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MAT 142 College Mathematics
Module #GM
Geometry
Terri L. Miller & Elizabeth E. K. Jones
revised April 20, 2011
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Perimeter and Area
!
"#!$%#!&'()&!*'!+*$%*!',%!+*,-.!'/!&#'0#*%.!1(*2!*1'3-(0#)+(')$4!/(&,%#+5!!
8#%(0#*#%!
We are going
to start our study of geometry with two-dimensional figures. We will look at
"#!1(44!4''6!$*!*2#!')#3-(0#)+(')$4!-(+*$)7#!$%',)-!*2#!/(&,%#!$)-!*2#!*1'3
the one-dimensional
distance around the figure and the two-dimensional space covered by
-(0#)+(')$4!+8$7#!7'9#%#-!:.!*2#!/(&,%#5!
the figure. !
;2#!8#%(0#*#%!'/!$!+2$8#!(+!-#/()#-!$+!*2#!-(+*$)7#!$%',)-!*2#!+2$8#5!!<()7#!
The perimeter of a shape is defined as the distance around the shape. Since we usually
1#!,+,$44.!-(+7,++!*2#!8#%(0#*#%!'/!8'4.&')+!=74'+#-!84$)#!/(&,%#+!12'+#!
discuss the +(-#+!$%#!+*%$(&2*!4()#!+#&0#)*>?!1#!$%#!$:4#!*'!7$47,4$*#!8#%(0#*#%!:.!@,+*!
perimeter of polygons (closed plane figures whose sides are straight line segment),
we are able$--()&!,8!*2#!4#)&*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
the perimeter of a circle, we call it by the
special name
of 7(%7,0/#%#)7#!
circumference
8#%(0#*#%!'/!$!7(%74#?!1#!7$44!(*!:.!*2#!+8#7($4!)$0#!'/!
7(%7,0/#%#)7#
5!!<()7#!
Since we dont
have straight sides to add up for the circumference (perimeter) of a circle, we
1#!-')A*!2$9#!+*%$(&2*!+(-#+!*'!$--!,8!/'%!*2#!7(%7,0/#%#)7#!=8#%(0#*#%>!'/!
$!7(%74#?!1#!2$9#!$!/'%0,4$!/'%!7$47,4$*()&!*2(+5!
have a formula
for calculating this.
!
!
Circumference
(Perimeter) of a Circle
J(%7,0/#%#)7#!=K#%(0#*#%>!'/!$!J(%74#!
!
!
JC= L=
π %2πr
!
r = radius
!
&(9(&-/,:*(#3(%;"(<,&<="(
>(
9(%;"(.:$?"&(%;-%(,*(-@@&#A,$-%"/(?'(BC7874DB(
!
π =the number that is approximated by 3.141593
!
BC$084#!DE!
Example 1.
Find the perimeter of the figure below.
F()-!*2#!8#%(0#*#%!'/!*2#!/(&,%#!:#4'1!
78(
8(
77(
6(
!
!
Solution:: It <'4,*(')E!
is tempting to just start adding of the numbers given together, but that will
G*!(+!*#08*()&!*'!@,+*!+*$%*!$--()&!'/!*2#!),0:#%+!&(9#)!*'&#*2#%?!
not give us the perimeter.
The reason that it will not is that this figure has SIX sides and
:,*!*2$*!1(44!)'*!&(9#!,+!*2#!8#%(0#*#%5!!;2#!%#$+')!*2$*!(*!1(44!)'*!
we are only given four
numbers. We must first determine the lengths of the two sides that
(+! *2$*!
2$+!perimeter.
<GH! +(-#+! $)-!
$%#! at
')4.!
&(9#)!
/',%!again to find the
are not labeled before
we *2(+!
can /(&,%#!
find the
Lets1#!look
the
figure
),0:#%+5! ! "#! 0,+*! /(%+*! -#*#%0()#! *2#! 4#)&*2+! '/! *2#! *1'! +(-#+!
lengths of the other sides.
*2$*!$%#!)'*!4$:#4#-!:#/'%#!1#!7$)!/()-!*2#!8#%(0#*#%5!!I#*A+!4''6!
Since our figure has$*!*2#!/(&,%#!$&$()!*'!/()-!*2#!4#)&*2+!'/!*2#!'*2#%!+(-#+5!
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
"#$%&!'()!*#+()&!,-.!-//!)#+,0!-$+/&.1!2&!-)&!-3/&!0'!4&0&)5#$&!0,&!
.0-)0!
2#0,! 0,&! ;&)0#%-/! .#4&.8! ! 9''<#$+! -0! 0,&! #5-+&! 3&/'21! 2&! %-$!
/&$+0,!
'*!
0,&!
.#4&.!#$4#%-0&4!
2,'.&! /&$+0,!
#.! $'0!
7)#$0&48!
! 9&0:.!
.&&!
0,-0!
0,&!
/&$+0,!
36! 0,&!
)&4!%())&$0/6!
3)-%<&0! #.!
0,&! .-5&!
-.!
.0-)0!
2#0,!
0,&!
;&)0#%-/!
.#4&.8!
!
9''<#$+!
-0!
0,&!
#5-+&!
3&/'21!
2&!
%-$!
0,&!/&$+0,!'*!0,&!;&)0#%-/!.#4&!2,'.&!/&$+0,!#.!=!($#0.8!!>,#.!5&-$.!
.&&! 0,-0!
0,&! /&$+0,!
#$4#%-0&4!
36! 0,&!'*!
)&4!
3)-%<&0!
0,&! .-5&!36!
-.!
0,-0!
2&! %-$!
%-/%(/-0&!
0,&!Geometry
/&$+0,!
0,&!
+)&&$!#.!.&+5&$0!
2
0,&!/&$+0,!'*!0,&!;&)0#%-/!.#4&!2,'.&!/&$+0,!#.!=!($#0.8!!>,#.!5&-$.!
.(30)-%0#$+! =! *)'5! ??8! ! >,#.! 5&-$.! 0,-0! 0,&! +)&&$! .&+5&$0! #.! @!
0,-0! length
2&! %-$!is%-/%(/-0&!
'*!that
0,&! we
+)&&$!
36! the length of
the vertical side whose
4 units. 0,&!
This/&$+0,!
means
can.&+5&$0!
calculate
($#0.8!
.(30)-%0#$+! =! *)'5! ??8! ! >,#.! 5&-$.! 0,-0! 0,&! +)&&$! .&+5&$0! #.! @!
green segment by subtracting 4 from 11. This 78(
means that the green segment is 7 units.
($#0.8!
8( 78(
8(
77(
8(
8(
77(
77(9(8(:(5(
77(9(8(:(5(
In
us
6(
!
!
6(
!
A$!-!.#5#/-)!5-$$&)1!2&!%-$!%-/%(/-0&!0,&!/&$+0,!'*!0,&!'0,&)!
!
5#..#$+!.#4&!(.#$+! ?= − C = B 8!!>,#.!+#;&.!(.!0,&!/&$+0,.!'*!-//!0,&!
a similar manner,
we can calculate the length of the other missing side
A$!-!.#5#/-)!5-$$&)1!2&!%-$!%-/%(/-0&!0,&!/&$+0,!'*!0,&!'0,&)!
.#4&.!-.!.,'2$!#$!0,&!*#+()&!3&/'28!
the lengths of all
the sides as shown in the figure below.
! 5#..#$+!.#4&!(.#$+! ?= − C = B 8!!>,#.!+#;&.!(.!0,&!/&$+0,.!'*!-//!0,&!
.#4&.!-.!.,'2$!#$!0,&!*#+()&!3&/'28!
78(
!
8(
8(
78(
;(
5(
;(
!"#$"%&'()#%"*(
+"&,$"%"&(-./(0&"-(
!
the
5(
using . This gives
77(
77(
+-1"(2(#3(45(
6(
D'2!0,-0!2&!,-;&!-//!0,&!/&$+0,.!'*!0,&!.#4&.1!2&!%-$!.#57/6!
Now that we"#!$%#!&'()&!*'!+*$%*!',%!+*,-.!'/!&#'0#*%.!1(*2!*1'3-(0#)+(')$4!/(&,%#+5!!
have all
the lengths of the sides, we can6(simply calculate the perimeter
by adding
%-/%(/-0&!0,&!7&)#5&0&)!36!-44#$+!0,&!/&$+0,.!0'+&0,&)!0'!+&0!
8#%(0#*#%!
D'2!0,-0!2&!,-;&!-//!0,&!/&$+0,.!'*!0,&!.#4&.1!2&!%-$!.#57/6!
the lengths "#!1(44!4''6!$*!*2#!')#3-(0#)+(')$4!-(+*$)7#!$%',)-!*2#!/(&,%#!$)-!*2#!*1'3
together
to
get
Since
perimeters
are
just
the
lengths
of
lines,
the
perimeter
would
= + ?= + ?? + C + @ + B = EF8 !!"#$%&!7&)#5&0&).!-)&!G(.0!0,&!/&$+0,.!
%-/%(/-0&!0,&!7&)#5&0&)!36!-44#$+!0,&!/&$+0,.!0'+&0,&)!0'!+&0!
be 50 units.-(0#)+(')$4!+8$7#!7'9#%#-!:.!*2#!/(&,%#5!
'*!/#$&.1!0,&!7&)#5&0&)!2'(/4!3&!EF!($#0.8!
! = + ?= + ?? + C + @ + B = EF8 !!"#$%&!7&)#5&0&).!-)&!G(.0!0,&!/&$+0,.!
!
! '*!/#$&.1!0,&!7&)#5&0&)!2'(/4!3&!EF!($#0.8!
8#%(0#*#%
The area of;2#!
a shape
is !'/!$!+2$8#!(+!-#/()#-!$+!*2#!-(+*$)7#!$%',)-!*2#!+2$8#5!!<()7#!
defined as the number of square units that cover a closed figure. For
!
-)&-!
most of the1#!,+,$44.!-(+7,++!*2#!8#%(0#*#%!'/!8'4.&')+!=74'+#-!84$)#!/(&,%#+!12'+#!
shape that we will be dealing with there is a formula for calculating
the area.
!
+(-#+!$%#!+*%$(&2*!4()#!+#&0#)*>?!1#!$%#!$:4#!*'!7$47,4$*#!8#%(0#*#%!:.!@,+*!
In some cases, our shapes will be made up of more than a single shape. In calculating
the
-)&-!
area of such$--()&!,8!*2#!4#)&*2+!'/!#$72!'/!*2#!+(-#+5!!"2#)!1#!*$46!$:',*!*2#!
shapes, we can just add the area of each of the single shapes together.
8#%(0#*#%!'/!$!7(%74#?!1#!7$44!(*!:.!*2#!+8#7($4!)$0#!'/!7(%7,0/#%#)7#5!!<()7#! 7(%7,0/#%#)7#!
We will start
with the formula for the area of a rectangle. Recall that a rectangle is a
1#!-')A*!2$9#!+*%$(&2*!+(-#+!*'!$--!,8!/'%!*2#!7(%7,0/#%#)7#!=8#%(0#*#%>!'/!
$!7(%74#?!1#!2$9#!$!/'%0,4$!/'%!7$47,4$*()&!*2(+5!
quadrilateral
with opposite sides parallel and right interior angles.
!
!
Area of a Rectangle
J(%7,0/#%#)7#!=K#%(0#*#%>!'/!$!J(%74#!
!
!
JA
= L=
π %bh
!
b = base of the rectangle
!
&(9(&-/,:*(#3(%;"(<,&<="(
>(9(%;"(.:$?"&(%;-%(,*(-@@&#A,$-%"/(?'(BC7874DB(
!
h = the height of the rectangle
!
BC$084#!DE!
Example 2.
Find the area of the figure below.
F()-!*2#!8#%(0#*#%!'/!*2#!/(&,%#!:#4'1!
78(
8(
77(
6(
!
!
<'4,*(')E!
G*!(+!*#08*()&!*'!@,+*!+*$%*!$--()&!'/!*2#!),0:#%+!&(9#)!*'&#*2#%?!
:,*!*2$*!1(44!)'*!&(9#!,+!*2#!8#%(0#*#%5!!;2#!%#$+')!*2$*!(*!1(44!)'*!
(+! *2$*! *2(+! /(&,%#! 2$+! <GH! +(-#+! $)-! 1#! $%#! ')4.! &(9#)! /',%!
),0:#%+5! ! "#! 0,+*! /(%+*! -#*#%0()#! *2#! 4#)&*2+! '/! *2#! *1'! +(-#+!
*2$*!$%#!)'*!4$:#4#-!:#/'%#!1#!7$)!/()-!*2#!8#%(0#*#%5!!I#*A+!4''6!
MAT 142 - Module
3
Solution: This figure is not a single rectangle. It can, however, be broken up into two
rectangles. We then will need to find the area of each of the rectangles and add them
together to calculate the area of the!"#$"%&'()#%"*(
whole figure.
+"&,$"%"&(-./(0&"-(
+-1"(2(#3(24(
!"#$"%&'()#%"*(
!"#$"%&'()#%"*(
There is more
than
one way to break
this figure into rectangles.+-1"(2(#3(24(
We will
only illustrate one
!"#$"%&'()#%"*(
+"&,$"%"&(-./(0&"-(
+"&,$"%"&(-./(0&"-(
+-1"(2(#3(24(
below.
67(
+"&,$"%"&(-./(0&"-(
67(
67(
67(
7(
7(
7(
+-1"(2(#3(24(
7(
66(
66(
66(
66(
5(
5(
5(
67(
7(
7(
7(
7(
67(
67(
67(
66(
66(
66(
5(
7(
7(
7(
66(
7(
5(
!
67(
!
!
!
67(
67(
67(
66(
66(
66(
66(
5(
5(
5(
5(
5(
"#!$%&#!'$()*!%+(&#!,$%,!)#!-%*!+.#%/!,$#!'$%0#!10!2*,(!%!.#3!
5(
5(
.#-,%*45#!67241.#!(*!5#7,8!%*3!%!4.##*!.#-,%*45#!67241.#!(*!.24$,89!!"#!
"#!$%&#!'$()*!%+(&#!,$%,!)#!-%*!+.#%/!,$#!'$%0#!10!2*,(!%!.#3!
"#!$%&#!'$()*!%+(&#!,$%,!)#!-%*!+.#%/!,$#!'$%0#!10!2*,(!%!.#3!
$%&#!,$#!5#*4,$'!(7!+(,$!'23#'!(7!,$#!.#3!.#-,%*45#9!!:,!3(#'!*(,!
We have shown .#-,%*45#!67241.#!(*!5#7,8!%*3!%!4.##*!.#-,%*45#!67241.#!(*!.24$,89!!"#!
above
that we can break the shape up into a red rectangle (figure on left) and
"#!$%&#!'$()*!%+(&#!,$%,!)#!-%*!+.#%/!,$#!'$%0#!10!2*,(!%!.#3!
.#-,%*45#!67241.#!(*!5#7,8!%*3!%!4.##*!.#-,%*45#!67241.#!(*!.24$,89!!"#!
;%,,#.!)$2-$!(*#!)#!-%55!,$#!+%'#!%*3!)$2-$!)#!-%55!,$#!$#24$,9!!<$#!
.#-,%*45#!67241.#!(*!5#7,8!%*3!%!4.##*!.#-,%*45#!67241.#!(*!.24$,89!!"#!
a green rectangle
(figure
on right). We have the lengths of both sides of the red rectangle.
$%&#!,$#!5#*4,$'!(7!+(,$!'23#'!(7!,$#!.#3!.#-,%*45#9!!:,!3(#'!*(,!
$%&#!,$#!5#*4,$'!(7!+(,$!'23#'!(7!,$#!.#3!.#-,%*45#9!!:,!3(#'!*(,!
%.#%!(7!,$#!.#3!.#-,%*45#!2'!
= +$ and
= @ ×which
?@ = => we
!
$%&#!,$#!5#*4,$'!(7!+(,$!'23#'!(7!,$#!.#3!.#-,%*45#9!!:,!3(#'!*(,!
It does not matter
which
one we call the Abase
call the height. The area of the
;%,,#.!)$2-$!(*#!)#!-%55!,$#!+%'#!%*3!)$2-$!)#!-%55!,$#!$#24$,9!!<$#!
;%,,#.!)$2-$!(*#!)#!-%55!,$#!+%'#!%*3!)$2-$!)#!-%55!,$#!$#24$,9!!<$#!
!
;%,,#.!)$2-$!(*#!)#!-%55!,$#!+%'#!%*3!)$2-$!)#!-%55!,$#!$#24$,9!!<$#!
%.#%!(7!,$#!.#3!.#-,%*45#!2'!
× ?@= =@=>
× ?@
! = => !
red rectangle is%.#%!(7!,$#!.#3!.#-,%*45#!2'!
A = bh
= 4 × 14 = 56. A = +$ A= =@ +$
"#!$%&#!,(!3(!%!52,,5#!;(.#!)(./!,(!72*3!,$#!%.#%!(7!,$#!4.##*!
A = +$ = @ × ?@ = => !
! %.#%!(7!,$#!.#3!.#-,%*45#!2'!
!
.#-,%*45#9!!"#!/*()!,$%,!,$#!5#*4,$!(7!(*#!(7!,$#!'23#'!2'!B!1*2,'9!!"#!
!little
We have to do a"#!$%&#!,(!3(!%!52,,5#!;(.#!)(./!,(!72*3!,$#!%.#%!(7!,$#!4.##*!
more work to find the area of the green rectangle. We know that the
"#!$%&#!,(!3(!%!52,,5#!;(.#!)(./!,(!72*3!,$#!%.#%!(7!,$#!4.##*!
$%3!,(!72*3!,$#!5#*4,$!(7!,$#!(,$#.!'23#!(7!,$#!4.##*!.#-,%*45#!)$#*!
