Embed Size (px)
DESCRIPTION
1.2 – Segments and Congruence. Positive length between two points. Segments with the same length. A. B. C. D. Rules of geometry that are not proved. The distance between 2 points is the absolute value of the difference between them. 1. 2. 3. 4. 5. 6. 7. 8. 7 – 2. = 5 u. 2 – 7. - PowerPoint PPT Presentation
Citation preview
1.2 – Segments and Congruence
Distance
Congruent Segments
Positive length between two points
Segments with the same length
CDAB CDmABm
CDAB
A B
C
D
PostulateRules of geometry that are not proved
Ruler Postulate
The distance between 2 points is the absolute value of the difference between them.
1 2 3 4 5 6 7 8
7 – 2 = 5u 2 – 7 = 5u
Segment Addition Postulate
Adding two pieces of a segment total the whole
A B C
AB + BC = AC
1. Measure the length of the segment to the nearest tenth of a centimeter.
AB = ________cm 3.5
1. Measure the length of the segment to the nearest tenth of a centimeter.
CD = ________cm 1.6
1. Measure the length of the segment to the nearest tenth of a centimeter.
EF = ________cm 2.7
2. Find the indicated length.
GJ = _______
16
16u
2. Find the indicated length.
ST = _______
37 – 18 = 19
19u
2. Find the indicated length.
YZ = _______
2x + 3 + 3x – 1 = 27
5x + 2 = 27 -2 -2
5x = 25x = 5
YZ = 3(5) – 1 YZ = 15 – 1 YZ = 14
14u
3. Name the congruent segments using proper notation.
CDAB
CBAD
4. Use the number line to find the indicated measure.
JK = _____ KL = _____
JL = _____ MK = _____
5u 4u
9u 10u
5. Plot the given points in a coordinate plane. Then determine whether the line segments named are congruent.
A(2, 2), B(2, –1), C(0, –2), D(3, –2)
CDAB and
A
BC D
AB = 3u
CD = 3u
CDAB
5. Plot the given points in a coordinate plane. Then determine whether the line segments named are congruent.
E(–3, 2), F(1, 2), G(2, 3), H(2, –2)
GHEF and EF = 4u
GH = 5uE F
G
H
Not congruent
6. In the diagram, points P, Q, R, and S are collinear, PS = 46, PR = 18, and PQ = QR. Find the indicated length.
PQ = QR =
QS =
RS =
46189
9
28
9u
9u
37u
28u
7. Point J is between H and K.. Use the given information to draw a diagram and write an equation in terms of x. Solve the equation. Then find HJ and JK.
HJ = 2x – 5 JK = 6x HK = 27
H J K2x – 5 6x
27
2x – 5 + 6x = 278x – 5 = 27
+5 +58x = 328 8
x = 4
HJ = 2x – 5 HJ = 2(4) – 5HJ = 8 – 5HJ = 3u
JK = 6xJK = 6(4)JK = 24u
HJ + JK = HK
7. Point J is between H and K.. Use the given information to draw a diagram and write an equation in terms of x. Solve the equation. Then find HJ and JK.
HJ = 4x + 7 JK = 5x – 8 HK = 53
H J K4x + 7 5x – 8
53
4x + 7 + 5x – 8 = 539x – 1 = 53
+1 +19x = 549 9
x = 6
HJ = 4x + 7HJ = 4(6) + 7HJ = 24 + 7HJ = 31u
JK = 5x – 8 JK = 5(6) – 8 JK = 30 – 8
HJ + JK = HK
JK = 22u
HW Problem
# 27
1.2 12-14 1-19 odd, 21-29, 41
Ans: D
Compass, ruler, and protractor, and calculator dueBook Cover and Syllabus Due Monday!!!!!!
