Section 2.2 Structure of Geometry. Definitions  Postulate(axiom)- is a statement accepted without proof  Theorem-is a statement that is proven by using.

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Presentation transcript:

Section 2.2 Structure of Geometry

Definitions  Postulate(axiom)- is a statement accepted without proof  Theorem-is a statement that is proven by using defs, postulates, or previously proven theorems

Postulates:  Ruler Postulate- the pts on a # line can be paired 1:1 with the set of reals #’s(coords) and the distance between any 2 pts = │ diff of their coords │ Ex: A B -6 4 d AB= │ -6 – 4 │ = 10 units

Example :  Find the distance from A to B: A B 3 8 AB = │ 8 – 3 │ = 5

 Segment Addition Postulate- Pts A, B, & C are collinear and B is in between A & C, then AB + BC = AC ( 2parts sum= whole) If B is between A and C, then AB + BC = AC A B C

S is between T and V. R is between S and T. T is between R and Q. QV = 18, QT= 6, TR=RS=SV. Make a sketch and answer.  Find 1. RS2. QS 3. TS4. TV

Suppose J is between H and K. Use the segment Addition Postulate to solve for x. Then find the length of each segment. 1. HJ = 2x HJ = 5x -3 JK = 3x + 3JK = 8x - 9 KH = 22KH = HJ = 2x + 1/3 JK = 5x + 2/3 KH = 12x - 4

 Protractor Postulate- the rays of a given angle can be paired 1:1 on the protractor and the measure is determined by: │ diffs of those real #’s │

 Angle addition Postulate- If C is in the interior of <AOD, then m<AOC + m<COD = m<AOD.

 Q is in the interior of <ROS. S is in the interior of <QOP. P is in the interior of <SOT. S is in the interior of <ROT and m<ROT= 160°. M<SOT = 100° and m<ROQ = m<QOS = m<POT. Make a sketch and answer the following: 1. find m<QOP 2. m<QOT3. m<ROQ

Let Q be in the interior of <POR. Use the angle addition Postulate to solve for x. Find the measure of each angle. 1. m<POQ = (X + 4)° 2. m<POQ = (3x + 7)° m<QOR = (2x – 2)° m<QOR = (5x – 2)° m<POR = 26° m<POR = 61°