"#!$%&#!,(!3(!%!52,,5#!;(.#!)(./!,(!72*3!,$#!%.#%!(7!,$#!4.##*!
length of one of.#-,%*45#9!!"#!/*()!,$%,!,$#!5#*4,$!(7!(*#!(7!,$#!'23#'!2'!B!1*2,'9!!"#!
the
sides
is 8 units. We had to find the length of the other side of the green
.#-,%*45#9!!"#!/*()!,$%,!,$#!5#*4,$!(7!(*#!(7!,$#!'23#'!2'!B!1*2,'9!!"#!
)#!-%5-15%,#3!,$#!0#.2;#,#.!2*!CD%;05#!?!%+(&#9!!:,'!5#*4,$!)%'!E!
.#-,%*45#9!!"#!/*()!,$%,!,$#!5#*4,$!(7!(*#!(7!,$#!'23#'!2'!B!1*2,'9!!"#!
rectangle when $%3!,(!72*3!,$#!5#*4,$!(7!,$#!(,$#.!'23#!(7!,$#!4.##*!.#-,%*45#!)$#*!
we calculated
the perimeter in Example 1 above. Its length was 7 units.
$%3!,(!72*3!,$#!5#*4,$!(7!,$#!(,$#.!'23#!(7!,$#!4.##*!.#-,%*45#!)$#*!
1*2,'9!
$%3!,(!72*3!,$#!5#*4,$!(7!,$#!(,$#.!'23#!(7!,$#!4.##*!.#-,%*45#!)$#*!
)#!-%5-15%,#3!,$#!0#.2;#,#.!2*!CD%;05#!?!%+(&#9!!:,'!5#*4,$!)%'!E!
)#!-%5-15%,#3!,$#!0#.2;#,#.!2*!CD%;05#!?!%+(&#9!!:,'!5#*4,$!)%'!E!
)#!-%5-15%,#3!,$#!0#.2;#,#.!2*!CD%;05#!?!%+(&#9!!:,'!5#*4,$!)%'!E!
67(
1*2,'9!
1*2,'9!
1*2,'9!
67(
67(
7(
7(
67(
66(
7(
7(
7(
7(
7(
7(
66(
66(
66(8(7(9(4(
66(
66(8(7(9(4(
66(8(7(9(4(
5(
66(8(7(9(4(
5( A = +$
5( = B × E = => 9!!<$1'!,$#!
<$1'!,$#!%.#%!(7!,$#!4.##*!.#-,%*45#!2'!
Thus the area of %.#%!(7!,$#!)$(5#!7241.#!2'!
the green rectangle is A = bh =5(8 × 7 = 56. Thus the area of the whole
<$1'!,$#!%.#%!(7!,$#!4.##*!.#-,%*45#!2'!
<$1'!,$#!%.#%!(7!,$#!4.##*!.#-,%*45#!2'!
A = +$ A= =B +$
× E == B=>
× 9!!<$1'!,$#!
E = => 9!!<$1'!,$#!
figure is
%.#% %.#%!(7!,$#!)$(5#!7241.#!2'!
(7 .#3 .#-,%*45# + %.#% (7 4.##* .#-,%*45#
??F 9!!:*!
<$1'!,$#!%.#%!(7!,$#!4.##*!.#-,%*45#!2'!
A = +$ ==B=>
× E+ ==>=>= 9!!<$1'!,$#!
%.#%!(7!,$#!)$(5#!7241.#!2'!
%.#%!(7!,$#!)$(5#!7241.#!2'!
area
red
rectangle
area
of (7
green
=
56
56
=9!!:*!
112.
%.#%of
(7%.#%
.#3 .#-,%*45#
(7
.#3 .#-,%*45#
++
%.#%
(7
+ %.#%
4.##*
.#-,%*45#
4.##*rectangle
.#-,%*45#
= => + =>
= =>
= ??F
++
=>9!!:*!
= ??F
%.#% (7 .#3 .#-,%*45# + %.#% (7 4.##* .#-,%*45# = => + => = ??F 9!!:*!
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.
6"-)$!"#$+&#+$'*$!"#$*.0-&#$.)$::;$)7-+&#$-/.!)5$$6".)$*.!)$2#,,$2.!"$!"#$
4#*./.!.'/$'*$+&#+$2".("$.)$!"#$/-3<#&$'*$)7-+&#$-/.!)$!"+!$2.,,$('=#&$+$
(,')#4$*.0-&#5$
$
4
Geometry
>-&$/#?!$*'&3-,+$2.,,$<#$*'&$!"#$+&#+$'*$+$%+&+,,#,'0&+35$$@$%+&+,,#,'0&+3$.)$
%+&+,,#,'0&+3$
+$7-+4&.,+!#&+,$2.!"$'%%').!#$).4#)$%+&+,,#,5$
Area of a Parallelogram
$
@&#+$'*$+$N+&+,,#,'0&+3$
$
A = bh
$
@ = <"
$
b = base of the parallelogram
$
7(8(%9"(7-*"(#3(%9"(:-&-;;";#1&-$(
9(h
8(%9"(9",19%(#3(%9"(:-&-;;";#1&-$(
= the height of the parallelogram
$
$
A'-$2.,,$/'!.(#$!"+!$!".)$.)$!"#$)+3#$+)$!"#$*'&3-,+$*'&$!"#$+&#+$'*$+$
You will notice that this is the same as the formula for the area of a rectangle. A rectangle
&#(!+/0,#5$$@$&#(!+/0,#$.)$B-)!$+$)%#(.+,$!C%#$'*$%+&+,,#,'0&+35$$6"#$"#.0"!$'*$
is just a special
type of parallelogram. The height of a parallelogram is a segment that
+$%+&+,,#,'0&+3$.)$+$)#03#/!$!"+!$('//#(!)$!"#$!'%$'*$!"#$%+&+,,#,'0&+3$+/4$
connects the!"#$<+)#$'*$!"#$%+&+,,#,'0&+3$+/4$.)$%#&%#/4.(-,+&$!'$<'!"$!"#$!'%$+/4$!"#$
top of the parallelogram and the base of the parallelogram and is perpendicular
to both the<+)#5$$D/$!"#$(+)#$'*$+$&#(!+/0,#1$!".)$.)$!"#$)+3#$+)$'/#$'*$!"#$).4#)$'*$!"#$
top and the base. In the case of a rectangle, this is the same as one of the sides
of the rectangle
that is perpendicular to the base.
&#(!+/0,#$!"+!$.)$%#&%#/4.(-,+&$!'$!"#$<+)#5$
$
Example 3.
Find the area of the figure below.
E?+3%,#$FG$
H./4$!"#$+&#+$'*$!"#$*.0-&#$<#,'2$
64(
2(
64(
$
$
Solution: In I',-!.'/G$
this figure, the base of the parallelogram is 15 units and the height is 6 units.
This mean that we D/$!".)$*.0-&#1$!"#$<+)#$'*$!"#$%+&+,,#,'0&+3$.)$:J$-/.!)$+/4$!"#$
only need to multiply to find the area of A = bh = 15 × 6 = 90 square
"#.0"!$.)$K$-/.!)5$$6".)$3#+/$!"+!$2#$'/,C$/##4$!'$3-,!.%,C$!'$*./4$
units.
!"#$+&#+$'*$ @ = <" = :J × K = LM $)7-+&#$-/.!)5$
$
You should notice that
we cannot find the perimeter of this figure since we do not have the
A'-$)"'-,4$/'!.(#$!"+!$2#$(+//'!$*./4$!"#$%#&.3#!#&$'*$!".)$*.0-&#$
lengths of all of the)./(#$2#$4'$/'!$"+=#$!"#$,#/0!")$'*$+,,$'*$!"#$).4#)1$+/4$2#$"+=#$/'$
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 = 12 (b1 + b2 )h
b1 = one base of the trapezoid
b2 = 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
%+"8,9&(1$(0$"$0,)6,/%$%-"%$<&//,<%0$%-,$&/,$7"0,$&'$%-,$%+"8,9&(1$"/1$%-,$
&%-,+$7"0,$&'$%-,$%+"8,9&(1$"/1$(0$8,+8,/1(<*."+$%&$7&%-$&'$%-,$7"0,03$$$
$
>5"68.,$?@$
MAT 142 - Module
A(/1$%-,$"+,"$&'$%-,$'()*+,$$
5
74(
68(
565(
$
$
Solution: For this
trapezoid, the bases are shown as the top and the bottom of the figure.
B&.*%(&/@$
The lengths of these sides
are 45 and 121 units. It does not matter which of these we say is
A&+$%-(0$%+"8,9&(1C$%-,$7"0,0$"+,$0-&!/$"0$%-,$%&8$"/1$%-,$7&%%&6$
b1 and which is b2 . The
height
of the trapezoid is 20 units. When we plug all this into the
&'$%-,$'()*+,3$$=-,$.,/)%-0$&'$%-,0,$0(1,0$"+,$?D$"/1$EFE$*/(%03$$G%$
1
(b1 + b2 )h = 12 (121 + 45)20 = 16607Esquare
units.
formula, we get A = 21&,0$/&%$6"%%,+$!-(<-$&'$%-,0,$!,$0"#$(0$
$"/1$!-(<-$(0$
7F3$$=-,$
Our next
will have
up which
-,()-%$&'$%-,$%+"8,9&(1$(0$FH$*/(%03$$I-,/$!,$8.*)$"..$%-(0$(/%&$%-,$
E
formulas will
be for finding
ofE a(EFE
triangle
(a three-sided
polygon). We
'&+6*."C$!,$),%$
: = the
(7E +area
7F )- =
+ ?D )FH = EJJH
0;*"+,$
F
F
more than one formula for this since there
are different situations that can come
*/(%03$
will require different formulas
$
Triangle Area Formulas
Area of a Triangle
for a triangle with a base and a height
A = 21 bh
b = the base of the triangle
h = the height of the triangle
Heron’s Formula
for a triangle with only sides
p
A = s(s − 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 = 12 (a + b + c)
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
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"(2(#3(45(
of the triangle. In this
case, we would use Herons formula. !"#$"%&'()#%"*(
Example
!"#$%&'()*(
+"&,$"%"&(-./(0&"-(
+,-.(/0'(#1'#(23(/0'(3,451'6(
5. !"#$%&'()*(
Find
the area of the figure
+,-.(/0'(#1'#(23(/0'(3,451'6(
+-1"(2(#3(45(
below
9(
:74(
9(
:74(
678(
(
(
678(
72&5/,2-*(
(
(
82/,9'(/0#/(,-(/0,:(3,451'(0#:(#(.#:0'.(&,-'(/0#/(,:(:02;-(/2(<'(
72&5/,2-*(
%'1%'-.,95&#1(/2(/0'(:,.'(/0#/(,:(=6>(5-,/:(,-(&'-4/06((?0,:(,:(02;(;'(
Solution: Notice
that
in this figure has a dashed line that is shown to be perpendicular to
82/,9'(/0#/(,-(/0,:(3,451'(0#:(#(.#:0'.(&,-'(/0#/(,:(:02;-(/2(<'(
,-.,9#/'(/0'(0',40/(23(/0'(/1,#-4&'(@/0'(.#:0'.(&,-'A(#-.(/0'(<#:'(
the side that is 8.2 units in length. This is how we indicate the height of the triangle (the
%'1%'-.,95&#1(/2(/0'(:,.'(/0#/(,:(=6>(5-,/:(,-(&'-4/06((?0,:(,:(02;(;'(
dashed line) and the 23(/0'(/1,#-4&'(@/0'(:,.'(/0#/(/0'(.#:0'.(&,-'(,:(%'1%'-.,95&#1(/2A6((
base
of the triangle (the side that the dashed line is perpendicular to).
,-.,9#/'(/0'(0',40/(23(/0'(/1,#-4&'(@/0'(.#:0'.(&,-'A(#-.(/0'(<#:'(
?0#/($'#-:(;'(0#B'(<2/0(/0'(0',40/(#-.(/0'(<#:'(23(/0,:(/1,#-4&'C(
That means we have both
the height and the base of this triangle, so we can just plug these
23(/0'(/1,#-4&'(@/0'(:,.'(/0#/(/0'(.#:0'.(&,-'(,:(%'1%'-.,95&#1(/2A6((
:2(;'(9#-(D5:/(%&54(/0':'(-5$<'1:(,-/2(/0'(321$5&#(/2(4'/(
numbers into the formula
to
get
?0#/($'#-:(;'(0#B'(<2/0(/0'(0',40/(#-.(/0'(<#:'(23(/0,:(/1,#-4&'C(
F
F
G = <0 = @=6>A@ E6)A = F=6E) (:H5#1'(5-,/:6(((
:2(;'(9#-(D5:/(%&54(/0':'(-5$<'1:(,-/2(/0'(321$5&#(/2(4'/(
>
>
1F
1
(G =AF =
= 18.45square units.
<0 = bh@==
6>A@ E(8.2)(4.5)
6)A = F=6E) (:H5#1'(5-,/:6(((
2>
2
>
82/,9'(/0#/(/0'(-5$<'1(I(,:(4,B'-(#:(/0'(&'-4/0(23(2-'(23(/0'(:,.':(
(
23(/0'(/1,#-4&'6((?0,:(:,.'(,:(-2/(#(0',40/(23(/0'(/1,#-4&'(:,-9'(,/(,:(
82/,9'(/0#/(/0'(-5$<'1(I(,:(4,B'-(#:(/0'(&'-4/0(23(2-'(23(/0'(:,.':(
-2/(%'1%'-.,95&#1(/2(#-2/0'1(:,.'(23(/0'(/1,#-4&'6((J/(,:(#&:2(-2/(#(
Notice that the number
6 is given as the length of one of the sides of the triangle. This side
23(/0'(/1,#-4&'6((?0,:(:,.'(,:(-2/(#(0',40/(23(/0'(/1,#-4&'(:,-9'(,/(,:(
<#:'(23(/0'(/1,#-4&'C(:,-9'(/0'1'(,:(-2(,-.,9#/,2-(23(/0'(
is not a height of the -2/(%'1%'-.,95&#1(/2(#-2/0'1(:,.'(23(/0'(/1,#-4&'6((J/(,:(#&:2(-2/(#(
triangle since it is not perpendicular to another side of the triangle. It
is also not a base of %'1%'-.,95&#1(.,:/#-9'(<'/;''-(/0#/(:,.'(#-.(/0'(2%%2:,/'(B'1/'"6((
the
triangle, since there is no indication of the perpendicular distance
<#:'(23(/0'(/1,#-4&'C(:,-9'(/0'1'(,:(-2(,-.,9#/,2-(23(/0'(
?0,:($'#-:(/0#/(,/(,:(-2/(5:'.(,-(/0'(9#&95&#/,2-(23(/0'(#1'#(23(/0'(
between that side and/1,#-4&'6(
the opposite vertex. This means that it is not used in the calculation
%'1%'-.,95&#1(.,:/#-9'(<'/;''-(/0#/(:,.'(#-.(/0'(2%%2:,/'(B'1/'"6((
of the area of( the triangle.
?0,:($'#-:(/0#/(,/(,:(-2/(5:'.(,-(/0'(9#&95&#/,2-(23(/0'(#1'#(23(/0'(
(/1,#-4&'6(
Example 6. !"#$%&'(I*(
(Find the area of the figure below
(
+,-.(/0'(#1'#(23(/0'(3,451'6(
!"#$%&'(I*(
+,-.(/0'(#1'#(23(/0'(3,451'6(
:74(
879(
:74(
879(
;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 = 21 bh = 12 (1.7)(2.6) = 21.21 square units.
Example 7. Find the area of the figure below
'$6;0/,)&'%")%+0)-"/6$#1)%").0%) @ =
>
;+ =
,F$1/0)$'&%,4)
)
MAT 142 - Module
G716<#0)9()
H&'3)%+0)1/01)"-)%+0)-&.$/04)
7(
>
E849 DE>4?D = >4>8 )
7
6(
8(
)
)
Solution: You!"#$%&"'()
should notice that we do not have a height for this triangle. This means
I"$),+"$#3)'"%&=0)%+1%)20)3")'"%)+1B0)1)+0&.+%)-"/)%+&,)%/&1'.#04))
that we cannot use the
formula that we have been using to find the area of this triangle.
:+&,)601',)%+1%)20)=1''"%)$,0)%+0)-"/6$#1)%+1%)20)+1B0)
We
do have the( length of all
three sides
of the triangle. This means that we can use Herons
K0/"'L,)H"/6$#1)
!"#$"%&'()#%"*(
;00')$,&'.)%")-&'3)%+0)1/01)"-)%+&,)%/&1'.#04))J0)3")+1B0)
8
Formula
to
find
the
area
of
this
triangle.
+-1"(22(#3(45(
@=
, E, − 1+"&,$"%"&(-./(0&"-(
DE, − ; DE, − = D %+0)#0'.%+)"-)1##)%+/00),&30,)"-)%+0)%/&1'.#04)):+&,)601',)
>
%+1%)20)=1')$,0)K0/"'L,)H"/6$#1)%")-&'3)%+0)1/01)"-)%+&,)
(
For this formula,
it does# not
matter# which side we label a, b, or c. For our purposes, we will
%/&1'.#04)
-(9(#."(*,/"(#4(%:"(%&,-.1;"(
"
=
%
+8.
