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AC Geometry/Trig.
Alan Grothues Semester 2
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Chapter 7: Ratios and Proportions
Ratio
Comparison between two numbers
ProportionEqual ratios
Ex. 1/2 = 2/4
1 4 Means Can change means or extremes and still have true statement
2 8 Extremes Reciprocal PropertyIf 1/2 = 2/4 then 2/1 = 4/2
Finding Ratio Values
!n a pol"gon# "ou can use a $ormula to $ind the measure o$ each angle i$ the ratios o$ theangles are %nown&
Ax + Bx + Cx + = Sum o all interior angles
Letterrepresents the part o$ the ratio
x represents value each must be multiplied b" to reach true value
Similar Pol!gons'ol"gons are similar i$$ their corresponding sides are proportional and theircorresponding angles are congruent
(hen identi$"ing similar pol"gons# ma%e sure to write out which ones aresimilar and the proportions that go along with that
A
"
)* 8 + 4
B C # F
, -
(hen solving for sides of similar triangles# pair up equal ratios with the un%nownvariable and solve
/C 0 1E2
AB/"# = BC/#F = AC/"F
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A$A$ Similarit! Postulate
If two angles of one triangle are congruent to two angles of another triangle, thentriangles are congruent
Bot% triangles are similar
&S$A$S Similarit! Postulate
If two sides of a triangle are proportional to corresponding sides of another triangle and
the included angles of the triangles are congruent, then triangles are similarS$S$S$ Similarit! Postulate
If three sides of a triangle are proportional to corresponding sides of another triangle,then the triangles are congruent
Side'Splitter (%eorem
If a line parallel to one side of a triangle intersects the other two sides, then it dividesthose sides proportionally
A B
A/C = B/"
C " 3ther combinations wor%# also
(riangle Angle'Bise)tor (%eorem
If a ray bisects an angle of a triangle, then it divides the opposite side into segmentsproportional to the other sides
Parallel *ine Segment Similarit! Corollar!
If three parallel lines intersect two transversals, then they divide the transversalsproportionally
i%e the 5ide5plitter 6heorem
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Chapter 8: Right Triangles and Trigonometry
eometri) ,ean
Equal means or extremesEx. )7 9 74
!n a right triangle with all three altitudes# all three triangles are similar
(ith all o$ these similarities# we are able to see two things&
If you have a right triangle, then the altitude to the hypotenuse is a geometric meanbetween the two parts of the hypotenuse
Part o -!pot$ Altitude to-!pot$
Altitude to -!pot$ Part o -!pot$
If you have a right triangle, then a leg of the triangle is a geometric mean between thewhole hypotenuse and the part closest to the leg
.%ole -!pot$ *eg o Rig%t (riangle
*eg o Rig%t (riangle Part o -!pot$ Closest to *eg
P!t%agorean (%eorem
If a triangle is a right triangle, then the sum of the squares of the legs is equal to thesquare of the hypotenuse
a2+ 2= )2
Can be used to $ind an" missing part o$ a right triangle
+: )9 ;
+ : )44 9 ;
+ ; ),< 9 ;
; 9 )-
)
=
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40o'40 o Rig%t (riangles
In a 45o45oright triangle, the hypotenuse is equal to one of the legs times !2
o'3 o Rig%t (riangles
In a "#o$#oright triangle, the smaller leg is half the hypotenuse and the longer leg isequal to the smaller leg times !"
