2 ) + 2π6)10) = 72π + 120π = 192π square units. This is an exact answer. An approximate answer is square units. Volume: In order to plug into the formula, we need to recall how to find the area of a circle the base 3-,/4@>0,4,4-7/-4:5@E-&&52,/-4:5,4A<I@;JKGF of the cylinder is 4H353-,/4@ a circle). We will the replace A in the formula with the formula for the area of a circle. V = Ah = πr 2 h = π6 2 )10) = 360π cubic inches. An approximation of this exact answer would be cubic inches. Our next set of formulas is going to be for spheres. A sphere is most easily thought of as a ball. The official definition of a sphere is a three-dimensional surface, all points of which are equidistant from a fixed point called the center of the sphere. A circle that runs along the surface of a sphere to that it cuts the sphere into two equal halves is called a great circle of

14 14 Geometry that sphere. A great circle of a sphere would have a diameter that is equal to the diameter of the sphere. Surface Area of a Sphere SA = 4πr 2 r = the radius of the circular base of the cylinder π = the number approximated by Volume of a Sphere V = 4 3 πr3 r = the radius of the circular; base of the cylinder ""&)"* π = the number approximated by ,-"./01-&2.3"4&". 5.6"7829: Example 14. "&)* Find the volume and surface area of the figure below +,-./012& /06,8359&2: < ),- ; =2&3/,2-* >0,4,4405?@58,1-/0//0.,/526/0405,4 < ),-704?@-./27&73&//05.,3426/0405/2 ; 7&73&//012& ?>05., ,4 0&626,/4.,/5?>0,4-4/0//05.,34,4 A A & < A & BD BD =2&3/,2-* 5 =. = ) = = = A? ;AB<,-704?@7--2:E34/ B B ; " B ; " AC >0,4,4405?@58,1-/0//0.,/526/0405,4 &38/0,4-395,-/2/06253&4/27&73&//012&3-. < ),-704?@-./27&73&//05.,3426/0405/ ? ; 7&73&//012& ?>05., ,4 F2&3* G ) G F = π5 = π A? ;AB< ) ) BG? DGA<H< 739,7,-704? 16 ) ) 0&626,/4.,/5?>0,4-4/0//05.,34,4 B B =3567I5* =I = Gπ5 = Gπ A?;AB< ) GA? B;BAD 4J35,-704? A A & < A & BD BD 5 =. = ) = = = A? ;AB<,-704?@7--2:E34/ B B ; " B ; " AC &38/0,4-395,-/2/06253&4/27&73&//012&3-. "&G* ? +,-./012&326/06,835 G F2&3* G ) F = π5 = π A? ;AB< ) ) BG? DGA<H< 739,7,-704? ) ) ; F2&326 B B =2&,.:,/0 =3567I5* =I = Gπ5 = Gπ A?;AB< ) GA? B;BAD 4J35,-704? < K/70,-8L4 -.>2 "&G* A< F = I0 +,-./012&326/06,835 Solution:This is a sphere. We are given that the diameter of the sphere is 3 5 inches. We 8 need to calculate the radius of the sphere to calculate the volume and surface area. The radius of a sphere is half of its diameter. This means that the radius is r = 1d = 13 5) = ) = 29 = inches. We can now just plug this number in to the formulas to calculate 2 8 the volume and surface area. Volume: V = 4 3 πr3 = 4 3 π ) cubic inches. Surface Area: SA = 4πr 2 = 4π ) square inches. Example 15. Find the volume of the figure F2&326 4;.&".2<" =2&,.:,/0 =.*"2<"*,>0 <;<">6<2<" K/70,-8L4 *,>0 -.>2 F = I0 ""&)"* +,-"./01-&2.3"4&". 5.6"7829: "&)* +,-./012& /06,8359&2: ; < A) < ),- Solution:This figure is a solid with the same shape base and top. The shape of the base and top is a trapezoid. Thus we will need to remember the formula for the area of a trapezoid. A< 4;.&".2<" =.*"2<"*,>0 <;<">6<2<" *,>0 A)

