GRAPHIC ORGANIZER: ANGLES IN A CIRCLE At the CENTER of the circle. ON the circle. INSIDE the circle. OUTSIDE the circle. Two Radii Two Chords, or A Chord and a Tangent, or A Chord and a Secant. Two chords Two Secants, or A Secant and a Tangent, or Two Tangents. The measure of the Intercepted Arc Half the measure of the Intercepted Arc Half the SUM of the measures of the Intercepted Arcs. Half the DIFFERENCE of the measures of the intercepted arcs. The angle EQUALS The angle is MADE BY The VERTEX is located
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GRAPHIC ORGANIZER:
ANGLES IN A CIRCLE
At the CENTER
of the circle.
ON
the circle.
INSIDE
the circle.
OUTSIDE
the circle.
Two Radii
Two Chords, or
A Chord and a Tangent, or
A Chord and a Secant.
Two chords Two Secants, or
A Secant and a Tangent, or
Two Tangents.
The measure of the
Intercepted Arc
Half the measure of the
Intercepted Arc
Half the SUM of the measures
of the Intercepted Arcs.
Half the DIFFERENCE of the
measures of the intercepted
arcs.
The angle EQUALS
The angle is MADE BY
The VERTEX is located
GRAPHIC ORGANIZER:
SEGMENTS IN A CIRCLE
The VERTEX is located
INSIDE OUTSIDE
the circle the circle
Two chords Two Secant Segments A Tangent and a Secant Segment
GRAPHIC ORGANIZER:
CLASSIFY QUADRILATERALS
NO Pairs of
Opposite Sides
Are Parallel
Exactly ONE Pair of
Opposite Sides
Are Parallel
BOTH Pair of
Opposite Sides
Are Parallel
GRAPHIC ORGANIZER:
POLYGONS – DEFINITION
GRAPHIC ORGANIZER:
SOME PROPERTIES
PROPERTY
ADDITION
MULTIPLICATION
COMMUTATIVE
A + B = B + A
A ! B = B ! A
ASSOCIATIVE
A + ( B + C ) = (A + B ) + C
A ! ( B ! C ) = ( A ! B ) ! C
IDENTITY
A + 0 = A
A ! 1 = A
INVERSE
A + B – B = A
A ! B ! (1/B) = A
PROPERTIES 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 A ! C = B ! C
Division Property of Equality:
If A = B and C " 0, then (A / C) = (B / C)
REFLEXIVE
A = A
DISTRIBUTIVE
A ! ( B + C ) = A ! B + A ! C
TRANSITIVE
If A = B And If B = C, Then A = C
Classifying Triangles
Find point C that makes !ABC :
a) equilateral
b) isosceles
c) right
d) acute
e) obtuse
f) scalene
Triangle Classifications
Given: AB is a side of !ABC
Draw the loci of all points for C so that
!ABC is:
a) equilateral
b) isosceles
c) right
d) acute
e) obtuse f) scalene
We Classify Triangles
By Angles
Obtuse – one angle is obtuse
Right – one angle is right
Acute – all 3 angles are acute
Equilangular – All 3 angles are 60°
AND
By Sides
Scalene – No sides are congruent
Isosceles – At leat two sides are congruent
Equilateral – All 3 sides are congruent
Problem:
CLASSIFYING TRIANGLES BY ANGLES AND SIDES
ISOSCELES
(at least two congruent sides) SCALENE
(no congruent sides)
RIG
HT
(one
rig
ht
ang
le)
45°45°
OB
TU
SE
(one
ob
tuse
an
gle
)
>90°
>90°
>90°
AC
UT
E
(all
acu
te a
ng
les)
<90°
<90°
EQUILATERAL (all congruent sides)
EQ
UIA
NG
UL
AR
(a
ll 6
0°
ang
les)
60° 60°
60°
Three angle-side combinations make no sense: Equiangular – Scalene Obtuse – Equilateral Right -- Equilateral
GRAPHIC ORGANIZER:
CLASSIFY NUMBERS
REAL NUMBERS
RATIONAL NUMBERS: … -1, … - !, … 0, … !, … 1, … (ratio of integers, no zero in the denominator)
INTEGERS: … -3, -2, -1, 0, 1, 2, 3, …
WHOLE NUMBERS: 0, 1, 2, 3, …
NATURAL (COUNTING) NUMBERS: 1, 2, 3, …
IRRATIONAL NUMBERS: (non-repeating decimals – cannot be written as a ratio of two integers)
Transcendental (can be multiplied by another irrational
number to make it rational): example 2 , 3 etc…
Non-Transcendental
Example: !, e …
IMAGINARY NUMBERS
Example: !1
GRAPHIC ORGANIZER:
ANGLE RELATIONSHIPS
CORRESPONDING ANGLES
ALTERNATE INTERIOR ANGLES
ALTERNATE EXTERIOR ANGLES
VERTICAL PAIR
Adjacent
Non-Adjacent
SAME-SIDE INTERIOR
SAME SIDE EXTERIOR
Adjacent (LINEAR PAIR)
Non-Adjacent
CONGRUENT:
!1" !2
COMPLEMENTARY:
!5 +!6 = 90°
SUPPLEMENTARY:
!3+!4 = 180°
Angle-Angle Similarity Postulate
(AA Sim)
If two angles of one triangle are congruent to two angles of a second triangle, then the two triangles are similar.
