geometry chapter 2 journal
DESCRIPTION
GEOMETRY CHAPTER 2 JOURNAL. VALERIA IBARGUEN 9-1. CONDITIONAL STATEMENT. This is a type of statement that can be written in a form of “ if p , then q ” P= Hypothesis Q= conclusion EXAMPLES: If mTRANSCRIPT
GEOMETRY
CHAPTER 2
JOURNALVALERIA IBARGUEN 9-1
CONDITIONAL STATEMENT
This is a type of statement that can be written in a form of “if p, then q”
P=Hypothesis Q=conclusion EXAMPLES:a) If m<A=195°, then <A is obtuseb) If an insect is a butterfly, then it has
four wingsc) If an angle is obtuse, then it has a
measure of 100°
COUNTER-EXAMPLES
A counter-example is a type of example that proves if a conjecture or statement is false. This could be a drawing, a statement or a number.
EXAMPLES:a) For any real number x, x2 > x5, 52 > 55, 25 > 5 b) Supplementary angles are adjecent
c) The radius of every planet in the solar system is less than 50,000 km. Mercury Venus Earth Mars Jupiter Satur
nUranus
Neptune
4880 12,100
12,800
6790 143,00
121,00
51 49,500
DEFINITION
This is a statement that tells or discribes a mathematical object and can be written as a true biconditional statement. A definition includes “if and only if”
EXAMPLES:a) A figure is a triangle if and only if it is a
three-sided polygon.b) A ray, segment or line is a segment
bisector if and only if it divides a segment into two congruent segments.
c) A traingle is straight if and only if it measures 180°.
BI-CONDITIONAL STATEMENTS
This is a statment that is written in the form “p if and only if q”. They are important. This is used when a conditional statement and its converse are combined together.
EXAMPLES:
a) Converse: If x=3, then 2x+5=11 Biconditional: 2x+5=11 if and only if x=3
b) Converse: If a point divides a segment into two congruent segments, then the point is a midpoint.
Biconditional: A point is a midpoint if and only if it divides the segments into two congruent segments.
c) Converse: If the dates is July 40th, then it Independence day. Biconditional: It is Independence day if and only if it is July 40th.
DEDUCTIVE REASONING This is the type of process in which we use logic to
draw conclusions of something. EXAMPLES:
a) If a team wins 10 games, the they play in the finals. If a team plays in the finals they they travel to Boston. The Reavens won 10 games. CONCLUSION:The Reavens will travel to Boston.
b) If two angles form a linear pair, then they are adjecent. If two angles are adjecent, then they share a side. <1 and <2 form a linear pair. CONCLUSION: <1 and<2 share a side.
c) If a polygon is a triangle, then it has three sides. If a polygon has three sides then it is not a quadrilateral. Polygon is a P triangle. CONCLUSION: A polygon is not a quadrilateral because ithas three sides.
LAWS OF LOGIC
Law of detachment:- If p-q is true we should assume if P is
true then Q must also be true
Law of Syllogism:- If P-Q is true and Q then R is true then if
P is true are must be true P and R is true.
LAW OF DETACHMENT Given: If two segments are congruent then they have the
same length. AB≅XY
Conjecture: AB=XY
hypothesis: two segments are congruent
conclusion: they have the same lenght
The given AB≅XY statements does match the hypothesis so the conjecture IS true.
Given: If you are 3 times tardy, you must go to detention. John is in detention.
Conjecture: John was tardy at least 3 times.
hypothesis: you are tardy 3 times
conclusion: you must go to detention.
The statement given to us matches the conclusion of a true conditiona, but the hypothesis is not true since John can be in detention for another reason so the conjecture is NOT valid.
LAW OF SYLLOGISM GIVEN: If m<A 90°, then <A is acute. If <A is acute then it is not a
right angle.
p= the measure of an angle is less then 90°
q= the angle is acute
r= the angle is not a right angle.
-This is trying to explain us that pq and qr is the conclusion of the first conditional and the hypothesis of the second conditional you can tell that at the en pr. So IT IS VALID
Given: If a number is divisible by 4 then it is divisible by 2. If a number is even, then it is divisible by 2.
Conjecture: If a number is divisible by 4, then it is even.
p= A number is divisible by 4
q= A number is divisible by 2
r= A number is even
-What this means is that pq and rq. The Law of Syllogism cannot be used to draw conclusions since q is the conlcusion of both conditional statements, even though pr is true the logic used to dra the conclusion is NOT VALID.
ALGEBRAIC PROOF
An algebraic proof is an argument that uses logic, definitions, properties. To do one, you have to do a 2 colum proof.