15 MAT Module 15 For this trapezoid, the lengths of the bases are 13 and 8 units. It does not matter which of these we say is b 1 and which is b 2. The height of the trapezoid is 5 units. The height of the solid is 15 units. We will start by plugging the information about the trapezoidal base into the formula for the area of a trapezoid. Once we have this area, we will plug that and the height of the solid into the volume formula. Area of the trapezoidal base: A = 1b b 2 )h = )5 = 52.5 square units. 2 Volume of trapezoidal solid: V = Ah = 52.5)15) = cubic units. Our next formulas will be for finding the volume of a cone or a pyramid. These two formulas are grouped together since they are very similar. Each is basically 1 times the area 3 of the base of the solid times the height of the solid. In the case of the cone, the base is a circle. In the case of the pyramid, we will have a base that is a rectangle. The height in both cases is the perpendicular distance from the apex to the plane which contains the base. A pyramid is a solid figure with a polygonal base in our case a rectangle) and triangular faces that meet at a common point the apex). A cone is the surface of a conic solid whose base is a circle. This is more easily thought of as a pointed ice-cream cone whose top is circular and level. Volume of a Rectangular Pyramid V = 1 3 lwh l = the length of the base of the pyramid w = the width of the base of the pyramid ""&)"* h = the perpendicular height of the pyramid +,-"./01-&2.3"4&". 5.6"78289 Volume of a Cone r = the radius of thei circular base of the cone π = &:;"&.0<-*2;"3<&3-,.&=.*"2;"3/" the number that is apporximated by Example 16. Find the volume of the figure. h = the perpendicular height of the cone "&)* +,-./012324/04,5367 H2324N2- V = 1 3 πr2 h C H = π D 6 0 >:;"/-="&;.<*.??&@<."0=ABCDC8EA ;:;"?"&?"/0<3-,.&;"<6;2;"3/" CD9 CE9 823/,2-* 8,-9/04,5360:9,6936;:-.22<:,<-,996 9-/,/6:7B02/06:36-/24CE9-/,/6:,:/0.,/624/09,6936;:7F/03:3:/993//06.,3:

16 16 Geometry ""&)"* +,-"./01-&2.3"4&". 5.6"7829: Solution: Since the figure has a circular base and looks like an ice cream cone, this must be a cone. In order to find the volume of a cone, we need the radius of the circular base and the height perpendicular "&)* height) of the cone. The height is given as 12 centimeters. The other +,-./012&324/04,5367 measurement of 10 centimeters is the diameter of the circular base. We thus must calculate the radius to get r = 1d = 1 10) = 5 centimeters. We are now ready to plug into the volume 2 2 of a cone formula. V = 1 3 πr2 h = 1 3 π52 ""&)"* )12) = 100π cubic centimeters is the exact volume. An approximation +,-"./01-&2.3"4&". of this volume would be cubic centimeters. 5.6"7829: "&)* +,-./012&324/04,5367 E,- C,- 82&3/,2-* 90:;24/0,;4,536,;6</-5&-./0;,.;24/04,536 AB,- 6/6,-5&;=/03;/0,;4,536,;6</-53&6>6,.7900,50/?6-.,<3&60,50/@,;AB,-<0;790&-5/024/0:;,;C,-<0;=-./0D,./024/0:;,;E,-<0;78,-<D01&&24 E,- /06/;426/012&34263&=D<-F3;/&35,-/2/012&3 246</-53&6>6,.4263&/25/ C,- 82&3/,2-* Example 17. Find the volume of the figure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olution: The base of this figure is a rectangle and the sides of the figure are triangles, thus this figure is a rectangular pyramid. The height perpendicular height) is 10 inches. The length of the base is 7 inches, and the width of the base is 5 inches. Since we have all of the parts for the volume formula, we can just plug into the volume of a rctangular pyramid formula to get V = 1lwh = 1 cubic inches. An approximation of this volume would be cubic inches. Example 18. Find the volume of the figure. Solution: The first thing that we need to do is figure out what type of figure this is. If we rotate the solid 90 degrees to the right, we get a figure that looks like this 3,=/0"& ;:;";"=6;2 ;"3,=/0"& "&5//95&9,:15&*.-&)")";&..5;&&*+3<+//.-,1)/ )"&*9,=/0))"&&+0&,/55/2,2)/-53>,)1&*+)/ &*./0))"?/50;/)"2,2)/-53"=)-,1)/./)"&& )/&+0/0)-",))"?/50;/)"5,+:15&*.&-&)"/0))" ;&&*+2,)3>:,*)"*&*.)"?/50;/)";,55:15&*. /;&&*+2,)3@&*,5518--&550=),:))"?/50;/)";,55 :15&*./;)"?/50;/)"5,+:15&*.)/+))"?/50; //0:0*)/5&.3 >-&55),)-&)")"5,+:15&*.3",.&0/)":&:05, This looks like a cylinder with the middle missing. A good way to think about this figure is a roll of paper towels. We are trying to find out the volume of the paper towels. The best way to do this is to figure out what the volume of the larger cylinder is without the missing