(rigonometr! ntrodu)tion
!n an" right triangle# there is alwa"s , characteristics o$ each side that can be expressed&
Sine 3pposite
="potenuse
Cosine d>acent
="potenuse
(angent 3pposite
d>acent
Cose)ant ="potenuse
3pposite
Se)ant ="potenuse
d>acent
Cotangent d>acent
3pposite
lso can be remembered b" 53=C=63
6ofind the side of any right triangle# >ust set up an equation using this $ormula?example using sine@&
%!p = leg52 ) = a52
-!p = 2a = a5
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Sin6degree o angle7 = sine ratio o triangle = a)tual sine ratio8888ound in )%art or )al)ulator88
7 x
36o
5
#x sin3 = 9/x = $0:9 9 = $0:9x x = 12
6ofind the measure of an angle in any right triangle# >ust set up an equation using this$ormula ?example using sine@&
sin'16sin;x7 = sin'16sine ratio o triangle7
3 5
xo
4
#x sin'16sin;x7 = sin'16/07 x = sin'16/07 x = 9o
mportant (rig Ratios
sin A 7
cos A 7
tan )
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40o
o
sin Bcos A- 7
tan A- 7 -
3o
sin A- 7
cos B
tan A-
Angle o #le
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)2= a2+ 2'26a767)osC
*a o Sines
sed when "ou have AAS ?to $ind a missing side and then the rest o$ the triangle@ orSSA?to $ind the missing angle then the rest o$ the triangle@
sinA = sinB = sinC
a )
Angles o Rotation
3n a coordinate graph# angles are measured $rom two ra"s&
Initial Ray the beginning ra" that starts at -&** and is alwa"s on the xaxisTerminal Ray the ending ra" o$ "our angle# goes wherever needed
!$ ra"s have the same terminal ra"s# the" are said to be conterminal
6his means that a certain angle can be named in more than one wa"
Positive ?initial ra" to terminal ra" in countercloc%wise@
Negative?initial ra" to terminal ra" in cloc%wise@
>360?initial ra" to terminal ra" in countercloc%wise with multiple loops
around Dadd -,* $or each rotation @ !360?initial ra" to terminal ra" in cloc%wise with multiple loops aroundDsubtract -,* $or each rotation @
)+o
-+o
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(rig Fun)tions in rap%s
(hen dealing with graphs# "ou can $ind the trig $unctions o$ a point on a graphrenaming the same parts o$ trig $unctions with names corresponding to the point location
opposite 9x
ad>acent 9 ! r = 5x2+ !2
h"potenuse 9 r
6o $ind the trig values o$ a given angle# "ou can alwa"s $ind it b" $inding the referenceangleF the angle $ormed $rom the terminal ra" to the closest xaxis
6rig ratios o$ the re$erence angle are alwa"s going to be the same as the originalangle measurement
Ge$erence angle +*o )-*o
>uadrant (rig Values!n each quadrant o$ a graph# the ?sine# cosine# or tangent@ ratio ma" be positive ornegative. 6o tell# use this tric% to tell i$ the" are positive&
S A All Students (a%e Calculus
V ( C
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Chapter 9: Cirles
Cir)le
ll points that are equidistant $rom a central point and coplanar
Radius
6he distance $rom the center to a point on the circle
segment whose endpoints are the center and a point on the circle
C%ord
segment whose endpoints are on the circle
"iameter
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chord that contains the center
*ine (!pes
"ecant Line line that goes through two points on a circle
Tangent Line line that goes through one point on a circle
'oint is calledpoint of tangency
Congruent Cir)les
Circles with equal radii
Con)entri) Cir)les
Circles with the same center
ns)ried
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(angent Corollar!
&angents to a circle from a point are congruent
AB = AC /
C
Ar)s
6he degree o$ the angle associated with the arc is the measure o$ that arc
%inor &rc an" arc smaller than )8*o
%a'or &rc an" arc larger than )8*o
"emicircle an arc whose measure is )8*o
Ar) (%eorems and Postulates
Ar) Addition Postulate
&he measure of the arc formed by two ad'acent arcs is the sum of the measuresof these two arcs
Congruent Ar) (%eorem
In the same circle or in congruent circles(
1) *ongruent arcs have congruent chords
2) *ongruent chords have congruent arcs
?C%ord'Center (%eorem
In the same circle or in congruent circles(
1) *hords equally distant from the center are congruent
2) *ongruent chords are equally distant from the center
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,ain (!pes o Angles in Cir)les
$entral &ngle angle whose vertex is the center
Inscri#ed &ngle angle whose vertex is on the circle
Interior "ecant &ngle angle whose vertex is inside the circle
(xterior "ecant &ngle angle whose vertex is outside the circle
ns)ried Angle (%eorems and Postulates
ns)ried Ar) (%eorem
&he measure of an inscribed angle is equal to half the measure of its intercepted
arc
b a m; = @ 6a7
Congruent ns)ried Angle Corollar!