& +Now
' = that
($#) = we
#*+,have
+!!-./!/0!123!4567!0809:;<=37!=3!;.!
let a be 6, b be 7, and
c
be
a, b, and c, we need to calculate s so that
$
!! $ )
<(9(-.#%:"&(*,/"(#4(%:"(%&,-.1;"(
we
can plug a, b, c , >09.3?"[email protected]!;.!@=3B!;<0!2902!.@!;<="!;9=23750!;.!C0!
and s H"/)%+&,)-"/6$#1A)&%)3"0,)'"%)61%%0/)2+&=+),&30)20)#1;0#)
into the formula. We get s = 12 (6 + 7 + 8) = 112A)(21) = 10.5. Now
=(9(%:"(%:,&/(*,/"(#4(%:"(
%&,-.1;"(
we
can plug everything in
Herons formula# to find the area
of;);0)9A)1'3)
this triangle
to be
# to
1);0)?A)
);0)
D=
";(A)"/)
" − 2=)(4))H"/)"$/)<$/<",0,A)20)2&##)#0%)
" − C)(" − 1) =
#*+,(#*+, − %)(#*+,
− &)(#*+,
− ') ==#*+#%%,'
*(9(>(?
p- O);)O)=D
p
!! $ M4))N"2)%+1%)20)+1B0)$1A);A)1'3)=A)20)'003)%")=1#=$#1%0),),")
A = s(s − a)(s − b)(s
− c) = 10.5(10.5 − 6)(10.5 − 7)(10.5 − 8) ≈ 20.33316 square units.
"E6290!63=;"+!
%+1%)20)=1')<#$.)1A);A)=)A)1'3),)&'%")%+0)-"/6$#14))J0).0%)
!
D3.;<[email protected]!;<2;!/0!290!=3;090";0B!=3!="!;<0!F:;<27.9023!G<0.90A+!!
Another formula
that we are interested in is the Pythagorean Theorem. This applies to only
G<="!2445=0"!;.!.35:!9=7<;!;9=23750"+!G<0!F:;<27.9023!G<0.90A!9052;0"!;<0!
right triangles.
The Pythagorean Theorem relates the lengths of the sides of a right triangle.
5037;<"!.@!;<0!"=B0"!.@!2!9=7<;!;9=23750+!
!
Pythagorean Theorem
!
F:;<27.9023!G<0.90A!
!
a2 + b2 != c2
!
$
$
$
!!2 + C = 1
! a = leg (one
(
side of the triangle that makes up the right angle)
!
-(9(:"1(;#."(*,/"(#3(%<"(%&,-.1:"(%<-%($-="*(>?(%<"(&,1<%(-.1:"@(
A(9(:"1(;-.#%<"&(*,/"(#3(%<"(%&,-.1:"(%<-%($-="*(>?(%<"(&,1<%(
b! = leg (another
side of the triangle that makes up the right angle
-.1:"@(
!
cB(9(<'?#%".>*"(;*,/"(#??#*,%"(%<"(&,1<%(-.1:"@(
= hypotenuse (side opposite the right angle)
!
!
H<03!6"=37!;<0!F:;<27.9023!G<0.90AI!=;!="!=A4.9;23;!;.!A2J0!"690!;<2;!/0!
When using the
Pythagorean Theorem, it is important to make sure that we always use the
25/2:"!6"0!;<0!507"!.@!;<0!;[email protected]!
2!23B!C!23B!;<0!<:4.;036"[email protected]!
1+!
legs of the triangle
for a and b and the hypotenuse
for c.
!
Example 8.KL2A450!'M!
Find the length of the third side of the triangle below.
N=3B!;<0!5037;<!.@!;<0!;<=9B!"=B0!.@!;<0!;9=23750!C05./+!
(
)
28(
67(
!
!
Solution: TheO.56;=.3M!
figure is a right triangle (as indicated by the box in one of the angles of
G<0!@=7690!="!2!9=7<;!;9=23750!(2"!=3B=12;0B!C:!;<0!C.L!=3!.30!.@!;<0!
the triangle). We need to decide what the side we are looking for is in terms of a leg or
23750"!.@!;<0!;9=23750)+!!H0!300B!;.!B01=B0!/<2;!;<0!"=B0!/0!290!
the hypotenuse of [email protected]!="!=3!;09A"!.@!2!507!.9!;<0!<:4.;036"0!.@!;<0!;9=23750+!!
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
G<0!<:4.;036"0!="!;<0!"=B0!.@!;<0!;9=23750!.44."=;0!;<0!9=7<;!23750+!!
formula. This meansG<2;!/.65B!C0!;<0!"=B0!;<2;!<2"!5037;<!$%!=3!.69!4=1;690+!!G<6"!
that the other two sides of the triangle are legs a and
1! b. We will let a
be 10, and we will thus be looking for b. When we plug all this into the formula, we get
8
Geometry
a2 + b 2 = c 2
102 + b2 = 262
100 + b2 = 676
We now need to solve this equation for b.
100 + b2 = 676
b2 = 576
√
b = 576 = 24
Since we are asked for a length, we have that the third side is 24 units. (Notice that this is
not square units since we are not finding an area).
We are now going to move on to circles. We already mentioned the perimeter (circumference)
of a circle in the perimeter sections. We also need a formula for finding the area of a circle.
Area of a Circle
A = πr2
r = radius of the circle
π = the number that is approximated by 3.141593
For circle problems we need to remember that the circumference (perimeter) of a circle is
only the curved part of the circle. It does not include either the radius or diameter of the
circle. In order to find the perimeter and area, we will need the radius of the circle. Recall
that the diameter of a circle is a segment from on point on the circle to another point
on the circle that passes through the!"#$"%&'()#%"*(
center of the circle. A radius of a circle is half of a
+-1"(23(#4(56(
diameter (i.e.+"&,$"%"&(-./(0&"-(
a segment that from one point on the circle to the center
of the circle).
Example 9.340&51(*67*
Find the perimeter and area of the circle below.
8")9*-:(*5(#"&(-(#*0)9*0#(0*/'*-:(*$"#$1(*;(1/<=*
22(
*
*
Solution: The>/1%-"/)7*
number 11 in the figure above is the length of the diameter of this circle.
We need the radius?:(*)%&;(#*@@*")*-:(*'"A%#(*0;/B(*"C*-:(*1()A-:*/'*-:(*9"0&(-(#*/'*
to be able to use the formulas. The radius is half of the diameter.
!"#$%&'(#()$(*
1
-:"C*$"#$1(=**D(*)((9*-:(*#09"%C*-/*;(*0;1(*-/*%C(*-:(*'/#&%10C=**?:(*
Thus
r = 2 (diameter) = 21 (11) = 5.5 . We are now ready to find both the perimeter
+,(#"&(-(#.*/'*0*
@
(circumference)
and#09"%C*"C*:01'*/'*-:(*9"0&(-(#=**?:%C*
area by plugging 5.5 into each# =formula
for =r.@ +@@.
Perimeter:
!"#$1(*
+9"0&(-(#.
= E=E =**
2
** 2
C = 2πr = 2(π)(5.5) = 11π ≈ 34.557519
D(*0#(*)/<*#(09F*-/*'")9*;/-:*-:(*5(#"&(-(#*+$"#$%&'(#()$(.*0)9*
! = 2π #
0#(0*;F*51%AA")A*E=E*")-/*(0$:*'/#&%10*'/#*#=*
,(#"&(-(#7***
&(7(&-/,8*(#4(%9"(
:,&:;"(
! = 2π# = 2+π .+E=E. = @@π ≈ GH=EEIE@6***
**
<(7(%9"(.8$="&(%9-%(
?:(*0)C<(#*@@π*%)"-C*"C*0)*(40$-*0)C<(#*'/#*-:(*5(#"&(-(#=**?:(*
,*(->>&#?,$-%"/(='(
0)C<(#*GH=EEIE@6*%)"-C*"C*0)*055#/4"&0-"/)=*
3@2A25B3(
*
2
** 2
D(*0#(*)/<*#(09F*-/*'")9*;/-:*-:(*5(#"&(-(#*+$"#$%&'(#()$(.*0)9*
! = 2π #
0#(0*;F*51%AA")A*E=E*")-/*(0$:*'/#&%10*'/#*#=*
,(#"&(-(#7***
&(7(&-/,8*(#4(%9"(
MAT 142 - Module
:,&:;"(
!
=
2
π#
=
2+
π
.+E=E.
= @@π ≈ GH=EEIE@6***
**
<(7(%9"(.8$="&(%9-%(
The
answer 11π units
is an exact
answer for the perimeter. The answer 34.557519
?:(*0)C<(#*@@
π*%)"-C*"C*0)*(40$-*0)C<(#*'/#*-:(*5(#"&(-(#=**?:(*
,*(->>&#?,$-%"/(='(
an3@2A25B3(
approximation. 0)C<(#*GH=EEIE@6*%)"-C*"C*0)*055#/4"&0-"/)=*
*
Area:
J#(07*
2
2
= 2πr
≈ 95.0331778
JA
= π#
= π +=
E=Eπ(5.5)
. 2 = GK=2E=
π 30.25π
≈ 6E=KGG@IIL
**?:(*0)C<(#*GK=2Eπ*
9
units is
CM%0#(*%)"-C*"C*0)*(40$-*0)C<(#*'/#*-:(*0#(0=**?:(*0)C<(#*
The answer 30.25π square
units is an exact answer for the area. The answer 95.0331778
6E=KGG@IIL*CM%0#(*%)"-C*"C*0)*055#/4"&0-"/)=*
square units is an approximation.!"#$"%&'()#%"*(
*
+"&,$"%"&(-./(0&"-(
+-1"(23(#4(56(
340&51(*@K7*
Example 10.
Find the perimeter and area of the semicircle below.
8")9*-:(*5(#"&(-(#*0)9*0#(0*/'*-:(*C(&"$"#$1(*;(1/<=*
!"#$%&"'()
*')%+&,)-."/#012)30)+450)4),01&6&.6#0)74)+4#8)"8)4)6&.6#09:));+0)
<&410%0.)"8)%+&,)+4#8)6&.6#0)&,)=>?:));+$,)%+0).4<&$,)&,)
=
. = 7=>?9 = @? :)
>
2CD(
*
* )
Solution: In thisA0.&10%0.()
problem, we have a semicircle (a half of a circle). The diameter of this
B0)'"3)64')8&'<)%+0)-0.&10%0.:))B0)3&##)8&'<)%+0)6&.6$180.0'60)
half circle is 120. Thus
the radius is r == 12 (120) = 60 .
"8)%+0)3+"#0)6&.6#0)4'<)%+0')<&5&<0)&%)/C)>),&'60)30)"'#C)+450)
Perimeter: We now +4#8)"8)4)6&.6#0:));+&,)3&##)D&50)$,)
can find the perimeter. We Ewill
circumference
of the whole circle
= >find
π 7@?9the
= =>?
π )4,)%+0)
and then divide it by
2 since we only have half of a circle. This will give us C = 2πr = 120π
6&.6$180.0'60)8".)%+0)3+"#0)6&.6#0)4'<)@?F)$'&%,)4,)%+0)
6&.6$180.0'60)"8)+4#8)"8)%+0)6&.6#0:)))
as the circumference for the whole circle and 60π units as the circumference of half of the
)
circle.
G'8".%$'4%0#C)%+&,)&,)'"%)%+0)4',30.)8".)%+0)-0.&10%0.)"8)%+0)
Unfortunately this is8&D$.0)D&50':));+0)6&.6$180.0'60)"8)4)6&.6#0)&,)%+0)6$.50<)-4.%:))
not the answer for the perimeter of the figure given. The circumference
of a circle is the curved
part. The straight line segment in our figure would not have been
;+0),%.4&D+%)#&'0),0D10'%)&')"$.)8&D$.0)3"$#<)'"%)+450)/00')
part of the circumference
of the whole circle, and thus it is not included as part of half of
-4.%)"8)%+0)6&.6$180.0'60)"8)%+0)3+"#0)6&.6#02)4'<)%+$,)&%)&,)'"%)
&'6#$<0<)4,)-4.%)"8)+4#8)"8)%+0)6&.6$180.0'60:))B0)"'#C)+450)
the circumference. We only have the red part shown in the picture below.
%+0).0<)-4.%),+"3')&')%+0)-&6%$.0)/0#"3:)
278(
B0),%&##)'00<)%")&'6#$<0)%+0),%.4&D+%),0D10'%)%+4%)&,)=>?)$'&%,)
We still need to include
the straight segment that is 120 units long. Thus the whole perimeter
#"'D:));+$,)%+0)3+"#0)-0.&10%0.)"8)%+0)8&D$.0)&,)
of the figure is
6$.50< -4.% + ,%.4&D+% -4.% = @?π + =>? ≈ K?L:HIJJJI $'&%,:)
)
curved
part + straight part = 60π + 120 ≈ 308.495559units.
M.04()
B0)64')4#,")8&'<)%+0)4.04:))N0.0)30)3&##)-#$D)@?)&')8".).)&')%+0)
We can also find the area. Here we will plug 60 in for r in the area formula for a
4.04)8".1$#4)8".)4)3+"#0)6&.6#0)4'<)%+0')<&5&<0)/C)>)8".)%+0)+4#8)
circle and then
divide by 2 for the half circle. This will give us A = π(60)2 = 3600π
6&.6#0:));+&,)3&##)D&50)$,) M = π 7@?9 > = K@??π ),O$4.0)$'&%,)8".)
Area:
whole
square units for the%+0)4.04)"8)%+0)3+"#0)6&.6#0)4'<)=L??F),O$4.0)$'&%,)8".)%+0)
area of the whole circle and 1800π square units for the area of the half
circle. Now are we 4.04)"8)%+0)+4#8)6&.6#0:))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?
If we think back to the definition of the area, (it is the number
".)&,)%+0.0)1".0)%+4%)30)'00<)%")<")#&Q0)30)<&<)&')8&'<&'D)%+0)
of square units needed
to cover the figure) we should see that there is nothing further to do.
-0.&10%0.R))*8)30)%+&'Q)/46Q)%")%+0)<08&'&%&"')"8)%+0)4.042)7&%)&,)
%+0)'$1/0.)"8),O$4.0)$'&%,)'00<0<)%")6"50.)%+0)8&D$.09)30)
The inside of the figure
was covered already. We just have half as much inside as we did
,+"$#<),00)%+4%)%+0.0)&,)'"%+&'D)8$.%+0.)%")<":));+0)&',&<0)"8)
before.
Example 11. Walk 10 yards south, then 12 west and then 3 yards south. How far from the
original starting point are you?
Solution: To solve this problem, it would help to have a picture.
10
5#2>%<?%3#$(!%!0-")@%")&'%<A%.&!"%#'(%")&'%B%3#$(!%!0-")4%%C0.%*#$%*$07%
")&%0$+,+'#2%!"#$"+',%;0+'"%#$&%30-D%
")&%*+,-$&%.#!%/01&$&(%#2$&#(34%%5&%6-!"%)#1&%)#2*%#!%7-/)%
%
+'!+(&%#!%.&%(+(%8&*0$&4%
E02-"+0'=%
Geometry
%
F0%!021&%")+!%;$082&7@%+"%.0-2(%)&2;%"0%)#1&%#%;+/"-$&4%
9:#7;2&%<<=%
!"#$"%
5#2>%<?%3#$(!%!0-")@%")&'%<A%.&!"%#'(%")&'%B%3#$(!%!0-")4%%C0.%*#$%*$07%
")&%0$+,+'#2%!"#$"+',%;0+'"%#$&%30-D%
%
E02-"+0'=%
F0%!021&%")+!%;$082&7@%+"%.0-2(%)&2;%"0%)#1&%#%;+/"-$&4%
26('-&/*(
!"#$"%
7('-&/*(
28('-&/*(
26('-&/*(
&'(%
G801&%.&%/#'%!&&%#%(&!/$+;"+0'%0*%)0.%.&%.#2>&(%+'%")&%;$082&74%%
28('-&/*(
5)#"%.&%#$&%#!>&(%"0%*+'(%+!%)0.%*#$%+"%+!%*$07%.)&$&%.&%!"#$"&(%
7('-&/*(
"0%.)&$&%.&%&'(&(4%%F)#"%(+!"#'/&%.0-2(%8&%$&;$&!&'"&(%83%#%
Above we can see a !"$#+,)"%2+'&%*$07%")&%!"#$"%"0%")&%*+'+!)4%%F)+!%.+22%8&%!)0.'%+'%")&%
description of how we walked in the problem. What we are asked to
&'(%
find is how far it is ;+/"-$&%8&20.%83%#%82#/>%(#!)&(%2+'&4%
from where we started to where we ended. That distance would be
!"#$"%
represented by a straight
line from the start to the finish.
This will be shown in the picture
%
G801&%.&%/#'%!&&%#%(&!/$+;"+0'%0*%)0.%.&%.#2>&(%+'%")&%;$082&74%%
%
below by a black dashed
line.
5)#"%.&%#$&%#!>&(%"0%*+'(%+!%)0.%*#$%+"%+!%*$07%.)&$&%.&%!"#$"&(%
%
"0%.)&$&%.&%&'(&(4%%F)#"%(+!"#'/&%.0-2(%8&%$&;$&!&'"&(%83%#%
%
!"$#+,)"%2+'&%*$07%")&%!"#$"%"0%")&%*+'+!)4%%F)+!%.+22%8&%!)0.'%+'%")&%
%
;+/"-$&%8&20.%83%#%82#/>%(#!)&(%2+'&4%
26('-&/*(
%
!"#$"%
%
%
%
%
%
%
%
28('-&/*(
%
7('-&/*(
%
26('-&/*(
%
%
&'(%
%
%
28('-&/*(
%
7('-&/*(
&'(%
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.
%
)&%#$&%*'%"/&%$,-/"%"$#:8%"/*.-/2%%</,!%,!%#%=6"/#-*$&#'%</&*$&3%
5$*41&32%%>+%0&%#((%#%:*.51&%*+%1,'&%!&-3&'"!?%0&%:#'%:$&#"&%#%
MAT 142 - Module
11
$,-/"%"$,#'-1&%0/*!&%/65*"&'.!&%,!%"/&%1,'&%0&%#$&%1**8,'-%+*$2%
%
!"#$"%
%
%
%
%
%
27('-&/*(
%
%
%
%
29('-&/*(
%
8('-&/*(
%
%
&'(%
%
@!%6*.%:#'%!&&%#4*7&?%"/&%*$#'-&%!&-3&'"!%#1*'-%0,"/%"/&%$&(%
As you can see above,
the orange segments along with the red segment that was 10 yards
!&-3&'"%"/#"%0#!%AB%6#$(!%#'(%"/&%(#!/&(%!&-3&'"%3#8&%.5%#%
and the dashed segment
make up a right triangle whose hypotenuse is the dashed segment.
