Pre-Algebra

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# Pre-Algebra - PowerPoint PPT Presentation

Learn to use properties of congruent figures to solve problems. 5-6. Congruence. Pre-Algebra. 5-6. Congruence. Pre-Algebra. A correspondence is a way of matching up two sets of objects. If two polygons are congruent, ALL of their corresponding sides and angles are congruent.

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5-6

Congruence

Pre-Algebra

5-6

Congruence

Pre-Algebra

A correspondence is a way of matching up two sets of objects.

If two polygons are congruent, ALL of their corresponding sides and angles are congruent.

In a congruence statement, the vertices in the second polygon are written in order of correspondence with the first polygon.

5-6

Congruence

55

55

Pre-Algebra

Example 1A: Writing Congruent Statements

Write a congruence statement for the pair of polygons.

The first triangle can be named triangle ABC. To complete the congruence statement, the vertices in the second triangle have to be written in order of the correspondence.

∠A≅∠Q, so ∠A corresponds to ∠Q.

∠B≅∠R, so ∠B corresponds to ∠R.

∠C≅∠P, so ∠C corresponds to ∠P.

The congruence statement is triangle ABC≅ triangle QRP.

5-6

Congruence

Pre-Algebra

Example 1B: Writing Congruent Statements

Write a congruence statement for the pair of polygons.

The vertices in the first pentagon are written in order around the pentagon starting at any vertex.

∠D≅∠ M, so ∠D corresponds to ∠M.

∠E≅∠ N, so ∠E corresponds to ∠N.

∠F≅∠ O, so ∠F corresponds to ∠O.

∠G≅∠ P, so ∠G corresponds to ∠P.

∠H≅∠Q, so ∠H corresponds to ∠Q.

The congruence statement is pentagon DEFGH≅ pentagon MNOPQ.

–8 –8

WX ≅ KL

a + 8 = 24

5-6

Congruence

a = 16

Pre-Algebra

Example 2A: Using Congruence Relationships to Find Unknown Values

A. Find a.

Subtract 8 from both sides.

6b = 30

ML ≅ YX

6b = 30

5-6

Congruence

6 6

Pre-Algebra

Example 2B: Using Congruence Relationships to Find Unknown Values

B. Find b.

Divide both sides by 6.

b = 5

5c = 85

∠J ≅∠V

5c = 85

5-6

Congruence

5 5

Pre-Algebra

Example 2C: Using Congruence Relationships to Find Unknown Values

C. Find c.

Divide both sides by 5.

c = 17

5-6

Congruence

Pre-Algebra

A correspondence is a way of matching up two sets of objects.

If two polygons are congruent, all of their corresponding sides and angles are congruent.

In a congruence statement, the vertices in the second polygon are written in order of correspondence with the first polygon.

5-6

Congruence

Pre-Algebra

Try This: Example 1A

Write a congruence statement for the pair of polygons.

The first trapezoid can be named trapezoid ABCD. To complete the congruence statement, the vertices in the second trapezoid have to be written in order of the correspondence.

A

B

|

60°

60°

||

||||

120°

120°

|||

D

C

∠A≅∠S, so ∠A corresponds to ∠S.

Q

R

|||

120°

120°

∠B≅∠T, so ∠B corresponds to ∠T.

||

||||

∠C≅∠Q, so ∠C corresponds to ∠Q.

60°

60°

|

∠D≅∠R, so ∠D corresponds to ∠R.

T

S

The congruence statement is trapezoid ABCD≅ trapezoid STQR.

5-6

Congruence

Pre-Algebra

Try This: Example 1B

Write a congruence statement for the pair of polygons.

The vertices in the first pentagon are written in order around the pentagon starting at any vertex.

110°

A

B

∠A≅∠M, so ∠A corresponds to ∠M.

110°

140°

140°

F

∠B≅∠N, so ∠B corresponds to ∠N.

C

110°

∠C≅∠O, so ∠C corresponds to ∠O.

E

110°

D

N

∠D≅∠P, so ∠D corresponds to ∠P.

110°

O

M

∠E≅∠Q, so ∠E corresponds to ∠Q.

140°

110°

110°

∠F≅∠L, so ∠F corresponds to ∠L.

P

140°

L

The congruence statement is hexagon ABCDEF≅ hexagon MNOPQL.

110°

Q

3a = 6

IH ≅ RS

3a = 6

5-6

Congruence

3 3

Pre-Algebra

Try This: Example 2A

A. Find a.

Divide both sides by 3.

3a

I

H

a = 2

6

4b°

S

R

120°

J

30°

Q

K

c + 10°

T

4b = 120

∠H ≅∠S

4b = 120

5-6

Congruence

4 4

Pre-Algebra

Try This: Example 2B

B. Find b.

Divide both sides by 4.

3a

I

H

b = 30°

6

4b°

S

R

120°

J

30°

Q

K

c + 10°

T

–10 –10

c + 10 = 30

∠K ≅∠T

c + 10 = 30

5-6

Congruence

Pre-Algebra

Try This: Example 2C

C. Find c.

Subtract 10 from both sides.

3a

I

H

c = 20°

6

90°

4b°

S

R

120°

90°

J

30°

c + 10°

Q

K

T