If two inscribed angles intercept the same arc, then the angles are congruent
ns)ried >uadrilateral Corollar!
If a quadrilateral is inscribed in a circle, then its opposite angles aresupplementary
(angent ns)ried Angle (%eorem
&he measure of an angle formed by a chord and a tangent is equal to half themeasure of the intercepted arc
nterior Se)ant Angle (%eorem
&he measure of an angle formed by two chords that intersect inside a circle +interiorsecant) is equal to half the sum of the measures of the intercepted arcs +average)
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b c
a
m; = @ 6a+)7
#xterior Se)ant Angle (%eorem
&he measure of an angle formed by two chords that intersect outside a circle +eteriorsecant) is equal to half the difference of the larger arc and smaller arc
a b c m;a = @ 6)'7
C%ord *engt% (%eorems
nterse)ting C%ord (%eorem
-hen two chords intersect inside a circle, the product of the segments of onechord equals the product of the segments of another chord
a b
c d a6d7 = 6)7
#xterior Se)ant C%ord (%eorem
-hen two secant segments are drawn to a circle from an eternal point, theproduct of one secant segment and its eternal segment equals the product of theother secant segment and its eternal segment
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/ 6a+7a = 6)+d7)
a
c d
Se)ant'(angent C%ord (%eorem
-hen a secant segment and a tangent segment are drawn to a circle form aneternal point, the product of a secant segment and its eternal segment is equalto the square of the tangent segment
a
b 6a+7 = )2
c
Chapter !!: Area
Area Postulates
&rea ")uare Postulate
&he area of a square is the square of the length of a side
$ongruent *igure Postulate
!$ two $igures are congruent# then the" have equal areas
Base -eig%t
+ase length o$ an" side o$ a parallelogram# the longest side o$ a triangle# or the twoparallel sides o$ a trapeHoid
,eight length o$ perpendicular segment $rom base to opposite side
Parallelogram Area (%eorem
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&he area of a parallelogram equals the product of its base and height
% A = ? %
(riangle Area (%eorem
&he area of a parallelogram equals the half of the product of the base and height
% A = @ 6 ? %7
R%omus Area (%eorem
&he area of a rhombus equals half of the product of the two diagonals
d1 A = @ 6d1? d27
d2
(rapeoid Area (%eorem
&he area of a trape.oid equals the average of the bases times the height
1
% A = @ 61+ 276%7
2
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Regular Pol!gon Area (%eorem
&he area of a regular polygon equals the number of sides times the radius +distance
from center to verte) times the sin+1#/n) times cos+1#/n)
r A = n ? r26sin1:/n76)os1:/n7
&he area of a regular polygon is equal to half of the product of the apothem
+perpendicular segment from center to side) and the perimeter
s
a A = @ 6a ? p7
Cir)le (%eorems
$ircle &rea Theorem
&he area of a circle is pi times the radius squared
r A = r2
$ircumference Theorem
&he circumference of a circle is the diameter times pi
d C = d
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Length of &rc Theorem
&he length of an arc is the quotient of the measure of the arc by "$# times twotimes pi times the radius
x
leng%tARC = 6mARC/3762r7
&rea of "ector Theorem
&he area of a sector is the quotient of the measure of the arc by "$#times pi times the radius squared
x Ase)tor= 6mARC/376r27
Area Ratios o Similar Pol!gons
!$ pol"gons are similar# then all corresponding parts# lengths# etc are proportional
/ut# the areas are di$$erent&
A1/ A2= 6Corresponding Part1/Corresponding Part272
Proailit!