17 MAT Module 17 part. We can then find the volume of the smaller cylinder or missing part. Finally, we will subtract the volume of the smaller cylinder from the volume of the larger cylinder to get the volume of our current solid. We will start with the larger cylinder. The radius of the circular base is given as 5 units. The height of the larger cylinder is 16 units. We can then calculate the volume of the larger cylinder to be V = Ah = πr 2 h = π5 2 )16) = 400π cubic units. Next we will calculate the volume of the smaller cylinder. The radius of the circular base of the smaller cylinder is 2 units. The height of the smaller cylinder is 16 units. We can calculate the volume of the smaller cylinder would be V = Ah = πr 2 h = π2 2 )16) = 64π ""&)"* cubic units. +,-"./01-&2.3"4&". ""&)"* 5.6"7829: +,-"./01-&2.3"4&". 5.6"7829: We now subtract the volume of the smaller cylinder from the volume of the larger cylinder "&)* to get the"&)* volume of our solid. Volume of larger cylinder volume of smaller cylinder +,-.)/ ,95-,9:;12.34<=>,3-,.,<, = 400π 64π = 336π cubic units. This is approximately cubic units. +,-.)/ ,95-,9:;12.34<=>,3-,.,<, 3>?->?.9?4,<>&?2.@8&<,8-&99>?-8-,.-.5-"/+2.9 3>?->?.9?4,<>&?2.@8&<,8-&99>?-8-,.-.5-"/+2.9 Example 19.?4A-&>.9<>,83,-8?45-"/ From an 8.5-inch by 11-inch piece of cardboard, 2-inch square corners are cut?4a-&>.9<>,83,-8?45-"/ out and the resulting flaps are folded up to form an open box. Find the volume and surface area of the box. B-&>?2-.* B-&>?2-.* Solution: For this+-,?42<,-5&:2?c2&&5,&&64&8>&?-d?45-" problem, it will be really helpful to make the box described above. You +-,?42<,-5&:2?C2&&5,&&64&8>&?-D?45-" start with a standard 9<3,2595-A/E-><?,?C2?4<?.9,923-8,/E-> piece of paper. You the cut out the dashed square indicated below 9<3,2595-A/E-><?,?C2?4<?.9,923-8,/E-> from each corner. You?43>?->??49<49<=>,2.923?95&-C8,-343-,.,/ make sure each side of the square is 2 inches in length.?43>?->??49<49<=>,2.923?95&-c8,-343-,.,/ E->D<>,34<29-8?4<=>,2<;2.34<2.&.@?4/ E->D<>,34<29-8?4<=>,2<;2.34<2.&.@?4/ What you are left with is the shape below. F4?6->,&8?C2?42<?4<45&-C/ F4?6->,&8?C2?42<?4<45&-C/ E->.-C8-&9>&-.@?49<49&2.<?-3,?5-"/G45-" E->.-C8-&9>&-.@?49<49&2.<?-3,?5-"/G45-"?4?C4A3,?92<,3?.@>&,<-&29/G42<5-"4<.-?-/?4?C4A3,?92<,3?.@>&,<-&29/G42<5-"4<.-?-/ H-?4A2.@?-C2&&.-?883??4A-&>-8?45-"/ H-?4A2.@?-C2&&.-?883??4A-&>-8?45-"/ F-.&6.9?-9?,2.?4&.@?4-8?45<-8?45-":?4 F-.&6.9?-9?,2.?4&.@?4-8?45<-8?45-":?4 C29?4-8?45<-8?45-":.9?442@4?-8?45-"/G4,9 C29?4-8?45<-8?45-":.9?442@4?-8?45-"/G4,9 9<49&2.<,,<.?<?4&.@?4-8?45<-8?45-"/G4 9<49&2.<,,<.?<?4&.@?4-8?45<-8?45-"/G4 -,2@2.&&.@?4-8?4,C<772.34</F,-A9;2.34< -,2@2.&&.@?4-8?4,C<772.34</F,-A9;2.34< 8,-?4?--8?4@.9C&<-,-A9;2.34<8,-?4 8,-?4?--8?4@.9C&<-,-A9;2.34<8,-?4 You now fold up along the dashed lines to create a box. The box that we have created is a rectangular solid. This box has no top. Not having a top will not affect the volume of the box. We only need to determine the length of the base of the box, the width of the base of the box, and the height of the box. The red dashed lines represent the length of the base of the box. The original length of the paper was 11 inches. We removed 2 inches from the top of the page and we also removed 2 inches from the bottom of the page. Thus each red dashed line is = 7 inches.