A
C
B
D
F
E
!ABC " !DEF
Extension problem: How would you draw two NON-congruent triangles that have 5 congruent parts? 8
PROVING TRIANGLES CONGRUENT
A
C
B
D
F
E
AB ! DE !A " !D
BC ! EF !B " !E
AC ! DF !C " !F
!ABC " !DEF CPCTC: Corresponding Parts of Congruent
Triangles are Congruent. 1
Side-Side-Side Triangle Congruence Postulate
(SSS)
If the sides of one triangle are congruent to the sides of second triangle, then the two triangles are congruent.
A
C
B
D
F
E
This is one of 4 “shortcuts” to prove triangles congruent that are shown in this brochure. 2
Side-Angle-Side Triangle Congruence Postulate
(SAS)
If two sides and the included angle of one triangle are congruent to two sides and the included angle of another triangle, then the two triangles are congruent.
A
C
B
D
F
E
Once you prove two triangles are congruent, you can justify that all the corresponding parts are congruent. How? CPCTC. 3
Angle-Side-Angle
Triangle Congruence Postulate (ASA)
If two angles and the included side of one triangle are congruent to two angles and the included side of second triangle, then the two triangles are congruent.
A
C
B
D
F
E
4
Angle-Angle-Side or Side-Angle-Angle
Triangle Congruence Theorem (AAS or SAA)
If two angles and a non-included side of one triangle are congruent to the corresponding two angles and side of a second triangle, then the two triangles are congruent.
A
C
B
D
F
E
Do you see the difference between this and the sketch on page 4?
5
Hypotenuse – Leg
Triangle Congruence Theorem (HL)
If the hypotenuse and a leg of one right triangle are congruent to the hypotenuse and corresponding leg of a second right triangle, then the two triangles are congruent.
C
BA
Notice how this is the exception or special case where SSA actually works. If the triangle is NOT a right triangle, you get the “Donkey Theorem.” (See page 7) 6
Side-Side-Angle or Angle-Side-Side (SSA or ASS)
also known as the “Donkey Theorem”
It is NOT a valid method to prove triangles are congruent to say that since two sides and a non-included angle of each are congruent that the triangles are congruent. It ain’t necessarily so! Check it out:
C
BA
BA
C
7
GRAPHIC ORGANIZER:
REMEMBER YOUR VOWELS
A + E + I + O = U AND Y
When submitting math work, always ask yourself:
A. Did I clearly identify and label or circle an ANSWER that makes sense for what I did?
E. Is my reasoning clearly and efficiently EXPLAINED?
I. Does my work reflect all the necessary INFORMATION I used to solve the problem?
Can others tell what letter, number and symbol represents?
O. Is my work well ORGANIZED with steps that follow a logical sequence?
U. If so, my work demonstrates that I UNDERSTAND the concepts involved.
… and always…
Y. Ask WHY.
Why did I select the strategies I did?
Why might others argue the benefits of a different strategy?
Why is the concept significant? …
GRAPHIC ORGANIZER:
RULES FOR PRESENTING
1. COME PREPARED.
2. VOLUNTEER TO PRESENT.
3. ADDRESS YOUR CLASSMATES.
4. SHOW RESPECT.
5. HELP FIND ERRORS.
6. CONTRIBUTE YOUR IDEAS.
7. POINT OUT CONNECTIONS.
8. CITE YOUR SOURCES.
GRAPHIC ORGANIZER:
PROBLEM SOLVING STRATEGY:
Identify the relationship. In math, how is what
is given and what you are asked to solve related?
Devise a plan. In math, write an equation.
Execute the plan. In math, solve the equation.
Answer the question that is asked. In math, this
may or may not be the solution to your
equation. Look at the question again.
GRAPHIC ORGANIZER:
STEPS FOR PROOF (shorthand used to explain your reasoning)
1. Write the GIVEN and PROVE statements.
(Identify and state the given information and what it is you are trying to prove).
2. Draw a SKETCH.
3. MARK the drawing. (the most important step)
A. First, mark only what is explicitly GIVEN.
B. Secondly, mark what is implied or implicitly true.
4. Set up a STATEMENT and REASON table.
5. Fill in the table. Justify each statement with a reason or rule.
SIX RULE CATEGORIES (used in this course)
1. Given Information
2. Definitions
3. Properties
4. Postulates
5. Theorems
6. Corollaries
IMPLICIT INFORMATION to watch for
• Vertical Pairs
• Common Side or Common Angle (Reflexive Property)
• Parallel Lines (and angle relationships formed by a transversal crossing parallel lines).