6he ratio o$ the target over the whole
Proailit! = Atarget/ A%ole
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Chapter !2: Area " #olume o$ Regular %& Shapes
Sura)e Area
Lateral "urface &rea?*$S$A@ area o$ the sides o$ the three dimensional shape ?notbases@
Total "urface &rea ?($S$A@ area o$ all o$ the sides o$ a three dimensional shape?including bases@
2ormula is alwa"s *$S$A + Aases
Volume
6hreedimensional area o$ a shape ?measured in unit-@
Prism
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6hreedimensional shape with congruent# parallel pol"gons $or bases
+ases congruent# parallel pol"gons
,eight altitude $rom one base to the other
6wo t"pes o$ prisms&
Right Prism all sides o$ prism are perpendicular
-#li)ue Prism sides are not perpendicular
L.".&. of Prism Theorem
&he lateral surface area of a prism equals the perimeter of the base times the
height
*$S$Aprism= 6Pase76%7
/olume of Prism Theorem
&he volume of a prism equals the area of the base times the height
Vprism= 6B76%7
P!ramid
6hreedimensional shape with a regular pol"gon base that $orms lateral $aces ?triangles@that meet at a vertex
+ase regular pol"gon $rom which the lateral $aces use each side as a base
Lateral face triangles going $rom the base o$ the p"ramid to the vertex
/ertex point that is perpendicular with the center o$ the base and in which alllateral $aces meet
,eight altitude $rom the center o$ the base to the vertex ?perpendicular@
"lant ,eight?l@ the height o$ a lateral $ace o$ the p"ramid
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L.".&. of Pyramid Theorem
&he lateral surface area of a pyramid equals half of the product of the perimetertimes the slant height
*$S$Ap!ramid= @ 6pl7
/olume of Pyramid Theorem
&he volume of a pyramid equals onethird of the area of the base times theheight
Vp!ramid= 1/ 6B76%7
C!linder
6hreedimensional shape with congruent# parallel# circle bases
L.".& of $ylinder Theorem
&he lateral surface area of a cylinder equals the product of the circumference ofthe base times the height
*$S$A)!linder= 6C76%7
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/olume of $ylinder Theorem
&he volume of a cylinder equals the area of the base times the height
V)!linder= 6B76%7
Cone
6hreedimensional shape with a circular base whose lateral $aces meet at a vertex
"lant ,eight?l@ height $rom point on the circle to the vertex
/ertex point that is perpendicular with the center o$ the base and in which alllateral $aces meet
L.".& of $one Theorem
&he lateral surface area of a cone equals the radius of the base times the slant
height times pi*$S$A)one= rl
/olume of $one Theorem
&he volume of a cone equals onethird of the area of the base times the height
V)one= 1/ 6B76%7
Sp%ere
6hreedimensional circle
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L.".& of "phere
&he lateral surface area of a sphere equals four times pi times the radiussquared
*$S$Asp%ere= 4r2
/olume of "phere
&he volume of a sphere equals fourthirds the product of pi times the radiuscubed
Vsp%ere= 4/r
Chapter !%: Coordinate Geometry
"istan)e Formula
1eals with the distance between two points on a coordinate graph
6hree di$$erent $ormulas that amount to the same distance $ormula
Pythagorean Theorem
a2+ 2= )2
()uation of a $ircle
r
2
= 6x'a7
2
+ 6!'7
2
istance *ormula
d = 56!2'!172+ 6x2'x172
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,idpoint Formula
6o $ind the midpoint between two given points# $ind the average o$ the x coordinates toreceive "our midpoint x# and the average o$ the " coordinates to get "ou midpoint "coordinates
, = 6 Dx1+x2E/2 D!1+!2E/2 7
Slope
6o $ind the slope o$ a line# use the change in " coordinates over the change in xcoordinates
slope 6m7= G
H
#Iuation o a *ine
6he equation o$ a line gives "ou the abilit" to determine an" point that is on that line
!t can be set up in a $ew di$$erent $ormats&
"tandard *orm& 0 y = b"lope!Intercept *orm& m+) 0 b = y
6o ma%e sure "ou set it up correctl" ever" time# use the I2in%e 2ormJ
G ! J
H x J a
Parallel and Perpendi)ular *ines
Parallel Lines parallel lines have congruent slopes
Perpendicular Lines perpendicular lines have opposite reciprocal slopes
9