$,-/"%"$,#'-1&%0/*!&%/65*"&'.!&%,!%"/&%(#!/&(%!&-3&'"2%%)&%:#'%
We can figure out
how long each of the legs of this triangle are fairly easily. The orange
+,-.$&%*."%/*0%1*'-%&#:/%*+%"/&%1&-!%*+%"/,!%"$,#'-1&%#$&%+#,$16%
segment that is an
extension
of the red segment is another 3 yards. Thus the vertical leg of
&#!,162%%</&%*$#'-&%!&-3&'"%"/#"%,!%#'%&C"&'!,*'%*+%"/&%$&(%
this triangle is 10!&-3&'"%,!%#'*"/&$%D%6#$(!2%%</.!%"/&%7&$",:#1%1&-%*+%"/,!%"$,#'-1&%
+ 3 = 13 yards long. The orange segment which is horizontal is the same
=6"/#-*$&#'%
,!%AB%E%D%F%AD%6#$(!%1*'-2%%</&%*$#'-&%!&-3&'"%0/,:/%,!%/*$,G*'"#1%,!%
length as the green
segment. This means that the horizontal leg of this triangle is 12 yards
</&*$&3%
"/&%!#3&%1&'-"/%#!%"/&%-$&&'%!&-3&'"2%%</,!%3&#'!%"/#"%"/&%
long. We can now use the Pythagorean Theorem to calculate how far away from the original
%
/*$,G*'"#1%1&-%*+%"/,!%"$,#'-1&%,!%AH%6#$(!%1*'-2%%)&%:#'%'*0%.!&%"/&%
starting
H
H
H point we are.
%%# + 4 = :
=6"/#-*$&#'%</&*$&3%"*%:#1:.1#"&%/*0%+#$%#0#6%+$*3%"/&%*$,-,'#1%
(
!"#$",'-%5*,'"%0&%#$&2%
-(:(;"1((
<(:(;"1((
=(:(>'?#%".@*"((
132 + 122 = c2
169 + 144 = c2
√
313 = c2
313 = c ≈ 17.6918
Here an exact answer does not really make sense. It is enough to approximate that we end
up 17.6918 yards from where we started.
Volume and Surface Area
We will continue our study of geometry by studying three-dimensional figures. We will look
at the two-dimensional aspect of the outside covering of the figure and also look at the
three-dimensional space that the figure encompasses.
The surface area of a figure is defined as the sum of the areas of the exposed sides of an
object. A good way to think about this would be as the area of the paper that it would take
to cover the outside of an object without any overlap. In most of our examples, the exposed
sides of our objects will polygons whose areas we learned how to find in the previous section.
When we talk about the surface area of a sphere, we will need a completely new formula.
The volume of an object is the amount of three-dimensional space an object takes up. It
can be thought of as the number of cubes that are one unit by one unit by one unit that it
takes to fill up an object. Hopefully this idea of cubes will help you remember that the units
for volume are cubic units.
12
Geometry
Surface Area of a Rectangular Solid
SA = 2(lw + lh + wh)
l = length of the base of the solid
w = width of the base of the solid
h = height of the solid
Volume of a Solid with a Matching Base and Top
V = Ah
A = area of the base of the solid
h = height of the solid
Volume of a Rectangular Solid
(specific type of solid with matching base and top)
V = lwh
l = length of the base of the solid
w = width of the base of the solid
!"#$"%&'()#%"*(
+#,-$"(./0(1-&2.3"(4&".(
h = height of the solid
5.6"(78(#2(9:(
!"#$%&'()*(
Example 12.
Find the volume and the surface area of the figure below
+,-.(/0'(12&3$'(#-.(/0'(4356#7'(#5'#(26(/0'(6,835'(9'&2:(
=<@($(
;<=($(
><?($(
(
(
Solution: ThisA2&3/,2-*(
figure is a box (officially called a rectangular prism). We are given the
B0,4(6,835'(,4(#(92"(C266,7,#&&D(7#&&'.(#(5'7/#-83&#5(%5,4$E<((F'(#5'(
lengths of each of the
length, width, and height of the box, thus we only need to plug into
the formula. Based 8,1'-(/0'(&'-8/04(26('#70(26(/0'(&'-8/0G(:,./0G(#-.(0',80/(26(/0'(
on the way our box is sitting, we can say that the length of the base
92"G(/034(:'(2-&D(-''.(/2(%&38(,-/2(/0'(625$3&#<((H#4'.(2-(/0'(:#D(
is 4.2 m; the width of
the base is 3.8 m; and the height of the solid is 2.7 m. Thus we can
235(92"(,4(4,//,-8G(:'(7#-(4#D(/0#/(/0'(&'-8/0(26(/0'(9#4'(,4(;<=($I(
quickly find the volume
of the box to be V = lwh = (4.2)(3.8)(2.7) = 43.092 cubic meters.
/0'(:,./0(26(/0'(9#4'(,4(><?($I(#-.(/0'(0',80/(26(/0'(42&,.(,4(=<@($<((
Although there is a formula
that we can use to find the surface area of this box, you should
B034(:'(7#-(J3,7K&D(6,-.(/0'(12&3$'(26(/0'(92"(/2(9'(
notice that each of Nthe
six= faces
surfaces)
of the box is a rectangle. Thus, the
= &:0
C ;<=EC><(outside
?EC=<@E = ;>
<LM= (739,7($'/'54<(
surface area is the sum
of the areas of each of these surfaces, and each of these areas is
(
O&/02380(/0'5'(,4(#(625$3&#(/0#/(:'(7#-(34'(/2(6,-.(/0'(4356#7'(
fairly straight-forward to calculate. We will use the formula in the problem. It will give us
#5'#(26(/0,4(92"G(D23(4023&.(-2/,7'(/0#/('#70(26(/0'(4,"(6#7'4(
SA = 2(lw + lh + wh)
= 2((4.2 × 3.8) + (4.2 × 2.7) + (3.8 × 2.7)) = 75.12 square meters.
A
of
C23/4,.'(4356#7'4E(26(/0'(92"(,4(#(5'7/#-8&'<((B034G(/0'(4356#7'(
#5'#(,4(/0'(43$(26(/0'(#5'#4(26('#70(26(/0'4'(4356#7'4G(#-.('#70(
cylinder is an object
with straight sides and circular ends of the same size. The volume
26(/0'4'(#5'#4(,4(6#,5&D(4/5#,80/P625:#5.(/2(7#&73&#/'<((F'(:,&&(34'(
a cylinder can be found in the same way you find the volume of a solid with a matching
/0'(625$3&#(,-(/0'(%529&'$<((Q/(:,&&(8,1'(34(
AO = =C&: + &0 + :0 E = =C ;<= S ><? + ;<= S =<@ + ><? S =<@E = @R<)= (
4J3#5'($'/'54<(
(
(
O(7D&,-.'5(,4(#-(29T'7/(:,/0(4/5#,80/(4,.'4(#-.(7,573&#5('-.4(26(/0'(4#$'(
4,U'<((B0'(12&3$'(26(#(7D&,-.'5(7#-(9'(623-.(,-(/0'(4#$'(:#D(D23(6,-.(/0'(
7D&,-.'5(
MAT 142 - 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.
!"#$"%&'()#%"*(
+#,-$"(./0(1-&2.3"(4&".(
Example
Volume of a Cylinder
5.6"(78(#2(9:(
!
V = Ah
L2'3%(!26!$!M?',-.(5!
!
!
A = area of the
L = base
E0 of the cylinder
!
!
h = height of the cylinder
4(;(%<"(.&".(#2(%<"(=.*"(#2(%<"(3',>/0"&(
!
<(;(%<"(<">6<%(#2(%<"(3',>/0"&(
!
Surface Area of a Cylinder
!
=356$7(!E5($!26!$!M?',-.(5!
!
SA = 2(πr2 ) + 2πrh
!
=E = )Cπ5 ) B + )π50
!
r = the radius of the circular base of the cylinder
!
&(;(%<"(&.0>-*(#2(%<"(3>&3-,.&(=.*"(#2(%<"(3',>/0"&(
h = height of the cylinder
<(;(%<"(<">6<%(#2(%<"(3',>/0"&(
!
?(;(%<"(/-$="&(%<.%(>*(.@@&#A>$.%"0(='(BC8D89EB(
!
π( = the number approximated by 3.141593
!
"#$%&'(!)*!
13. Find the volume and surface area of the figure below
+,-.!/0(!12'3%(!$-.!4356$7(!$5($!26!/0(!6,835(!9('2:!
;<!,-!
;)!,-!
!
!
Solution: This
figure is a cylinder. The diameter of its circular base is 12 inches. This
=2'3/,2-*!
means that the >0,4!6,835(!,4!$!7?',-.(5@!!>0(!.,$%(/(5!26!,/4!7,573'$5!9$4(!,4!;)!
radius of the circular base is r = 21 d = 12 (12) = 6 inches. The height of the
cylinderi s 10 inches.
To calculate the volume and surface area, we simply need to plug into
,-70(4@!!>0,4!%($-4!/0$/!/0(!5$.,34!26!/0(!7,573'$5!9$4(!,4!
the formulas. 5 = ; . = ; C;)B = A !,-70(4@!!>0(!0(,80/!26!/0(!7?',-.(5,!4!;<!,-70(4@!!
)
)
Surface Area:
2
>2!7$'73'$/(!/0(!12'3%(!$-.!4356$7(!$5($D!:(!4,%&'?!-((.!/2!&'38!,-/2!
SA = 2(πr ) + 2πrh = 2(π × 62 ) + 2π(6)(10) = 72π + 120π = 192π square units. This is an
/0(!625%3'$4@!
exact answer. An
approximate answer is 603.18579 square units.
=356$7(!E5($*!!
Volume:
=E = )Cπ5 ) B + )π50 = )Cπ ⋅ A) B + )π CABC;<B = G)π + ;)<π = ;F)π 4H3$5(!
In order to plug 3-,/4@!!>0,4!,4!$-!(#$7/!$-4:(5@!!E-!$&&52#,%$/(!$-4:(5!,4!A<I@;JKGF!
into the formula, we need to recall how to find the area of a circle (the base
of the cylinder is
a circle). We will the replace A in the formula with the formula for the
4H3$5(!3-,/4@!
area of a circle. V = Ah = πr2 h = π(62 )(10) = 360π cubic inches. An approximation of this
exact answer would be 1130.97336 cubic inches.
Our next set of formulas is going to be for spheres. A sphere is most easily thought of as a
ball. The official definition of a sphere is a three-dimensional surface, all points of which are
equidistant from a fixed point called the center of the sphere. A circle that runs along the
surface of a sphere to that it cuts the sphere into two equal halves is called a great circle of
14
Geometry
that sphere. A great circle of a sphere would have a diameter that is equal to the diameter
of the sphere.
Surface Area of a Sphere
SA = 4πr2
r = the radius of the circular base of the cylinder
π = the number
approximated by 3.141593
!"#$"%&'()#%"*(
+#,-$"(./0(1-&2.3"(4&".(
5.6"(78(#2(9:(
Volume of a Sphere
!"#$%&'()*(
+,-.(/0'(12&3$'(#-.(4356#7'(#5'#(26(/0'(6,835'(9'&2:(
V = 4 πr3
3
)
<
,-
r = the radius of the circular; base of the cylinder
!"#$"%&'()#%"*(
+#,-$"(./0(1-&2.3"(4&".(
5.6"(78(#2(9:(
π = the number approximated by 3.141593
!"#$%&'()*(
Example 14.
Find the volume and surface area of the figure below
+,-.(/0'(12&3$'(#-.(4356#7'(#5'#(26(/0'(6,835'(9'&2:(
)
<
,;
(
(
=2&3/,2-*(
>0,4(,4(#(4%0'5'?((@'(#5'(8,1'-(/0#/(/0'(.,#$'/'5(26(/0'(4%0'5'(,4(
<
) (,-70'4?((@'(-''.(/2(7#&73&#/'(/0'(5#.,34(26(/0'(4%0'5'(/2(
;
7#&73&#/'(/0'(12&3$'(#-.(4356#7'(#5'#?((>0'(5#.,34(26(#(4%0'5'(,4(
(
0#&6(26(,/4(.,#$'/'5?((>0,4($'#-4(/0#/(/0'(5#.,34(,4(
(
A
A & < # A & BD # BD
5 = . = $) ! = $ ! =
= A?;AB< (,-70'4?((@'(7#-(-2:(E34/(
% ;given
" B% ;
" AC
Solution:This=2&3/,2-*(
is a sphere.B We Bare
that
the diameter of the sphere is 3 58 inches. We
>0,4(,4(#(4%0'5'?((@'(#5'(8,1'-(/0#/(/0'(.,#$'/'5(26(/0'(4%0'5'(,4(
%&38(/0,4(-3$9'5(,-(/2(/0'(625$3&#4(/2(7#&73&#/'(/0'(12&3$'(#-.(
need to calculate the< radius
of the sphere to calculate the volume and surface area. The
)
(,-70'4?((@'(-''.(/2(7#&73&#/'(/0'(5#.,34(26(/0'(4%0'5'(/2(
4356#7'(#5'#?(
radius of a sphere is ;half
of its diameter. This means that the radius is r = 12 d = 12 (3 85 ) =
G ) G
1 29
) ≈ BGnumber
F2&3$'*((
= π
5 = just
π (A?plug
;AB< )this
?DGA<H< (739,7(,-70'4?(
( ) = 29
= 1.81257#&73&#/'(/0'(12&3$'(#-.(4356#7'(#5'#?((>0'(5#.,34(26(#(4%0'5'(,4(
inches.
WeF can
now
in to the formulas to calculate
2 8
16
)
)
0#&6(26(,/4(.,#$'/'5?((>0,4($'#-4(/0#/(/0'(5#.,34(,4(
B
B
the volume and surface =356#7'(I5'#*((
area.
=I = Gπ5 = Gπ (A?;AB< ) ≈ GA?B;BAD (4J3#5'(,-70'4?(
4A
3
A & <3#) ≈A &24.941505
BD # BD cubic inches.
Volume: V = 34 πr
=
π(1.8125
5 =3 . = $ ) ! = $ ! =
= A?;AB< (,-70'4?((@'(7#-(-2:(E34/(
(
B 2 B % ; " B % ;2 " AC
Surface Area: SA
( = 4πr = 4π(1.8125 ) ≈ 41.28219 square inches.
%&38(/0,4(-3$9'5(,-(/2(/0'(625$3&#4(/2(7#&73&#/'(/0'(12&3$'(#-.(
!"#$%&'(G*(
4356#7'(#5'#?(
Example 15. Find
the volume of the figure
+,-.(/0'(12&3$'(26(/0'(6,835'((
G
G
F2&3$'*((F = π5 ) = π (A?;AB<;() ) ≈ BG?DGA<H< (739,7(,-70'4?(
)
)
F2&3$'(26(#(
=356#7'(I5'#*(( =I = Gπ5 B = Gπ (A?;AB< B ) ≈ GA?B;BAD (4J3#5'(,-70'4?(
=2&,.(:,/0(#(
<(
(
K#/70,-8(L#4'(
(
#-.(>2%(
!"#$%&'(G*(
A<(
F = I0 +,-.(/0'(12&3$'(26(/0'(6,835'((
4;(.&".(#2(%<"(
F2&3$'(26(#(
=.*"(#2(%<"(*#,>0(
=2&,.(:,/0(#(
<(;(<">6<%(#2(%<"(
K#/70,-8(L#4'(
*#,>0(
Solution:This
figure
#-.(>2%(
;(
<(
A)(
(
is a solid with the same shape base and top. The shape of the base and
top is a trapezoid. Thus we will need to remember
the formula for the area of a trapezoid.
A<(
F = I0
4;(.&".(#2(%<"(
=.*"(#2(%<"(*#,>0(
<(;(<">6<%(#2(%<"(
*#,>0(
A)(
(
MAT 142 - Module
15
For this trapezoid, the lengths of the bases are 13 and 8 units. It does not matter which of
these we say is b1 and which is b2 . The height of the trapezoid is 5 units. The height of the
solid is 15 units. We will start by plugging the information about the trapezoidal base into
the formula for the area of a trapezoid. Once we have this area, we will plug that and the
height of the solid into the volume formula.
Area of the trapezoidal base:
A = 12 (b1 + b2 )h = 21 (13 + 8)5 = 52.5 square units.
Volume of trapezoidal solid:
V = Ah = (52.5)(15) = 787.5 cubic units.
Our next formulas will be for finding the volume of a cone or a pyramid. These two
formulas are grouped together since they are very similar. Each is basically 13 times the area
of the base of the solid times the height of the solid. In the case of the cone, the base is a
circle. In the case of the pyramid, we will have a base that is a rectangle. The height in both
cases is the perpendicular distance from the apex to the plane which contains the base.
A pyramid is a solid figure with a polygonal base (in our case a rectangle) and triangular
faces that meet at a common point (the apex). A cone is the surface of a conic solid whose
base is a circle. This is more easily thought of as a pointed ice-cream cone whose top is
circular and level.
Volume of a Rectangular Pyramid
V = 13 lwh
l = the length of the base of the pyramid
w = the width of the base of the pyramid
!"#$"%&'()#%"*(
h = the perpendicular height of the pyramid
5.6"(78(#2(89(
+#,-$"(./0(1-&2.3"(4&".(
Example
!
