lesson 1 - ddtwo.org

Segments and Congruence

Friday, January 31, 2020

Lesson 1.2

Segments and their Measure

The Ruler Postulate

The points on a line can be matched one to one with the real numbers.

·A

·B

The real number that corresponds to a point is the coordinate of the point

What is the coordinate of:

1) Point A

2) Point B

= 1

= 4.5


Distance between point A and B in symbol is AB

AB = | coordinate of A - coordinate of B|

Example: Find the distance between the given points

·A

·B1.)

Solution: AB = | coordinate of A - coordinate of B|

AB = | |1 - 4.5 = |-3.5 | = 3.5




Example: Find the distance between the given points

2.)

Find: AG

GR

AE

=| |2 - -7 = |9| = 9

876543210-1-2-3-4-5-6-7-8

● ● ●● ●G R A C E

=| |-7 - -3 = |-4| = 4

=| |2 - 7 = |-5| = 5




Try it yourself: Find the distance between the given points

3.)

Find: CG

ER

EG

=| |6 - -7 = |13| = 13

876543210-1-2-3-4-5-6-7-8

● ● ●● ●G R A C E

=| |7 - -3 = |11| = 11

=| |7 - -7 = |14| = 14


Segment Addition Postulate

·A ·C·B

If B is between A and C

Then

AB + BC = AC

•EXAMLE 3

Example 1: Use the diagram to find GH.

Use the Segment Addition Postulate to write an equation. Then solve the equation to find GH.

SOLUTION

Segment Addition Postulate.

Substitute 36 for FH and 21 for FG.

Subtract 21 from each side.

21 + GH=36

FG + GH=FH

=15 GH



SOLUTION

Example 2: Maps

The cities shown on the map lie approximately in a straight line. Use the given distances to find the distance from Lubbock, Texas, to St. Louis, Missouri.

Because Tulsa, Oklahoma, lies between Lubbock and St. Louis, you can apply the Segment Addition Postulate.

LS = LT + TS = 380 + 360 = 740

The distance from Lubbock to St. Louis is about 740 miles.

ANSWER

Segments and their Measure Segment Addition Postulate



Example 3:

In the figure: AD = 20, BD = 15, CD = 4

A CB· ·· D·Find: AB

BC

AC



Example 4:

RS = 2x + 10, ST = x - 4, RT= 21

·R

·T

·S

Find x

RS

Connecting with Algebra

ST



More Examples



Try it yourself! Book p.18 #29

Midpoint and BisectorLesson 1.2 continued

Congruent Segment

Midpoint

Bisector


Congruent Segment• Congruent segments are segments with

equal length.

Congruence Symbol:

EXAMPLE:

1. If AB = ED THEN AB ED

2. Use the number line to determine whether the statement is true or false. Explain why.

876543210-1-2-3-4-5-6-7-8

● ● ●● ●G R A C E

a) AR AE b) GR AEtrue false

Midpoint of Segment and Bisector

• The midpoint of a segment is a point that divides the segment into two congruent segments.

ADE● ● ●

In the Figure Point D is the midpoint of EA if

1. D is between E and A

2. ED = DA


Bisect means to separate into two

congruent parts.

A segment bisector can be a point,

line, ray, segment or plane.

A B● ●●

P


Solution: 5x – 6 = 2x

5x – 2x = 6

3x = 6

x = 2

AB = 8

BCA● ● ●

5x – 6 2x

Example 1: Algebra Problem

In the Figure Point C is the midpoint of AB

Find: x , AB


In the skateboard design, VW bisects XY at point T,

and XT = 39.9 cm. Find XY.

Example 2: Skateboard

SOLUTION

EXAMPLE 1

Point T is the midpoint of XY .

So, XT = TY = 39.9 cm.

XY = XT + TY

= 39.9 + 39.9

= 79.8 cm


Substitute.

Add.


Midpoint and Distance FormulaLesson 1.4

Midpoint Formula

Distance Formula


Midpoint in the Coordinate Plane

Midpoint of AB =

y

x

·A

(x1, y1)

·B

(x2, y2)

•Midpoint

2,

2

1212 yyxxMidpoint Formula

In the coordinate plane,

B with coordinates (x2, y2)

whose endpoints are A with

coordinate (x1, y1)

the midpoint of a segment

is given by the

and


FIND MIDPOINT The endpoints of RS are R(1,–3)

and S(4, 2). Find the coordinates of the midpoint

M.

Example 1 Using the Midpoint Formula

M =

2,

2

1212 yyxx

SOLUTION

M =

R( 1, -3 ) S( 4, 2 ) x1

y1 x2y2

4 + 12

2 + -3

2,

M = 52

-1

2,


Example 2

Missing one Endpoint given a Midpoint

Given one endpoint (x1, y1) and a midpoint (x, y)

Missing Endpoint= ( 2x - x1, 2y - y1)

Find the endpoint K of JK if J is (1,

4) and the midpoint is M(2, 1)

SOLUTION J( 1, 4 ) M( 2, 1 ) x1

y1 x y

K = ( 2x- x1, 2y - y1)

K = (2)- 1 (1) - 4,2 2

K = 3 -2,


Try it yourself!

Midpoint Formula

Missing Endpoint = ( 2x - x1, 2y - y1)

2

yy,

2

xx 1212

Guided Practice

M = 4 5,

V = -6 -8,


Try it yourself!

Midpoint Formula

Missing Endpoint = ( 2x - x1, 2y - y1)

2

yy,

2

xx 1212

More Practice

Find the midpoint between

Find the missing endpoint with the given

Distance in the Coordinate Plane

·A

(x1, y1)

·B(x2, y2)

In the coordinate plane

B with coordinates (x2, y2)

whose endpoints are A with coordinate (x1, y1) and

the distance of a segment

is denoted by

2

12

2

12 )()( yyxx AB =

Distance AB

Distance Formula



1. What is the approximate length of RS with

endpoints R (2, 3) and S (4, -1).

Example Using the Distance Formula

RS =

SOLUTION

RS =

R( 2, 3 ) S( 4, -1 ) x1

y1 x2y2

4 – 22

-1 – 32

+

2

12

2

12 )()( yyxx

RS = 22

-4 2+

RS = 20 = 2 5


Try it yourself!

Distance Formula 2

12

2

12 )yy()xx(


Practice Work

Distance Formula 2

12

2

12 )yy()xx(

Find the distance between the 2 given points.

1. (2, 4) and (-1, 3) 2. (-3, 4 ) and (2, -1)

10D 2D 5

Pythagorean TheoremLesson 1.5

Solving for One missing side of a right triangle


The Pythagorean TheoremThe Pythagorean Theorem

In a right triangle, the square of the length of the

hypotenuse is equal to the sum of the squares of

the lengths of the two sides.

c

a

b

c2= a2 + b2

zx

y

z2 = x2 + y2

Note: Pythagorean Theorem is only use in RIGHT TRIANGLES to find one missing side.

Missing Hypotenuse

12

x5

SOLUTION

x = x = 13

In the right triangle, find x

22 12 5

x = 144 25

x = 169

THE PYTHAGOREAN THEOREM

52 + 122 = x2

52 + 122=x2

Missing side

Find the missing side of the right triangle.

x

14

7SOLUTION

x = 147

x =


37

x2 + 72 = 142

x2 + 𝟒𝟗 = 196

x2 = 196 - 49

x2 = 147


x = 240 x ≈ 15.492 Answer: D


Practice Problems

x

3

6

14x

Solve the missing side of each right triangle

1) 2)

37

lesson 1 - ddtwo.org

Documents