EXAMPLES:a)Prove: x=2 ifGiven: 2x-6=4x-10
STATEMENT REASON
2x-6=4x-10 Given
+10 +10 Addition Property
2x+4=4x Simplify
-2x -2x Subtraction Property
4x=2x Simplify
2 2 Divsion Property
2=x Simplify
X=2 Symetric property
ALGEBRAIC PROOF
b)-5=3n+1 c)sr=3.6STATEMENT REASON
-5=3n+1 Given
-1 -1 Subtraction Property
-6=3n Simplify
3 3 Division Property
-2=n Symplify
N=-2 Symmetric Property
STATEMENT REASON
sr=3.6S=75 km/hR=6 pixels per meter
GIVEN
(75)(6)=3.6p Substitution property
450=3.6p Simplify
3.6 3.6 Division Property
125=p Simplify
P=125 pixels Symmetric Property
SEGMENT AND ANGLE PROPERTIES OF CONGRUENCE AND EQUALITY
PROPERTY OF EQUALITY:ADDITION PROPERTY OF EQUALITY
If a=b, then a+c=b+c
SUBTRACTION PROPERTY OF EQUALITY
If a=b, then a-c=b-c
MULTIPLICATION PROPERTY OF EQUALITY
If a=b, then ac=bc
DIVISION PROPERTY OF EQUALITY
If a=b then c≠= then a/c=b/c
REFLEXIVE PROPERTY OF EQUALITY
A=A
SYMMETRIC PROPERTY OF EQUALITY
If a=b, then b=a
TRANSITIVE PROPERTY OF EQUALITY
If a=b and b=c, then a=c
SUBSTITUTION PROPERTY OF EQUALITY
If a=b then b can be substituted for a in any expression.
SEGMENT AND ANGLE PROPERTIES OF CONGRUENCE AND EQUALITY
PROPERTIES OF CONGRUENCE:Reflexive Property of CongruenceFigure A≅ figure A
− −EF≅EF
Symmetric Property of CongruenceIf figure A≅ figure B then figure B≅A
IF <1 ≅ <2 then <2 ≅ <1
Transitive Property of CongruenceIf figure A ≅ figure B and figure B ≅ figure C then figure A ≅ figure C
If <1 ≅ <2 and <2 ≅ <3 then <1 ≅ <3
TWO-COLUM PROOFS
To do a two colum proofs you have to list each step of how you found your answer.
EXAMPLES:STATEMENT REASON
C=9f+90C=102
Given
102=9f+90 Substitution
-90 -90 Subtraction
12=9f Simplify
9 9 Division
1.3=f Simplify
F=1.3 Symmetric Property
TWO-COLUM PROOFS
STATEMENT REASON
C=$5.75+$0.89mC=$11.98
Given
$11.98=$5.75+$0.89m
Substitution
-$5.75 -$5.75
Subtraction Property
$6.23=$0.89m
Simplify
$0.89m $0.89m
Division
7=m Simplification
M=7 Symmetric Property
STATEMENTE REASON
Y+1=5 GIVEN
-1 -1 Substitution
Y=4 Simplify
Y=4 Symmetric Property
LINEAR PAIR POSTULATE (LPP)
This is when all linear pairs are linear postulates, SUPPLEMENTARY
EXAMPLES:
Given: angle<1 and < 2 are linear pair
Prove: <1 and <2 supplementary.
<1 and <2 are linear pair
Given
<1 and <2 form a linear pair
Definition of linear pair
M<1+M<2= 180° Striaght angle’s definition
<1 and <2 are supplementary
Deffintion of supplementary
LINEAR PAIR POSTULATE
Given: <1 and <2 are supplementary <3 and <4 are supplementary.
Prove:<1≅<4
STATEMENT REASON
<1 and <2 are supplementary <3 and <4 are supplemtary
GIVEN
<1 + <2 = 180°, <3 + <4=180° Deffinition of supplementary angles
M<1+M<2=M<3+M<4 Substitution
<2≅<3 Given
M<2=M<3 Deffinition of congruent andgles
M<1=M<4 Subtraction property of steps 3 and 5
<1≅<4 Definition of Congruent angles
LINEAR PAIR POSTULATE
Given: BE ≅ CE, DE ≅ AEProve: AB ≅ CD
SATEMENT REASON
BE ≅ CE, DE ≅ AB Given
BE + AE= AB Segment Addition Postulate
CE + DE= CD Segment Addition Postulate
CE + DE= AB Substitution
CD=AB Substitution
CONGRUENT COMPLEMENTS AND SUPPLEMENTS THEOREMS
CONGRUENT COMPLEMENT THEOREM:
THEOREM HYPOTHESIS CONCLUSION
If two angles are complementary to the same angle (or to two congruent angles) then the two angles are congruent
<1 and <2 are complementary<2 and <3 are complementary
<1 ≅ <3
CONGRUENT COMPLEMENTS AND SUPPLEMENTS THEOREMS
CONGRUENT SUPPLEMENT THEOREM:
THEOREM HYPOTHESIS CONCLUSION
If two angles are supplemtnary to the same angle (or two congruent angle) then the two angles are congruent
<1 and <2 are supplementary<2 and <3 are supplementary
<1 ≅ <3
VERTICAL ANGLES THEOREM
VERTICAL ANGLE THEOREM:THEOREM HYPOTHESIS CONCLUSION
Vertical angles are congruent
<A and <B are verical angles
<A ≅ <B
COMMON SEGMENTS THEOREM
COMMON SEGMENTS THEOREM:THEOREM HYPOTHES
ISCONCLUSIO
N
Given a collinear points A,B, C and D arranged as showns if AB≅CD then AC≅BD
AB≅CD AC≅BD