Volume
of a Cone
!
H2'3%(!24!$!N2-(!!
!
V =! 13 πr2 h
!
C
H = π6 D0
!
r = the radius of theIcircular base of the cone
!
!
π =&(:(%;"(&.0<-*(#2(%;"(3<&3-,.&(=.*"(#2(%;"(3#/"(
the number that is apporximated by 3.141593
>(:(%;"(/-$="&(%;.%(<*(.??&#@<$.%"0(='(ABCDC8EA(
!
;(:(%;"(?"&?"/0<3-,.&(;"<6;%(#2(%;"(3#/"(
h = the perpendicular height of the cone
!
(
!
16.
Find the volume of the figure.
"#$%&'(!)*!
+,-.!/0(!12'3%(!24!/0(!4,536(7!
CD!9%!
!
CE!9%!
!
82'3/,2-*!
8,-9(!/0(!4,536(!0$:!$!9,693'$6!;$:(!$-.!'22<:!',<(!$-!,9(!96($%!
92-(=!/0,:!%3:/!;(!$!92-(7!!>-!26.(6!/2!4,-.!/0(!12'3%(!24!$!92-(=!
?(!-((.!/0(!6$.,3:!24!/0(!9,693'$6!;$:(!$-.!/0(!0(,50/!
@&(6&(-.,93'$6!0(,50/A!24!/0(!92-(7!!B0(!0(,50/!,:!5,1(-!$:!CD!
9(-/,%(/(6:7!!B0(!2/0(6!%($:36(%(-/!24!CE!9(-/,%(/(6:!,:!/0(!
.,$%(/(6!24!/0(!9,693'$6!;$:(7!!F(!/03:!%3:/!9$'93'$/(!/0(!6$.,3:!
16
Geometry
!"#$"%&'()#%"*(
Solution: Since
the figure has a circular base and looks like an ice 5.6"(78(#2(9:(
cream cone, this must be
+#,-$"(./0(1-&2.3"(4&".(
a cone. In order to find the volume of a cone, we need the radius of the circular base and the
!"#$%&'()*(height) of the cone. The height is given as 12 centimeters. The other
height (perpendicular
measurement of+,-.(/0'(12&3$'(24(/0'(4,536'7(
10 centimeters is the diameter of the circular base. We thus must calculate
(
the radius to get r = 21 d = 21 (10) = 5 centimeters. We are now ready to plug into the volume
(
2
of a cone formula.
V = 13 πr2 h = 13 π(5!"#$"%&'()#%"*(
)(12) = 100π cubic centimeters is the exact volume.
(
An approximation
of this volume would be 314.159266 cubic centimeters.
+#,-$"(./0(1-&2.3"(4&".(
5.6"(78(#2(9:(
(
(
(
+,-.(/0'(12&3$'(24(/0'(4,536'7(
(
(
(
(
E(,-(
(
(
(
C(,-(
(
82&3/,2-*(
(
90'(:#;'(24(/0,;(4,536'(,;(#(6'</#-5&'(#-.(/0'(;,.';(24(/0'(4,536'(
AB(,-(
(
#6'(/6,#-5&';=(/03;(/0,;(4,536'(,;(#(6'</#-53&#6(%>6#$,.7((90'(0',50/(
(
?%'6%'-.,<3&#6(0',50/@(,;(AB(,-<0';7((90'(&'-5/0(24(/0'(:#;'(,;(C(
(
,-<0';=(#-.(/0'(D,./0(24(/0'(:#;'(,;(E(,-<0';7((8,-<'(D'(0#1'(#&&(24(
E(,-(
(
/0'(%#6/;(426(/0'(12&3$'(426$3&#=(D'(<#-(F3;/(%&35(,-/2(/0'(12&3$'(
(
24(#(6</#-53&#6(%>6#$,.(426$3&#(/2(5'/(
C(,-(
82&3/,2-*(
A
A
HEB
G90'(:#;'(24(/0,;(4,536'(,;(#(6'</#-5&'(#-.(/0'(;,.';(24(/0'(4,536'(
= &D0 = ?C@?E@?AB@ =
(<3:,<(,-<0';7((J-(#%%62",$#/,2-(24(
Solution: The base of
this
and the sides of the figure are triangles, thus
H figure
H is a rectangle
I
((#6'(/6,#-5&';=(/03;(/0,;(4,536'(,;(#(6'</#-53&#6(%>6#$,.7((90'(0',50/(
/0,;(12&3$'(D23&.(:'(AA)7))))C(<3:,<(,-<0';7(
this figure is a rectangular
pyramid.
The
height
(perpendicular height) is 10 inches. The
?%'6%'-.,<3&#6(0',50/@(,;(AB(,-<0';7((90'(&'-5/0(24(/0'(:#;'(,;(C(
(
length of the base is ,-<0';=(#-.(/0'(D,./0(24(/0'(:#;'(,;(E(,-<0';7((8,-<'(D'(0#1'(#&&(24(
7 inches, and the width of the base is 5 inches. Since we have all of
the parts for K3;/(&,L'(D,/0(#6'#;=(D'(<#-(#..(#-.(;3:/6#</(12&3$';(24(.,44'6'-/(;2&,.;(/2(
the volume
formula, we can just plug into the volume of a rctangular pyramid
/0'(%#6/;(426(/0'(12&3$'(426$3&#=(D'(<#-(F3;/(%&35(,-/2(/0'(12&3$'(
4,-.(/0'(12&3$'(24(#(;2&,.(/0#/(,;(#(<2$:,-#/,2-(24($26'(/0#-(2-'(;2&,.(26(
1
formula to get V = 3 lwh
= 13 (7)(5)(10) = 350
cubic inches. An approximation of this volume
24(#(6</#-53&#6(%>6#$,.(426$3&#(/2(5'/(
3
/0#/(0#1'(2-'(;2&,.(6'$21'.(462$(#-2/0'67(
would be 116.66667
cubic
A inches.
A
HEB
(
G = &D0 = ?C@?E@?AB@ =
(<3:,<(,-<0';7((J-(#%%62",$#/,2-(24(
H
I
(( H
Example 18.!"#$%&'(C*(
Find the
volume
of
the
figure.
/0,;(12&3$'(D23&.(:'(AA)7))))C(<3:,<(,-<0';7(
+,-.(/0'(12&3$'(24(/0'(4,536'7(
(
K3;/(&,L'(D,/0(#6'#;=(D'(<#-(#..(#-.(;3:/6#</(12&3$';(24(.,44'6'-/(;2&,.;(/2(
I(
4,-.(/0'(12&3$'(24(#(;2&,.(/0#/(,;(#(<2$:,-#/,2-(24($26'(/0#-(2-'(;2&,.(26(
E(
!"#$"%&'()#%"*(
/0#/(0#1'(2-'(;2&,.(6'$21'.(462$(#-2/0'67(
A)(
5.6"(78(#2(98(
( +#,-$"(./0(1-&2.3"(4&".(
(
(
!"#$%&'(C*(
!"#$%&'()$)"&*+$)",)$-#$*##.$)/$./$&($%&+0'#$/0)$-",)$)12#$/%$%&+0'#$
Solution: The 82&3/,2-*(
first
thing that we need to do is figure out what type of figure this is. If we
+,-.(/0'(12&3$'(24(/0'(4,536'7(
)"&($&(3$$4%$-#$'/),)#$)"#$(/5&.$67$.#+'##($)/$)"#$'&+")8$-#$+#)$,$%&+0'#$
rotateD/50;#$/%$,$
the solid 90 degrees to the right, we get a figure that looks like this
)",)$5//9($5&9#$)"&($
O15&*.#'$
I(
$
E(
D = <"
$
A)(
$
(
4(:(%;"(.&".(#2(
$
(
%;"(<.*"(#2(%;"(
$
82&3/,2-*(
3',=/0"&(
;(:(%;"(;"=6;%(#2( $
%;"(3',=/0"&(
$
$
$
!"&($5//9($5&9#$,$:15&*.#'$-&)"$)"#$;&..5#$;&((&*+3$$<$+//.$-,1$)/$
This looks like a cylinder
with the middle missing. A good way to think about this figure is
)"&*9$,=/0)$)"&($%&+0'#$&($,$'/55$/%$2,2#'$)/-#5(3$$>#$,'#$)'1&*+$)/$
a roll of paper towels. %&*.$/0)$)"#$?/50;#$/%$)"#$2,2#'$)/-#5(3$$!"#$=#()$-,1$)/$./$)"&($&($
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
)/$%&+0'#$/0)$-",)$)"#$?/50;#$/%$)"#$5,'+#'$:15&*.#'$&($-&)"/0)$)"#$
;&((&*+$2,')3$$>#$:,*$)"#*$%&*.$)"#$?/50;#$/%$)"#$(;,55#'$:15&*.#'$
/'$;&((&*+$2,')3$$@&*,5518$-#$-&55$(0=)',:)$)"#$?/50;#$/%$)"#$(;,55#'$
:15&*.#'$%'/;$)"#$?/50;#$/%$)"#$5,'+#'$:15&*.#'$)/$+#)$)"#$?/50;#$
/%$/0'$:0''#*)$(/5&.3$
$
>#$-&55$(),')$-&)"$)"#$5,'+#'$:15&*.#'3$$!"#$',.&0($/%$)"#$:&':05,'$
AB(,-(
Example 17.!"#$%&'()*(
Find the volume of the figure.
MAT 142 - 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 = πr2 h = π(52 )(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 = πr2 h = π(22 )(16) = 64π
!"#$"%&'()#%"*(
cubic units.+#,-$"(./0(1-&2.3"(4&".(
!"#$"%&'()#%"*(
5.6"(78(#2(9:(
+#,-$"(./0(1-&2.3"(4&".(
5.6"(78(#2(9:(
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
+,-$(#.()/012.34(56(7712.34(%2'3'(-8(3#,95-#,9:(;12.34(<=>#,'(3-,.',<(#,'(
= 400π − 64π+,-$(#.()/012.34(56(7712.34(%2'3'(-8(3#,95-#,9:(;12.34(<=>#,'(3-,.',<(#,'(
= 336π cubic units. This is approximately 1055.57513 cubic units.
3>?(->?(#.9(?4'(,'<>&?2.@(8&#%<(#,'(8-&9'9(>%(?-(8-,$(#.(-%'.(5-"/((+2.9(
3>?(->?(#.9(?4'(,'<>&?2.@(8&#%<(#,'(8-&9'9(>%(?-(8-,$(#.(-%'.(5-"/((+2.9(
?4'(A-&>$'(#.9(<>,8#3'(#,'#(-8(?4'(5-"/(
Example 19.
From an 8.5-inch by 11-inch piece of cardboard, 2-inch square corners are cut
?4'(A-&>$'(#.9(<>,8#3'(#,'#(-8(?4'(5-"/(
(
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'(,'#&&6(4'&%8>&(?-($#D'(?4'(5-"(
problem, it will be really helpful to make the box described above. You
+-,(?42<(%,-5&'$:(2?(C2&&(5'(,'#&&6(4'&%8>&(?-($#D'(?4'(5-"(
9'<3,25'9(#5-A'/((E->(<?#,?(C2?4(#(<?#.9#,9(%2'3'(-8(%#%',/((E->(
start with a standard
piece of paper. You the cut out the dashed square indicated below
9'<3,25'9(#5-A'/((E->(<?#,?(C2?4(#(<?#.9#,9(%2'3'(-8(%#%',/((E->(
?4'(3>?(->?(?4'(9#<4'9(<=>#,'(2.923#?'9(5'&-C(8,-$('#34(3-,.',/((
from each corner. ?4'(3>?(->?(?4'(9#<4'9(<=>#,'(2.923#?'9(5'&-C(8,-$('#34(3-,.',/((
You make sure each side of the square is 2 inches in length.
E->($#D'(<>,'('#34(<29'(-8(?4'(<=>#,'(2<(;(2.34'<(2.(&'.@?4/(
E->($#D'(<>,'('#34(<29'(-8(?4'(<=>#,'(2<(;(2.34'<(2.(&'.@?4/(
(
(
(
((
((
((
((
((
((
(
F4#?(6->(#,'(&'8?(C2?4(2<(?4'(<4#%'(5'&-C/(
What you are left F4#?(6->(#,'(&'8?(C2?4(2<(?4'(<4#%'(5'&-C/(
with
is the shape below.
(
((
((
((
((
((
((
((
( E->(.-C(8-&9(>%(#&-.@(?4'(9#<4'9(&2.'<(?-(3,'#?'(#(5-"/((G4'(5-"(
E->(.-C(8-&9(>%(#&-.@(?4'(9#<4'9(&2.'<(?-(3,'#?'(#(5-"/((G4'(5-"(
?4#?(C'(4#A'(3,'#?'9(2<(#(,'3?#.@>&#,(<-&29/((G42<(5-"(4#<(.-(?-%/((
You now fold up along
the dashed lines to create a box. The box that we have created is a
?4#?(C'(4#A'(3,'#?'9(2<(#(,'3?#.@>&#,(<-&29/((G42<(5-"(4#<(.-(?-%/((
H-?(4#A2.@(#(?-%(C2&&(.-?(#88'3?(?4'(A-&>$'(-8(?4'(5-"/(((
rectangular solid. H-?(4#A2.@(#(?-%(C2&&(.-?(#88'3?(?4'(A-&>$'(-8(?4'(5-"/(((
This
box has no top. Not having a top will not affect the volume of the
(
box.
( F'(-.&6(.''9(?-(9'?',$2.'(?4'(&'.@?4(-8(?4'(5#<'(-8(?4'(5-":(?4'(
F'(-.&6(.''9(?-(9'?',$2.'(?4'(&'.@?4(-8(?4'(5#<'(-8(?4'(5-":(?4'(
C29?4(-8(?4'(5#<'(-8(?4'(5-":(#.9(?4'(4'2@4?(-8(?4'(5-"/((G4'(,'9(
We only need to determine
the length of the base of the box, the width of the base of the
C29?4(-8(?4'(5#<'(-8(?4'(5-":(#.9(?4'(4'2@4?(-8(?4'(5-"/((G4'(,'9(
9#<4'9(&2.'<(,'%,'<'.?<(?4'(&'.@?4(-8(?4'(5#<'(-8(?4'(5-"/((G4'(
box, and the height
of the box. The red dashed lines represent the length of the base of the
9#<4'9(&2.'<(,'%,'<'.?<(?4'(&'.@?4(-8(?4'(5#<'(-8(?4'(5-"/((G4'(
-,2@2.#&(&'.@?4(-8(?4'(%#%',(C#<(77(2.34'</((F'(,'$-A'9(;(2.34'<(
box. The original -,2@2.#&(&'.@?4(-8(?4'(%#%',(C#<(77(2.34'</((F'(,'$-A'9(;(2.34'<(
length of the paper was 11 inches. We removed 2 inches from the top of
8,-$(?4'(?-%(-8(?4'(%#@'(#.9(C'(#&<-(,'$-A'9(;(2.34'<(8,-$(?4'(
the page and we also
removed 2 inches from the bottom of the page. Thus each red dashed
8,-$(?4'(?-%(-8(?4'(%#@'(#.9(C'(#&<-(,'$-A'9(;(2.34'<(8,-$(?4'(
line is 11 − 2 − 2 = 7 inches.
18
Geometry
!"#$"%&'()#%"*(
The green +#,-$"(./0(1-&2.3"(4&".(
dashed lines represent the width of the base of the box.
The original width of the
5.6"(78(#2(9:(
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 8.5 − 2 − 2 = 4.5
+,-$(#.()/012.34(56(7712.34(%2'3'(-8(3#,95-#,9:(;12.34(<=>#,'(3-,.',<(#,'(
inches.
3>?(->?(#.9(?4'(,'<>&?2.@(8&#%<(#,'(8-&9'9(>%(?-(8-,$(#.(-%'.(5-"/((+2.9(
We now need?4'(A-&>$'(#.9(<>,8#3'(#,'#(-8(?4'(5-"/(
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-.*(
Now we are ready +-,(?42<(%,-5&'$:(2?(C2&&(5'(,'#&&6(4'&%8>&(?-($#D'(?4'(5-"(
of calculate the volumes V = lwh = (7)(4.5)(2) = 63 cubic inches.
9'<3,25'9(#5-A'/((E->(<?#,?(C2?4(#(<?#.9#,9(%2'3'(-8(%#%',/((E->(
Now we need to calculate
the surface area of our box. Since there is no top to this box, we
?4'(3>?(->?(?4'(9#<4'9(<=>#,'(2.923#?'9(5'&-C(8,-$('#34(3-,.',/((
can start formula for the surface area of a box. We will then need to subtract off the area of
the top of the box. E->($#D'(<>,'('#34(<29'(-8(?4'(<=>#,'(2<(;(2.34'<(2.(&'.@?4/(
This will give us SA = 2(lw+wh+lh) = 2((7×4.5)+(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 = 109 − 31.5 = 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.
F4#?(6->(#,'(&'8?(C2?4(2<(?4'(<4#%'(5'&-C/(
(
(
!"#$"%&'()#%"*(
(
+#,-$"(./0(1-&2.3"(4&".(
5.6"(78(#2(9:(
(
(
!
(
!
(
!
(
"#$%&'(!)*!
E->(.-C(8-&9(>%(#&-.@(?4'(9#<4'9(&2.'<(?-(3,'#?'(#(5-"/((G4'(5-"(
Example 20. A propane
gas tank consists of a cylinder with a hemisphere at each end. Find
?4#?(C'(4#A'(3,'#?'9(2<(#(,'3?#.@>&#,(<-&29/((G42<(5-"(4#<(.-(?-%/((
+!&,-&$.(!/$0!1$.2!3-.04010!-5!$!36'4.7(,!8419!$!9(%40&9(,(!$1!($39!(.7:!!
the volume of ;4.7!19(!<-'=%(!-5!19(!1$.2!45!19(!-<(,$''!'(./19!40!>?!5((1!$.7!19(!74$%(1(,!
the tank
if the overall length is 20 feet and the diameter of the cylinder is 6 feet.