18 18 Geometry ""&)"* +,-"./01-&2.3"4&". 5.6"7829: The green dashed lines represent the width of the base of the box. The original width of the paper was 8.5 inches. We removed 2 inches from the left side of the page and also removed 2 inches from "&)* the right side of the page. Thus each green dashed line is = 4.5 +,-.)/ ,95-,9:;12.34<=>,3-,.,<, inches. 3>?->?.9?4,<>&?2.@8&<,8-&99>?-8-,.-.5-"/+2.9 We now need?4a-&>.9<>,83,-8?45-"/ to think about the height of the box. Since we have folded up the sides for form the height of the box, we just need to determine how tall those sides are. Since they were made by cutting out 2-inch square from each corner, these sides must be 2 inches high. B-&>?2-.* +-,?42<,-5&:2?C2&&5,&&64&8>&?-D?45-" Now we are ready of calculate the volumes V = lwh = 7)4.5)2) = 63 cubic inches. Now we need to calculate 9<3,2595-A/E-><?,?C2?4<?.9,923-8,/E-> the surface area of our box. Since there is no top to this box, we can start formula for?43>?->??49<49<=>,2.923?95&-c8,-343-,.,/ the surface area of a box. We will then need to subtract off the area of E->D<>,34<29-8?4<=>,2<;2.34<2.&.@?4/ the top of the box. This will give us SA = 2lw+wh+lh) = )+4.5 2)+7)) = 109 square inches for the box with the top included. The top would have the same area as the base of the box. This would be A = lw = 7)4.5) = 31.5 square inches. Thus the surface area of our figure is total surface area area of the top = = 77.5 square inches. There is another way to calculate the surface area of this box. The surface area is the amount of paper it would take to cover the box without overlap. You should notice that this is the same as the amount of paper we used to make the box. Thus, it is enough to calculate the area of the paper as shown here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xample 20. A propane gas tank consists of a cylinder with a hemisphere at each end. Find the volume of the tank if the overall length is 20 feet and the diameter of the cylinder is 6 feet. Solution: We are told this tank consists of a cylinder one its side) with a hemisphere at each end. A hemisphere is half of a sphere. To find the volume, we need to find the volume of the cylinder and the volumes of each hemisphere and then adding them together. Lets start with the hemisphere sections. The diameter of the circular base of the cylinder is indicated to be 6 feet. This would also be the diameter of the hemispheres at each end of the cylinder. We need the radius of the sphere to find its volume. Once we calculate the volume of the whole sphere, we multiply it by 1 to find the volume of the hemisphere half 2 of a sphere). We calculate the radius to be r = 1d = 1 6) = 3 feet. Now we can calculate 2 2 the volume of the two hemispheres at the ends of the tank. J.3833=119<-= &9,F8=14&641 I6K <-= &9,C95-50&9,D:B P-=-5 A&9, 33=119,74=01-I, = L > 7 = L C@D = M51:N-883. > 33=119<-= &9, : G51O40&9,*

19 MAT Module 19 Left Hemisphere: The volume of a whole sphere is V = 4 3 πr3 = 4 3 π33 ) = 36π cubic feet. We now multiply this volume by 1 to find the volume of the hemisphere to get 1V = 1 36π) = 18π cubic feet Right Hemisphere: The volume of a whole sphere is V = 4 3 πr3 = 4 3 π33 ) = 36π cubic feet. We now multiply this volume by 1 to find the volume of the hemisphere to get 1V = 1 36π) = 18π cubic feet You might notice that we went through exactly the same process with exactly the same number for each hemisphere. We could have shortened this process by realizing that putting together the two hemispheres on each end, which were of the same diameter, would create a whole sphere. We could just have calculated the volume of this whole sphere. Now we need to calculate the volume of the cylinder. We need the radius of the cylinder and the height of the cylinder to find its volume. The radius of the cylinder is the same as the radius of the hemispheres at each end. Thus the radius of the cylinder is 3 feet. It may look like the height of the cylinder is 20 feet. It turns out that this is not the case. The 20 feet includes the hemispheres at each end. We need to subtract the part of the 20 feet that represents the hemispheres. radius of hemisphere = 3 ft 6 ft radius of hemisphere = 3 ft 20 ft height of cylinder Looking at the figure above, we see that the distance from the end of the left hemisphere to the left end of the cylinder is the radius of the hemisphere. The radius of the hemisphere is 3 feet. Similarly, the distance from the right end of the cylinder to the end of the right cylinder is also the radius of the hemisphere. The radius of the hemisphere is 3 feet. If we now subtract these two distances from the overall length of the tanks, we will have the height of the cylinder to be h = = 14 feet. We can now calculate the volume of the cylinder. Cylinder: V = Ah = πr 2 h = π3 2 )14) = 126π cubic feet. We can now find the volume of the tank by adding together the volumes of the cylinder, the right hemisphere, and the left hemisphere. We get the volume to be V = 126π +18π +18π = 162π cubic feet. This is approximated by cubic feet. Example 21. A regulation baseball hardball) has a great circle circumference of 9 inches; a regulation softball has a great circle circumference of 12 inches. a. Find the volumes of the two types of balls. b. Find the surface areas of the two types of balls.