H-?(4#A2.@(#(?-%(C2&&(.-?(#88'3?(?4'(A-&>$'(-8(?4'(5-"/(((
(
-5!19(!36'4.7(,!40!@!5((1:!!
F'(-.&6(.''9(?-(9'?',$2.'(?4'(&'.@?4(-8(?4'(5#<'(-8(?4'(5-":(?4'(
C29?4(-8(?4'(5#<'(-8(?4'(5-":(#.9(?4'(4'2@4?(-8(?4'(5-"/((G4'(,'9(
!
@!51!
9#<4'9(&2.'<(,'%,'<'.?<(?4'(&'.@?4(-8(?4'(5#<'(-8(?4'(5-"/((G4'(
!
-,2@2.#&(&'.@?4(-8(?4'(%#%',(C#<(77(2.34'</((F'(,'$-A'9(;(2.34'<(
8,-$(?4'(?-%(-8(?4'(%#@'(#.9(C'(#&<-(,'$-A'9(;(2.34'<(8,-$(?4'(
>?!51!
!
!
Solution: We are
told this tank consists of a cylinder (one its side) with a hemisphere at
A-'=14-.*!
each end. A hemisphere
is half of a sphere. To find the volume, we need to find the volume
B(!$,(!1-'7!1940!1$.2!3-.04010!-5!$!36'4.7(,!C-.(!410!047(D!8419!$!
of the cylinder and the
volumes of each hemisphere and then adding them together.
9(%40&9(,(!$1!($39!(.7:!!+!9(%40&9(,(!40!9$'5!-5!$!0&9(,(:!!E-!54.7!
19(!<-'=%(F!8(!.((7!1-!54.7!19(!<-'=%(!-5!19(!36'4.7(,!$.7!19(!
Lets start with the hemisphere sections. The diameter of the circular base of the cylinder
<-'=%(0!-5!($39!9(%40&9(,(!$.7!19(.!$774./!19(%!1-/(19(,:!
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
G(1H0!01$,1!8419!19(!9(%40&9(,(!0(314-.0:!!E9(!74$%(1(,!-5!19(!
volume of the whole sphere,
we multiply it by 12 to find the volume of the hemisphere (half
34,3='$,!I$0(!-5!19(!36'4.7(,!40!4.743$1(7!1-!I(!@!5((1:!!E940!8-='7!
$'0-!I(!19(!74$%(1(,!-5!19(!9(%40&9(,(0!$1!($39!(.7!-5!19(!
of a sphere). We calculate
the radius to be r = 12 d = 12 (6) = 3 feet. Now we can calculate
the volume of the two36'4.7(,:!!B(!.((7!19(!,$74=0!-5!19(!0&9(,(!1-!54.7!410!<-'=%(:!!
hemispheres at the ends of the tank.
J.3(!8(!3$'3='$1(!19(!<-'=%(!-5!19(!89-'(!0&9(,(F!8(!%='14&'6!41!
I6!K!1-!54.7!19(!<-'=%(!-5!19(!9(%40&9(,(!C9$'5!-5!$!0&9(,(D:!!B(!
P-'=%(!-5!$!
A&9(,(!
L
L
3$'3='$1(!19(!,$74=0!1-!I(! , = 7 = C@D = M !5((1:!!N-8!8(!3$.!
>
>
!!
3$'3='$1(!19(!<-'=%(!-5!19(!18-!9(%40&9(,(0!$1!19(!(.70!-5!19(!
1$.2:!
G(51!O(%40&9(,(*!
MAT 142 - Module
19
Left Hemisphere:
The volume of a whole sphere is V = 43 πr3 = 34 π(33 ) = 36π cubic feet. We now multiply this
volume by 21 to find the volume of the hemisphere to get 12 V = 12 (36π) = 18π cubic feet.
Right Hemisphere:
The volume of a whole sphere is V = 43 πr3 = 34 π(33 ) = 36π cubic feet. We now multiply this
volume by 21 to find the volume of the hemisphere to get 12 V = 12 (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
radius of
hemisphere = 3 ft
6 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 = 20−3−3 = 14 feet. We can now calculate the volume of the cylinder.
Cylinder:
V = Ah = πr2 h = π(32 )(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 508.938 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.
20
Geometry
Solution:
Part a:
In order to find the volume of a sphere, we need the radius of the sphere. In this problem,
we are not given the radius. Instead we are given the circumference of a great circle of the
sphere. From this information, we can calculate the radius of the great circle, which is the
radius of the sphere.
Radius of the baseball:
We calculate the radius of the baseball by plugging in the circumference of the great circle of
the baseball into the formula for the circumference of the circle and solve for r (the radius).
C = 2πr
9 = 2πr
9
=r
2π
Now that we have the radius of the baseball, we can calculate the volume, by plugging the
radius into the formula for the volume of a sphere.
4
V = πr3
3
3
4
9
V = π
3
2π
4 × 93 × π
V =
3 × 23 × π 3
2916π
V =
24π 3
243
V = 2 cubic inches
2π
This is approximately 12.3105 cubic inches.
Radius of the softball:
We will go through the same process with the softball to calculate the radius of the softball.
C = 2πr
12 = 2πr
12
=r
2π
6
=r
π
Now that we have the radius of the softball, we can calculate the volume, by plugging the
radius into the formula for the volume of a sphere.
MAT 142 - Module
21
4
V = πr3
3
3
4
6
V = π
3
π
3
4×6 ×π
V =
3 × π3
864π
V =
3π 3
288
V = 2 cubic inches
π
This is approximately 29.1805 cubic inches.
Part b:
In order to find the surface area of a sphere, we need the radius of the sphere. We calculated
the radius for each type of ball in part a. We only need to plug this information in to the
formula for surface area of a sphere.
Surface area of a baseball:
2
9
SA = 4π
2π
4π(9)2
SA = 2 2
2π
324π
SA =
4π 2
81
SA =
square inches
π
This is approximately 25.7831 square inches.
Surface area of a softball:
2
6
SA = 4π
π
4π(6)2
SA =
π2
144π
SA =
π2
144
SA =
square inches
π
This is approximately 45.83662 square inches.
Our final example is an application problem. We will need to be able to use dimensional
analysis, volumes, and common sense in order to be able to answer the question.
22
Geometry
Example 22. Mike Jones bought an older house and wants to put in a new concrete driveway. The driveway will be 30 feet long, 10 feet wide, and 9 inches thick. Concrete (a mixture
of sand, gravel, and cement) is measured by the cubic yard. One sack of dry cement mix
costs $7.30, and it takes four sacks to mix up 1 cubic yard of concrete. How much will it
cost Mike to buy the cement?
Solution: The driveway that is being poured will be a rectangular solid or box. Thus
in order to answer this question, we will first need to find the volume of this box (the
amount of cubic units it will take to fill this box). The problem tells us that concrete is
measured by the cubic yard. This lets us know that those are the units we will want to
calculate with. None of the dimensions of the driveway are given in yards. We will need to
use our dimensional analysis (unit conversion) to convert all of the measurements to yards.
This could be done at a later point, but this is the easiest place to take care of the conversion.
Convert length:
30 feet 1 yard
×
= 10yards
1
3 feet
Convert width:
10 feet 1 yard
10
×
= yards. We are not going to estimate this value since the approximation
1
3 feet
3
with introduce error. As in the finance section, we dont want to round until the end of the
problem or in a place where it is absolutely necessary.
Convert height:
9 inches
1 yard
1
×
= yards = .25yards. Since the decimal representation of this number
1
36 inches
4
is a terminating decimal, we can use this representation in our calculations.
Volume of driveway:
)(.25) = 25
We are now ready to calculate the volume of the driveway. V = lwh = (10)( 10
3
3
cubic yards.
Now, we are told that it takes 4 bags of cement to make one cubic yard of concrete. So
we will now calculate how many bags of cement to buy. Mikes driveway is 25
cubic yards
3
25
and each of those cubic yards requires 4 bags of cement. Thus Mike will need 3 × 4 = 100
3
bags or approximately 33.333 bags. Here is where common sense needs to come in. There is
no store that will sell .333 bags of cement mix. Stores only sell whole bags of cement mix.
Thus we will need to round up to the next whole bag (we cant round down or we will not
have enough cement mix to complete the driveway). This means that Mike will need to buy
34 bags of cement mix. Each of these bags will cost $7.30. This means that Mike will pay
7.30 × 34 = $248.20 for the cement mix for this driveway.
Similar Triangles
We will continue our study of geometry by looking at similar triangles. Two triangles are
similar if their corresponding angles are congruent (or have the same measurement).
MAT 142 - Module
23
!
In the picture above, the corresponding angles are indicated in the two triangles by the same
number of hash marks. In other words, the angle with one hash mark in the smaller triangle
corresponds to the angle with one hash mark in the larger triangle. These angles have the
same measure or are congruent.
There are also corresponding sides in similar triangles. In the triangles above side a corresponds to side d (for example). These sides do not necessarily have the same measure. They
do, however, form a ratio that is the same no matter which pair of corresponding sides the
ratio is made from. Thus we can write the following equation
!"#$"%&'()#%"*(
+,$,-.&(/&,.01-"*(
2.1"(34(#5(67(
a
b
c
= =
d
e & f$
"
= = '
%
#
!
triangle are always
Notice that the sides of one particular
written on top of the fractions
()*+"#'*,&*'*,#'-+%#-')!').#'/&0*+"12&0'*0+&.32#'&0#'&24&5-'40+**#.').'*)/')!'
and the sides of the other triangle are always written on the bottom of the fractions. It does
*,#'!0&"*+).-'&.%'*,#'-+%#-')!'*,#')*,#0'*0+&.32#'&0#'&24&5-'40+**#.').'*,#'
not matter which triangle
is put in which part of the fraction as longs as we are consistent
$)**)6')!'*,#'!0&"*+).-7''8*'%)#-'.)*'6&**#0'4,+",'*0+&.32#'+-'/1*'+.'4,+",'
within a problem. /&0*')!'*,#'!0&"*+).'&-'2).3-'&-'4#'&0#'").-+-*#.*'4+*,+.'&'/0)$2#67'
'
E+6+2&0'F0+&.32#'H&*+)-'
Similar Triangles
Ratios
'
'
a
= eb =& fc= $ = "
d
'
% # !
'
'
%'
!'
'
&'
"'
'
!
'
#'
$'
'
'
It should be noted that
although our triangles are in the same relative position, this is not
8*'-,)12%'$#'.)*#%'*,&*'&2*,)13,')10'*0+&.32#-'&0#'+.'*,#'-&6#'0#2&*+9#'
needed for triangles /)-+*+).:'*,+-'+-'.)*'.##%#%'!)0'*0+&.32#-'*)'$#'-+6+2&07'';.#')!'*,#'*0+&.32#-'
to be similar. One of the triangles can be rotated or reflected.
"&.'$#'0)*&*#%')0'0#!2#"*#%7'
'
Please note that pictures below are not drawn to scale.
<2#&-#'.)*#'*,&*'/+"*10#-'$#2)4'&0#'.)*'%0&4.'*)'-"&2#7'
'
Example 23. Given=>&6/2#'?@'
that the triangles are similar, find the lengths of the missing sides.
A+9#.'*,&*'*,#'*0+&.32#-'&0#'-+6+2&0:'!+.%'*,#'2#.3*,-')!'*,#'6+--+.3'-+%#-7'
>'
?C'
?CC'
BC'
''
D'
5'
'
'
Solution: There is one' side missing in the triangle on the left. This side is labeled x. There
is also one side missingE)21*+).@'
from the triangle on the right. This side is labeled y. The side x in
the triangle on the left corresponds
to the side labeled 10 in the triangle on the right. We
F,#0#'+-').#'-+%#'6+--+.3'+.'*,#'*0+&.32#').'*,#'2#!*7''F,+-'-+%#'+-'
2&$#2#%'>7''F,#0#'+-'&2-)').#'-+%#'6+--+.3'!0)6'*,#'*0+&.32#').'*,#'
0+3,*7''F,+-'-+%#'+-'2&$#2#%'57''F,#'-+%#'>'+.'*,#'*0+&.32#').'*,#'2#!*'
")00#-/).%-'*)'*,#'-+%#'2&$#2#%'?C'+.'*,#'*0+&.32#').'*,#'0+3,*7''G#'
24
Geometry
know this because these sides connect the angle with one hash mark to the angle with three
hash marks in each of the triangles. The side 100 in the triangle on the left corresponds to
the side labeled 8 in the triangle on the right. Finally the side 90 in the triangle on the left
corresponds to the side labeled y in the triangle on the right. We will need this information
to find values for x and y.
Find x :
We will start by finding the value for x. We will need to form a ratio with the side labeled
x and is corresponding side in the triangle on the right. We will also need a pair of corresponding side both of whose measurements we have and make a ratio out of those. Once
we make an equation using these ratios, we will just need to solve the equation for x. I am
going to choose to place the sides of the triangle on the left on the top of each of the ratios.
x
100
=
10
8
8x = 10(100)
8x = 1000
x = 125units
Find y:
We will continue by finding the value for y. We will need to form a ratio with the side
labeled y and is corresponding side in the triangle on the right. We will also need a pair
of corresponding side both of whose measurements we have and make a ratio out of those.
Although we know the measurements of all the other sides of the triangle, it is best to avoid
using a side we just calculate to make further calculations if possible. The reason for this
is that if we have made a mistake in our previous calculation, we would end up with an
incorrect answer here as well. Once we make an equation using these ratios, we will just
need to solve the equation for y. I am going to choose to place the sides of the triangle on
the left on the top of each of the ratios.
!"#$"%&'()#%"*(
+,$,-.&(/&,.01-"*(
2.1"(34(#5(67(
+,,
90/, = 100
"= 0
y
8
0./,- = " .+,,- !
8(90) = 100y
*(, = +,,"
720*)(=#$%&'
100y
! = !"
7.2units = y
Example
!
!
24.
Given that the triangles are similar, find the lengths of the missing
1234567!(8!
9%:7$!&;3&!&;7!&<%3$=67'!3<7!'%4%63<>!?%$@!&;7!67$=&;'!A?!&;7!4%''%$=!'%@7')!
DC)
D!
"!
B)(!
()0!
!!!!!!!
2!
B*)(C!
!
!
!
EA6#&%A$8!
F#<!?%<'&!GAH!;7<7!%'!&A!@7&7<4%$7!I;%J;!'%@7'!%$!&;7!&<%3$=67!A$!
&;7!67?&!JA<<7'5A$@!&A!I;%J;!'%@7'!%$!&;7!&<%3$=67!A$!&;7!<%=;&)!!
K;7!'%@7!%$!&;7!&<%3$=67!A$!&;7!67?&!63H767@!"!GA%$'!&;7!3$=67!I%&;!
A$7!;3';!43<L!3$@!&;7!3$=67!I%&;!&IA!;3'!43<L')!!K;7!'%@7!%$!&;7!
sides.
MAT 142 - Module
25
Solution: Our first job here is to determine which sides in the triangle on the left correspond
to which sides in the triangle on the right. The side in the triangle on the left labeled y
joins the angle with one hash mark and the angle with two has marks. The side in the triangle on the right that does this is the side labeled 37.25. Thus these are corresponding sides.
In a similar manner, we can see that the side labeled 3.2 in the triangle on the left corresponds to the side labeled x on the triangle on the right. Finally the side labeled 2.8 in the
triangle on the left corresponds to the side labeled 45 in the triangle on the right.
Find x :
We will start by finding the value for x. We will need to form a ratio with the side labeled
x and is corresponding side in the triangle on the right. We will also need a pair of corresponding side both of whose measurements we have and make a ratio out of those. Once
we make an equation using these ratios, we will just need to solve the equation for x. I am
going to choose to place the sides of the triangle on the left on the top of each of the ratios.
3.2
2.8
=
x
45
3.2(45) = x(2.8)
144 = 2.8x
144
units = x
2.8
This is approximately 51.42857 units.
Find y:
We will continue by finding the value for y. We will need to form a ratio with the side
labeled y and is corresponding side in the triangle on the right. We will also need a pair
of corresponding side both of whose measurements we have and make a ratio out of those.
Although we know the measurements of all the other sides of the triangle, it is best to avoid
using a side we just calculate to make further calculations if possible. The reason for this
is that if we have made a mistake in our previous calculation, we would end up with an
incorrect answer here as well. Once we make an equation using these ratios, we will just
need to solve the equation for y. I am going to choose to place the sides of the triangle on
the left on the top of each of the ratios.
y
2.8
=
37.25
45
y(45) = 37.25(2.8)
45y = 104.3
104.3
y=
units
45
This is approximately 3.21778 units.
Our final example is an application of similar triangles.
26
Geometry
Example 25. A 3.6-foot tall child casts a shadow of 4.7 feet at the same instant that a
telephone pole casts a shadow of 15 feet. How tall is the telephone pole?
Solution: For this problem, it helps!"#$"%&'()#%"*(
to have a picture.
+,$,-.&(/&,.01-"*(
/"01$'
2*3',--.'
2.1"(34(#5(67(
6%1-'
!"#$"%&'()#%"*(
+,-.%(/&,01-2"(/&,-#1#$"%&'(
30-"(45(#6(78(
!"#$%&'(')*+',--.''
!"#$%&'('45',--.'
+9!:/(/+9;)!<=(/+9!>)>?=/+@(
!
When we look"#$%&'()%*+!
at the' picture above, we see that we have two triangles that are similar. We
7"-8'&-'1%%9'#.'."-'60/.:;-'#<%=->'&-'!--'."#.'&-'"#=-'.&%'
,-'%.!&/012%'(34!'5(*!*%&'(/36!7/8!*5/829!#%!:#2%!'/!9/!'5%!-/22/;(34+!
are looking for the
height
of the pole. The pole side of the triangle on the right corresponds
.;0#8?1-!'."#.'#;-'!0@01#;*''7-'#;-'1%%908?',%;'."-'"-0?".'%,'."-'
•
<:2&82:'%!'5%!2%34'5*!/-!*(9%*!:39!:342%*!/-!:!.(45'!'.(:342%!8*(34!
to the side of the child
side of the triangle on the left. The two shadow sides of the triangles
6%1-*''A"-'6%1-'!0$-'%,'."-'.;0#8?1-'%8'."-';0?".'/%;;-!6%8$!'.%'."-'
'.(4/3/0%'.(&!.:'(/*=!
also correspond to each
other. Thus we can write an equation of ratios to find the height of
!0$-'%,'."-'/"01$'!0$-'%,'."-'.;0#8?1-'%8'."-'1-,.*''A"-'.&%'!"#$%&'
• >/2)%!;/.9!1./#2%0*!(3)/2)(34!.(45'!'.(:342%*!:39!'.(4/3/0%'.(&!.:'(/*=!
the pole.
!0$-!'%,'."-'.;0#8?1-!'#1!%'/%;;-!6%8$'.%'-#/"'%."-;*''A":!'&-'/#8'
!
&;0.-'#8'-B:#.0%8'%,';#.0%!'.%',08$'."-'"-0?".'%,'."-'6%1-*'
?/&:#82:.7+!
2*3
)*+
,*!7/8!.%:96!7/8!*5/829!#%!2//@(34!-/.!'5%!-/22/;(34!)/&:#82:.7!;/.9*!:39!
=
3.6
6%1454.7
'5%(.!9%-(3('(/3*+!
=
Dpole
2*3CD45C =15
6%1-D )*+ C
• 571/'%38*%!
'
5)
=
)
*
+
6%1• :9$:&%3'!*(9%!
3.6(15) = pole(4.7)
5)
• /11/*('%!*(9%!
6%154 ==4.7pole
!
)*+
54
A/.082:*+!
A":!'."-'6%1-'0!'#66;%E0@#.-1F'44*)GH23',--.'"0?"*'
= pole
'
B/8!*5/829!#%!2//@(34!-/.!'5%!-/22/;(34!-/.082:*!:*!7/8!.%:9+!
4.7
.:'(/!-/.!*(3C,D!;5%.%!,!(*!:!3/3E.(45'!:342%!(3!.(45'!'.(:342%!
Thus the pole is •approximately
11.48936 feet high.
• .:'(/!-/.!&/*C,D!;5%.%!,!(*!:!3/3E.(45'!:342%!(3!.(45'!'.(:342%!
• .:'(/!-/.!':3C,D!;5%.%!,!(*!:!3/3E.(45'!:342%!(3!.(45'!'.(:342%!
!
Right Triangle Trigonometry
!
!
We will complete
our study with a further study of right triangles. We will look at trigonoF%!;(22!&/012%'%!/8.!*'897!;('5!:!-8.'5%.!*'897!/-!.(45'!'.(:342%*=!!F%!;(22!
metric value as
defined by ratios of the sides of a right triangle.
2//@!:'!'.(4/3/0%'.(&!):28%!:*!9%-(3%9!#7!.:'(/*!/-!'5%!*(9%*!/-!:!.(45'!
'.(:342%=!
/11/*('%!
571/'%38*%!
:9$:&%3'!
,!
!
G3!'5%!-(48.%!:#/)%6!7/8!&:3!*%%!'5%!*(9%*!/-!:!.(45'!'.(:342%!2:#%2%9=!!H5%!
571/'%38*%!
571/'%38*%
In the figure *(9%!2:#%2%9!
above, you
can see!(*!:2;:7*!/11/*('%!'5%!.(45'!:342%!/-!'5%!.(45'!
the sides of a right triangle labeled. The side labeled
'.(:342%=!!H5%!3:0%*!/-!'5%!/'5%.!';/!*(9%*!/-!'5%!.(45'!'.(:342%!:.%!
hypotenuse is
always opposite the right angle of the right triangle. The names of the other
9%'%.0(3%9!#7!'5%!:342%!'5:'!(*!#%(34!9(*&8**%*=!!G3!/8.!&:*%6!;%!;(22!#%!
two sides of the right triangle are determined by the angle that is being discusses. In our
9(*&8*(34!'5%!*(9%*!(3!'%.0*!/-!'5%!:342%!2:#%2%9!,=!!H5%!:342%!,!(*!-/.0!#7!
case, we will be
discusing the sides in terms of the angle labeled A. The angle A is form by
'5%!571/'%38*%!/-!'5%!.(45'!'.(:342%!:39!'5%!*(9%!/-!'5%!.(45'!'.(:342%!'5:'!
the hypotenuse of the right triangle and the side of the right triangle that is called adjacent.
The adjacent side will always make up part of the angle that is being discussed and not be
the hypotenuse. The side of the right triangle that does not form part of angle A is called
the opposite side. The opposite side will never form part of the angle being discussed.
%()%$'*+#
!"#$%&&'(#%()%$'*+,##-.'#%()%$'*+#"!('#/!&&#%&/%0"#1%2'#34#4%5+#67#+.'#%*8&'#
+.%+#!"#9'!*8#(!"$3""'(#%*(#*6+#9'#+.'#.046+'*3"',##-.'#"!('#67#+.'#5!8.+#
+5!%*8&'#+.%+#(6'"#*6+#7651#4%5+#67#%*8&'#:#!"#$%&&'(#+.'#6446"!+'#"!(',##-.'#
6446"!+'#
6446"!+'#"!('#/!&&#*';'5#7651#4%5+#67#+.'#%*8&'#9'!*8#(!"$3""'(,#
MAT 142 - Module
27
#
The trigonometric
function values of a particular value can be as the ratio of a particular
-.'#+5!86*61'+5!$#73*$+!6*#;%&3'"#67#%#4%5+!$3&%5#;%&3'#$%*#9'#%"#+.'#5%+!6#67#
pair of sides of%#4%5+!$3&%5#4%!5#67#"!('"#67#%#5!8.+#+5!%*8&'#$6*+%!*!*8#%*#%*8&'#67#+.%+#
a right triangle containing an angle of that measure. We will look at three
1'%"35',##<'#/!&&#&662#%+#+.5''#4%5+!$3&%5#+5!86*61'+5!$#5%+!6*",###
particular trigonometric
ratios.
#
-5!86*61'+5!$#F%+!6"#
#
Trigonometric Ratios
#
6446"!+'
#
"!*H: G=
= opposite
sin(A)
hypotenuse
.046+'*3"'
#
adjacent
cos(A) = hypotenuse
#
%()%$'*+
#
$6"H: G = opposite
tan(A) =.046+'*3"'
adjacent
#
#
6446"!+'
+%*H: G =
#
%()%$'*+
#
sin = shortened form of the sine function
#
cos*,1(8(*.#&%"1"9(5#&$(#5(*,1"(5:1;%,#1(
= shortened form of the cosine function
#
;#*(8(*.#&%"1"9(5#&$(#5(;#*,1"(5:1;%,#1(
%01(8(*.#&%"1"9(5#&$(#5(%01-"1%(5:1;%,#1(
#
tan<(8(%."(01-2"(=02:"(
= shortened from of the tangent function
#
#
A = the angle value
#
<'#/!&&#3"'#+.'"'#5%+!6"#+6#%*"/'5#=3'"+!6*"#%963+#+5!%*8&'"#9'&6/#%*(#+.'*#
We will use these
ratios to answer questions about triangles below and then we will go
/'#/!&&#86#+.5638.#%#$634&'#67#%44&!$%+!6*#4569&'1",#
through a couple
of application problems.
#
Example 26.>?%14&'#@A#
Use Trigonometric ratios to find the unknown sides and angles in the right
triangles below: B"'#-5!86*61'+5!$#5%+!6"#+6#7!*(#+.'#3*2*6/*#"!('"#%*(#%*8&'"#!*#+.'#
5!8.+#+5!%*8&'"#9'&6/A#
E#
?#
CD°#
@D#
0#
#
Solution: We are being asked to find values for x, y, and B. We will do the angle B first.
Angle B :
We can find the measure of angle B without using any trigonometric ratios. What we need
to remember to find this value is that the sum of the three angles of a triangle will always
add up to 180 degrees. It does not matter the size or shape of the triangle. The sum of
the three angles will always be 180 degrees. We know that one angle is a right angle. Its
measure is 90 degrees. The measure of the other angle is given to be 60 degrees. Thus we
just need to calculate
90 + 60 + B = 180
150 + B = 180
B = 30
Thus the measure of angle B is 30 degrees.
28
Geometry
Side x :
We will now work to find the length of side x. We need to start by determining which angle
we are going to use for our problem. As in the past, it is best to use an angle that is given.
Thus we will be using the angle labeled 60 degrees. Our next step is to determine which side
x is relative to the angle labeled 60 degrees. Since the side labeled x is opposite the right
angle, it is the hypotenuse. There are two trigonometric ratios that include the hypotenuse.
Thus, we need to determine which one to use. This will be determined by the other side of
the triangle whose measure we know. This is the side labeled 10. Since this side is not one of
the sides of the triangle that makes up the angle labeled 60 degrees, it is the opposite side.
This means that we need to use the trigonometric ratio that has both the hypotenuse and
the opposite side. That ratio is the sine ratio. We will plug into that equation and solve for
x.
opposite
hypotenuse
10
sin(60) =
x
sin(60)
10
=
1
x
x sin(60) = 1(10)
10
x=
sin(60)
sin(A) =
Through out this solution, we have left sin(60) in this form. This saves us from needing to
round until the end of the problem. sin(60) is just a number that at the end of the problem
can be calculated by our calculator. Our answer is approximately x ≈ 11.54701.
Side y:
We will finish by finding the length of side y. As in the previous par of the problem, it is best
to use the angle labeled 60 degrees. Our next step is to determine which side y is relative to
the angle labeled 60 degrees. Since the side labeled y is forms part of the angle labeled 60
degrees, it is the adjacent side. There are two trigonometric ratios that include the adjacent
side. Since the other side that is given is the side labeled 10 and this side is the opposite
side (see explanation above), we will need to use the trigonometric ratio that has both the
opposite side and the adjacent side. That ratio is the tangent ratio. We will plug into that
equation and solve for y.
MAT 142 - Module
29
opposite
adjacent
10
tan(60) =
y
tan(60)
10
=
1
y
y tan(60) = 1(10)
!"#$"%&'()#%"*(
10
y=
tan(60)
tan(A) =
+,-.%(/&,01-2"(/&,-#1#$"%&'(
30-"(45(#6(75(
!"#$%&"'$%('(")*'*$+%()$,-'./'"01/'+/2('(0,3456'),'(")*'2$#78''!")*'
*01/*'%*'2#$7',//9),&'($'#$%,9'%,()+'("/'/,9'$2'("/':#$;+/78''
Through out this solution,
we have left tan(60) in this form. This saves us from needing to
(0,3456')*'<%*('0',%7;/#'("0('0('("/'/,9'$2'("/':#$;+/7'=0,';/'
round until the end of =0+=%+0(/9';>'$%#'=0+=%+0($#8''?%#'0,*./#')*'0::#$@)70(/+>'
the problem. tan(60) is just a number that at the
>'A'end of the problem
B8CCDB5D8'
can be calculated by our
calculator. Our answer is approximately y ≈ 5.773503.
'
'
Example 27. Use Trigonometric
ratios to find the unknown sides and angles in the right
E@07:+/'FG'
triangles below:
H*/'!#)&$,$7/(#)='#0()$*'($'2),9'("/'%,I,$.,'*)9/*'0,9'0,&+/*'),'("/'
#)&"('(#)0,&+/*';/+$.G'
KB°'
>'
@'
J'
D'
'
'
Solution: As we L$+%()$,G'
solve this problem, we will leave out the explanations of how we determine
the names of the sides J*'./'*$+1/'(")*':#$;+/7-'./'.)++'+/01/'$%('("/'/@:+0,0()$,*'$2'"$.'
of the triangle.
./'9/(/#7),/'("/',07/*'$2'("/'*)9/*'$2'("/'(#)0,&+/8'
Angle A:
'
As with the previous problem,
J,&+/'JG' the sum of the angles of a triangle is 180 degrees. Thus we
J*'.)("'("/':#/1)$%*':#$;+/7-'("/'*%7'$2'("/'0,&+/*'$2'0'(#)0,&+/')*'
calculate
MN5'9/&#//*8''!"%*'./'=0+=%+0(/'''
KB=
+ J180
= MN5
90 + 45O5++ A
MDB + J = MN5
'
135 + A = 180
J = KB
A=
45
!"%*'("/'7/0*%#/'$2'0,&+/'
J')*'KB'9/&#//*8'
Thus the measure of
'
L)9/'@G'
L)9/'@A
'2$#7*':0#('$2'("/'0,&+/'("0(')*'+0;/+/9'($';/'KB'9/&#//*-'
angle
is 45 degrees.
("%*'(")*')*'("/'09<0=/,('*)9/8''P/'0#/'0+*$'&)1/,'("/'7/0*%#/'$2'
("/'*)9/'$::$*)(/'("/'0,&+/'($';/'D8''!"%*'./'.)++'.0,('($'%*/'("/'
(0,&/,('#0()$8'
Side x :
Side x forms part of the angle that is labeled to be 45 degrees, thus this is the adjacent side.
We are also given the measure of the side opposite the angle to be 3. Thus we will want to
use the tangent ratio.
30
Geometry
3
+,-.%(/&,01-2"(/&,-#1#$"%&'(
tan(45) =
!"#$"%&'()#%"*(
x
30-"(45(#6(78(
3
tan(45)
=
1 "#! = (x
$%&'
*
x tan(45) = 1(3)
$%&' "#! (
3= * +
x= )
tan(45)
* $%&'
"#! = )'(!
* =
Our answer is x = 3.
(
$%&' "#!
,-.+%&/01.+2/+*+3+(4+
We did not need to use
trigonometric ratios to find x. We could have used the fact that our
+
triangle has two angles
that are equal. Such a triangle is an isosceles
triangle. We should
51+626+&7$+&116+$7+-/1+$.287&791$.2:+.%$27/+$7+;2&6+
*4++51+:7-<6+
recall that the sides =%>1+-/16+$=1+;%:$+$=%$+7-.+$.2%&8<1+=%/+$07+%&8<1/+$=%$+%.1+1?-%<4++
of an isosceles triangle opposite the equal angles are equal in length.
Thus since one of the@-:=+%+$.2%&8<1+2/+%&+2/7/:1<1/+$.2%&8<14++51+/=7-<6+.1:%<<+$=%$+$=1+
side was length 3, the side labeled x is also of length 3.
/261/+7;+%&+2/7/:1<1/+$.2%&8<1+7AA7/2$1+$=1+1?-%<+%&8<1/+%.1+1?-%<+2&+
<1&8$=4++B=-/+/2&:1+7&1+7;+$=1+/261+0%/+<1&8$=+(C+$=1+/261+<%D1<16+*+
Side y:
2/+%</7+7;+<1&8$=+(4+
Side y is opposite the right angle of the triangle and thus is the hypotenuse.
+
given opposite side to
be 3. Thus to find y, we will need to use the sine ratio.
@261+EF+
@261+E+2/+7AA7/2$1+$=1+.28=$+%&8<1+7;+$=1+$.2%&8<1+%&6+$=-/+2/+$=1+
=EA7$1&-/14++51+%</7+=%>1+82>1&+7AA7/2$1+/261+$7+D1+(4++B=-/+$7+
3
sin(45) =
;2&6+EC+01+02<<+&116+$7+-/1+$=1+/2&1+.%$274+
y
(
/2&' "#! =
We also have
sin(45) E3
=
/2&'
1 "#! = (y
) = 1(3)
E
y sin(45)
E /2&' "#! = )'(!
3
(
y=
E =sin(45)
/2&' "#!
Our answer is approximately
y ≈ 4.24264. E+3+"4G"GH"4+
,-.+%&/01.+2/+%AA.7*29%$1<E+
+
Example 28.
Use Trigonometric ratios to find the unknown sides and angles in the right
I*%9A<1+(F+
triangles below:J/1+B.287&791$.2:+.%$27/+$7+;2&6+$=1+-&K&70&+/261/+%&6+%&8<1/+2&+$=1+
.28=$+$.2%&8<1/+D1<70F+
%+
:+
M+
#4H+
#(4"L+
+
Solution:
Angle B :
As with the previous problem, the sum of the angles of a triangle is 180 degrees. Thus we
calculate
90 + 53.4 + B = 180
143.4 + B = 180
B = 36.6
MAT 142 - Module
31
Thus the measure of angle B is 36.6 degrees.
Side c:
We will use the given angle labeled 53.4 degrees as our angle for solving this problem. Side
c forms part of the triangle that is opposite the right angle. Thus it is the hypotenuse. We
are also given the measure of the side adjacent to the angle we are using to be 5.6. Thus we
will want to use the cosine ratio.
5.6
cos(53.4) =
c
cos(53.4)
5.6
=
1
c
c cos(53.4) = 1(5.6)
5.6
c=
cos(53.4)
Our answer is approximately c ≈ 9.39243.
Side a:
We will use the given angle labeled 53.4 degrees as our angle for solving this problem. Side
a does not form the angle we are using. Thus it is the opposite side. We are also given the
measure of the side adjacent to the angle we are using to be 5.6. Thus we will want to use
the tangent ratio.
a
5.6
tan(53.4)
a
=
1
5.6
5.6 tan(53.4) = 1(a)
tan(53.4) =
5.6 tan(53.4) = a
!"#$"%&'()#%"*(
Our answer is+,-.%(/&,01-2"(/&,-#1#$"%&'(
approximately a ≈ 7.54041.
30-"(45(#6(47(
Example 29.
Use Trigonometric ratios to find the unknown sides and angles in the right
!"#$%&'()*(
triangles below:+,'(-./0121$'3./4(.#3/1,(31(5/26(37'(82921:2(,/6',(#26(#20&',(/2(37'(
./073(3./#20&',(;'&1:*(
?=>(
4(
A(
<=>(
@(
(
Solution: In this
problem, we are not given any angle to use. Instead we will need to change
B1&83/12*(
the labels of our sidesC2(37/,(%.1;&'$D(:'(#.'(213(0/E'2(#2F(#20&'(31(8,'=((C2,3'#6(:'(:/&&(
as we solve each of the angles in turn. We will start by finding side c.
2''6(31(47#20'(37'(&#;'&,(15(18.(,/6',(#,(:'(,1&E'('#47(15(37'(
Since this is a right triangle,
we can use the Pythagorean theorem to find the length of c.
#20&',(/2(38.2=((G'(:/&&(,3#.3(;F(5/26/20(,/6'(4=((B/24'(37/,(/,(#(./073(
Side c:
3./#20&'D(:'(4#2(8,'(37'(HF37#01.'#2(37'1.'$(31(5/26(37'(&'2037(15(
The legs (a and b) are
4=( given to be 7.0 and 8.0. It does not matter which we label a and
which we label b.
(
B/6'(4*(
The approximate length
of side c is 10.63015.
-7'(&'0,(I#(#26(;J(#.'(0/E'2(31(;'(?=>(#26(<=>=((C3(61',(213($#33'.(
:7/47(:'(&#;'&(#(#26(:7/47(:'(&#;'&(;=(
#O + ;O = 4O
?=> O + <=> O = 4 O
)N + M) = 4 O
KKL = 4 O
(
32
Geometry
Angle A:
As we solve for angle A, we need to label the sides whose measures are given relative to
angle A. The side labeled 8.0 forms part of the angle A. Thus it is the adjacent side. The
side labeled 7.0 does not form any part of the angle A. Thus it is the opposite side. The
trigonometric ratio that includes both the adjacent and opposite sides is the tangent ratio.
We will fill in the information to that equation and solve for A.
We need to know how to solve for A in this equation. As in our section on exponential
functions and their inverses, there is an inverse function (a functions that undoes) for the
tangent function. On the calculator it is labeled tan−1 . Thus we can finally solve for A by
calculating
Thus the measure of angle A is approximately A ≈ 41.18593 degrees.
Angle B :
As we solve for angle B, we need to relabel the sides whose measures are given relative to
angle B. The side labeled 7.0 forms part of the angle B. Thus it is the adjacent side. The
side labeled 8.0 does not form any part of the angle B. Thus it is the opposite side. The
trigonometric ratio that includes both the adjacent and opposite sides is the tangent ratio.
We will fill in the information to that equation and solve for A.
We will once again use the inverse function of the tangent function. On the calculator it is
labeled tan−1 . Thus we can finally solve for B by calculating
!"#$"%&'()#%"*(
Thus the measure
of angle B is approximately B ≈ 48.81407 degrees.
+,-.%(/&,01-2"(/&,-#1#$"%&'(
30-"(45(#6(47(
Example 30.
Use Trigonometric ratios to find the unknown sides and angles in the right
!"#$%&'()*(
triangles below:+,'(-./0121$'3./4(.#3/1,(31(5/26(37'(82921:2(,/6',(#26(#20&',(/2(37'(
./073(3./#20&',(;'&1:*(
#( @A<BC(
<=>(
?(
;(
(
Solution: (
D1&83/12*(
Angle A:
?*(
As with the previous?20&'(
problem,
the sum of the angles of a triangle is 180 degrees. Thus we
?,(:/37(37'(%.'E/18,(%.1;&'$F(37'(,8$(15(37'(#20&',(15(#(3./#20&'(/,(
calculate
G=H(6'0.'',<((-78,(:'(4#&48&#3'(((
BH + @A<B + ? = G=H
GAA<B + ? = G=H
(
90 + 43.9 + A = 180
? = @><G
133.9?+
A = 180
-78,(37'($'#,8.'(15(#20&'(
(/,(@><G(6'0.'',<(
(
A = 46.1
D/6'(#*(
Thus the measure ofI'(:/&&(8,'(37'(0/E'2(#20&'(&#;'&'6(@A<B(6'0.'',(#,(18.(#20&'(51.(
angle A is 46.1 degrees.
,1&E/20(37/,(%.1;&'$<((D/6'(#(51.$,(%#.3(15(37'(#20&'(&#;'&'6(@A<B(
6'0.'',<((-78,(/3(/,(37'(#6J#4'23(,/6'<((I'(#.'(#&,1(0/E'2(37'(
Side a:
We will use the given$'#,8.'(15(37'(7K%13'28,'(31(;'(<=><((-78,(:'(:/&&(:#23(31(8,'(37'(
angle labeled 43.9 degrees as our angle for solving this problem. Side a
41,/2'(.#3/1<(
forms part of the angle
labeled 43.9 degrees. Thus it is the adjacent side. We are also given
#
= will want to use the cosine ratio.
@A<BLwe
the measure of the hypotenuse to be .86.41,M
Thus
<=>
41,M @A<BL
#
=
(
G
<=>
<=> 41,M @A<BL = GM# L
<=> 41,M @A<BL = #
N8.(#2,:'.(/,(#%%.1"/$#3'&K(#(O(H<>GB>P@<(
(
MAT 142 - Module
33
a
.86
a
cos(43.9)
=
1
.86
.86 cos(43.9) = 1(a)
cos(43.9) =
.86 cos(43.9) = a
Our answer is approximately a ≈ 0.619674.
Side b:
We will use the given angle labeled 43.9 degrees as our angle for solving this problem. Side
b does not form the angle we are using. Thus it is the opposite side. We are also given the
measure of the hypotenuse to be .86. Thus we will want to use the sine ratio.
!"#$"%&'()#%"*(
+,-.%(/&,01-2"(/&,-#1#$"%&'(
b
.86
!
sin(43.9)
'()* %&$#" =b
= $+,
1'()* %&$#" .86!
=
.86 sin(43.9)
- = 1(b)
$+,
30-"(45(#6(47(
sin(43.9) =
.
$+, ⋅ '()* %&
.86 sin(43.9)
=$#b" = -*! "
$+, ⋅ '()* %&$#" = !
Our answer is approximately
b ≈ 0.596326.
/01.2)'341.('.255167(8294:;.
!.<.=$>#,&?,$.
.
@4.3(::.A()('B.!;.:66C()D.29.'684.255:(E29(6).516!:48'.A61.601.1(DB9.91(2)D:4.
We will finish by
looking at some application problems for our right triangle trigonometric
91(D6)68491(E.129(6'$.
ratios.
.
Example 31. F7285:4.,G.
A support cable runs from the top of the telephone pole to a point on the
H.'055619.E2!:4.10)'.A168.9B4.965.6A.9B4.94:45B6)4.56:4.96.2.56()9.6).9B4.
ground 47.2 feet from
its base. If the cable makes an angle of 28.7◦ with the ground, find
D160)I.%J$?.A449.A168.(9'.!2'4$..KA.9B4.E2!:4.82C4'.2).2)D:4.6A.?+$J°.
(rounding to the nearest
tenth of a foot)
3(9B.9B4.D160)IL.A()I.*160)I()D.96.9B4.)4214'9.94)9B.6A.2.A669".
56:4.
?+$J°
%J$?.A9.
.
2$. 9B4.B4(DB9.6A.9B4.56:4.
a. the height of the
pole
!$. 9B4.:4)D9B.6A.9B4.E2!:4.
b. the length of the
. cable
M6:09(6)G.
Solution:
NB4.5(E9014.2!6O4.'B63'.9B29.34.B2O4.2.1(DB9.91(2)D:4.'(9029(6)$...
The picture above shows
. that we have a right triangle situation.
Part a:
P219.2G.
The pole is opposite the
angle of 28.7 degrees that is given. The other side of the triangle
NB4.56:4.('.6556'(94.9B4.2)D:4.6A.?+$J.I4D144'.9B29.('.D(O4)$..NB4.
that we know to be 47.269B41.'(I4.6A.9B4.91(2)D:4.9B29.34.C)63.96.!4.%J$?.A449.A618'.5219.
feet forms part of the angle of 28.7 degrees. Thus it is the adjacent
6A.9B4.2)D:4.6A.?+$J.I4D144'$..NB0'.(9.('.9B4.2IQ2E4)9.'([email protected](::.
side. We will be able to
use the tangent ratio to solve this problem since it includes both
!4.2!:4.96.0'4.9B4.92)D4)9.129(6.96.'6:O4.9B('.516!:48.'()E4.(9.
the opposite and adjacent sides.
()E:0I4'.!69B.9B4.6556'(94.2)I.2IQ2E4)9.'(I4'$.
34
Geometry
opposite
adjacent
pole
tan(28.7) =
47.2
pole
tan(28.7)
=
1
47.2
47.2 tan(28.7) = 1(pole)
tan(A) =
47.2 tan(28.7) = pole
Thus the pole is approximately 25.8 feet tall.
Part b:
The cable is opposite the right angle of triangle and thus is the hypotenuse. We still know
that the adjacent side is 47.2 feet. We will be able to use the cosine ratio to solve this
problem since it includes both the hypotenuse and adjacent sides.
adjacent
hypotenuse
47.2
cos(28.7) =
cable
47.2
cos(28.7)
=
1
cable
(cable) cos(28.7) = 1(47.2)
47.2
cable =
cos(28.7)
tan(A) =
Thus the cable is approximately 53.8 feet tall.
Example 32. You are hiking along a river and see a tall tree on the opposite bank. You
measure the angle of elevation of the top of the tree and find it to be 62.0◦ . You then
walk 45 feet directly away from the tree
and measure the angle of elevation. If the second
!"#$"%&'()#%"*(
measurement +,-.%(/&,01-2"(/&,-#1#$"%&'(
is 48.5◦ , how tall is the tree? Round your answer to 30-"(44(#5(46(
the nearest foot.
%,*&° ()*+°
!"##$
%&$'!$
$
-./0!1.23$
4516$7".8/#9$:1//$"#;01"#$<$/1!!/#$9."#$</=#8"<$!5<2$!5#$7"#>1.06$
Solution:
7".8/#96*$$?#$:1//$6!<"!$8@$/..A12=$<!$!5#$!:.$B1''#"#2!$!"1<2=/#6$
This problem will require
a little more algebra than the previous problems. We will start
!5<!$:#$5<>#$<2B$:"1!12=$!"1=.2.9#!"1C$"<!1.6$!5<!$12C/0B#$!5#$!"##$
by looking at the two'."$#<C5$.'$.0"$!"1<2=/#6*$$?#$:1//$6!<"!$8@$/..A12=$<!$!5#$81==#"$
different triangles that we have and writing trigonometric ratios that
!"1<2=/#6$D65.:2$12$"#B$8#/.:E*$
!"##$
include
(shown
%&$'!$
$
-./0!1.23$
4516$7".8/#9$:1//$"#;01"#$<$/1!!/#$9."#$</=#8"<$!5<2$!5#$7"#>1.06$
MAT 142 - Module
7".8/#96*$$?#$:1//$6!<"!$8@$/..A12=$<!$!5#$!:.$B1''#"#2!$!"1<2=/#6$
the tree for!5<!$:#$5<>#$<2B$:"1!12=$!"1=.2.9#!"1C$"<!1.6$!5<!$12C/0B#$!5#$!"##$
each of our triangles. We will start by looking at the
'."$#<C5$.'$.0"$!"1<2=/#6*$$?#$:1//$6!<"!$8@$/..A12=$<!$!5#$81==#"$
in red below).
!"1<2=/#6$D65.:2$12$"#B$8#/.:E*$
%,*&°
35
bigger triangles
!"##$
()*+°
F$'!$
%&$'!$
$
G2$!516$"#B$!"1<2=/#H$!5#$!"##$16$.77.61!#$!5#$<2=/#$!5<!$16$=1>#2$!.$
In this red triangle,8#$%,*&$B#="##6*$$?#$<"#$</6.$=1>#2$!5#$/#2=!5$.'$%&$'##!$<6$7<"!$
the tree is opposite the angle that is given to be 48.5 degrees. We are
.'$!5#$61B#$!5<!$16$<BI<C#2!$!.$!5#$<2=/#$=1>#2$!.$8#$%,*&$B#="##6*$$
also given the length
of 45 feet as part of the side that is adjacent to the angle given to be
-12C#$:#$5<>#$7<"!$.'$!516$<BI<C#2!$61B#H$:#$<"#$=.12=$!.$/<8#/$!5#$
48.5 degrees. Since we have part of this adjacent side, we are going to label the other part
F$'##!*$$4506$!5#$:5./#$<BI<C#2!$61B#$
of the side to be x .!5#"$7<"!$.'$!5#$61B#$!.$8#$
feet. Thus the whole adjacent
side is 45 + x feet long. Since we have the
16$ %& + F $'##!$/.2=*$$-12C#$:#$5<>#$!5#$.77.61!#$<2B$!5#$<BI<C#2!$
opposite and the adjacent sides, we can use the tangent ratio. For the red triangle, we have
61B#6H$:#$C<2$06#$!5#$!<2=#2!$"<!1.*$$J."$!5#$"#B$!"1<2=/#H$:#$5<>#$
.77.61!#
!<2DF E =
<BI<C#2!
$ tan(x) = opposite
!"##
adjacent
!<2D %,*&E =
%& + F
tree
$
tan(48.5) =
45 + x
?#$:1//$C.2!120#$8@$/..A12=$<!$!5#$69<//#"$"#C!<2=/#$D65.:2$12$8/0#$
!"#$"%&'()#%"*(
8#/.:E$
We will continue by looking
at the smaller triangle (shown in blue below)
+,-.%(/&,01-2"(/&,-#1#$"%&'(
%,*&°
30-"(45(#6(47(
!"##$
()*+°
-$'!$
$
./$!012$345#$!"16/74#8$!0#$!"##$12$9::921!#$!0#$6/74#$!06!$12$71;#/$!9$
In this blue triangle, the tree is opposite
the angle that is given to be 62 degrees. Based on
3#$()$<#7"##2*$$=62#<$9/$>06!$>#$<1<$'9"$!0#$"#<$!"16/74#8$>#$?/9>$
what we did for the red triangle, !06!$!0#$4#/7!0$9'$!0#$21<#$6<@6A#/!$!9$!0#$6/74#$71;#/$!9$3#$()$
we know that the length of the side adjacent to the angle
<#7"##2$12$
given to be 62 degrees is x feet long.
Since-$'##!$49/7*$$B1/A#$>#$06;#$!0#$9::921!#$6/<$!0#$
we have the opposite and the adjacent sides, we
6<@6A#/!$21<#28$>#$A6/$52#$!0#$!6/7#/!$"6!19*$$C9"$!0#$"#<$!"16/74#8$
can use the tangent ratio. For the blue triangle, we have
>#$06;#$
9::921!#
!6/E- D =
opposite
6<@6A#/!
tan(x) = $
!"## adjacent
!6/E()D =
%&$'!$
-
tree
tan(62) =
$
x
F9>$>#$6"#$"#6<G$'9"$!0#$647#3"6*$$H#$06;#$!>9$#I56!19/2$
!"##
!"##
tree
Now we are ready for the algebra.!6/E
We
have
equations
= 45+x
and tan(62) =
%,*&
D = two $6/<$
!6/E()D tan(48.5)
=
$>1!0$!>9$5/?/9>/2$
%&
+
tree
with
two
unknowns
(variable).
We
can
use
the
substitution
method
(solve
one of the
E;6"1634#D*$$H#$A6/$52#$!0#$2532!1!5!19/$J#!09<$E294;#$9/#$9'$!0#$
x
equations for one of the variable and
then plug that in to the other equation) to determine
#I56!19/2$'9"$9/#$9'$!0#$;6"1634#$6/<$!0#/$:457$!06!$1/$!9$!0#$9!0#"$
the height of the tree. One way to#I56!19/D$!9$<#!#"J1/#$!0#$0#170!$9'$!0#$!"##*$$K/#$>6G$!9$79$0#"#$
go here is to solve the equation tan(62) = tree
for x. This
x
!"##
will give us
12$!9$294;#$!0#$#I56!19/$ !6/E()D =
$'9"$-*$$L012$>144$71;#$52$$
-
!6/E()D =
!"##
-
!6/E()D !"##
=
M
$
- !6/E()D = ME!"##D
!"##
36
Geometry
tree
x
tan(62)
tree
=
1
x
x tan(62) = 1(tree)
tree
x=
tan(62)
tan(62) =
We can now plug this expression for x into the equation tan(48.5) =
height of the tree.
tan(48.5) =
tree
45+x
and solve for the
tree
45 + x
tan(48.5)
tree
=
1
45 + x
(45 + x) tan(48.5) = tree
45 tan(48.5) + x tan(48.5) = tree
Write both sides as fractions.
Cross multiply.
Apply distributive property.
tree
45 tan(48.5) +
tan(48.5) = tree
Substitute in expression for x.
tan(62)
tree
45 tan(48.5) = tree −
tan(48.5)
Collect variable on the same side.
tan(62)
1
45 tan(48.5) = tree 1 −
tan(48.5)
Factor out the variable tree.
tan(62)
45 tan(48.5)
1
1 − tan(62) tan(48.5)
Divide both sides by value in parentheses.
We now just need to plug this expression into our calculator to find out the height of the
tree. The tree is approximately 127 feet tall.