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Similar Polygons - Section 4.6

Similar Polygons - Section 4.6

Author: Michele Harris
Description:

The student will be able to prove, both formally and informally, if polygons are similar.

Similar Polygons, high school geometry

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Tutorial

Definitions

Similar polygons - 2 polygons are similar ( ~ ) if their corresponding angles are = and their corresponding sides are proportional.

​Corresponding angles - 2 = angles in 2 similar polygons ( one angle in each polygon), in the same position

​Corresponding sides - 2 sides in 3 similar polygons (one side in each polygon) which join the vertices of 2 pairs of corresponding angles.

*The following 2 conditions are necessary if 2 polygons are to be similar:

​1.  The corresponding angles must be equal

2.  The corresponding sides must be proportional

Ratio of similitude - the ratio of 2 corresponding sides of 2 similar polygons

*No new theorems or corollaries

Now, it's time to try some problems.  Be sure to look at what the problem has given you and draw your diagrams before beginning the formal proofs.  Write your answers on paper before looking at the answers.  You will have a hard time understanding the questions if you don't try them first :)  The first 4 exercises are informal.

Source: ABEKA Plane Geometry 2006

Exercise #1

​  Are 2 congruent polygons necessarily similar?  Why/  If similar, what would be the ration of similtude?

 

Source: ABEKA Plane Geometry 2006

Answer for #1

​Yes.  Their angles and sides are equal respectively.  This is the definition of congruency.  Therefore,  the corresponding sides form the constant ratio of similtude, 1.

Source: ABEKA Plane Geometry 2006

Exercise #2

​Are any two similar polygons necessarily congruent?  Why?

Source: ABEKA Plane Geometry 2006

Answer for #2

​No.  The angles are equal respectively, but the sides are not necessarily equal respectively.  Each side in one polygon might be twice the size of the corresponding side in the other polygon.

Source: ABEKA Plane Geometry 2006

Exercise #3

Show that 2 similar polygons must have the same number of sides.  This can be done in informal proof form.

 

                       

Source: ABEKA Plane Geometry 2006

Answer for #3

​If one polygon had one more side than the other, there could be no corresponding side to form a ratio with this side.  Also, there would be no angle in the second polygon to correspond to the extra angle in the first polygon.

Source: ABEKA Plane Geometry 2006

Exercise #4

Are any 2 regular polygons similar?  Explain your answer.  Would 2 regular hexagons be similar?

Source: ABEKA Plane Geometry 2006

Answer for #4

​Only if they have the same number of sides.  In this problem, the angles are = respectively and the ratios of corresponding sides would be = (according to axiom 5).  2 regular hexagons would be similar.

Source: ABEKA Plane Geometry 2006

Exercise #5

Prove that two similar polygons are congruent if their ratio of similtude is 1.

Given:  Polygon ABCD.... ~ Polygon A'B'C'D'......

              ratio of simultudes = 1

Statements for Proof:

1.  Polygon ABCD.... ~ Polygon A'B'C'D'......

2.  AB: A'B' = BC:B'C' = CD:C'D'........

3.  AB:A'B' = 1,  BC:B'C' = 1, CD:C'D' = 1.......

4.  AB = A'B';  BC = B'C';  CD = C'D'.....

5.  <A = <A';   <B = >B';   <C = <C';  <D = <D'

6.  Polygon ABCD..... is congruent to Polygon A'B'C'D'........

Source: ABEKA Plane Geometry 2006

Reasons for proof, exercise #5

​1.  given ( of course!)

​2.  definition of ~ polygons

​3.  definition of ratio of similitude

​4.  In any proportion the product of the means equals the products of the extremes

​5.  definition of ~ polygons

​6.  definition of congruent polygons

Source: ABEKA Plane Geometry 2006

Exercise #6

Prove that any two equilateral triangles are similar.

*don't forget to draw your diagrams

Given:  Equilateral triangles ABC and DEF

​Prove:  the triangles are ~

​statements for proof:

​1.  

​2.  <A, <B, <C, <D, <E, <F = 60 degrees

​3.  <A = <D;  <B=<E; <C=<F

​4.  AB = BC = CA;  DE = EF = FD

​5.  AB:DE = BC:EF = CA:FD

​6.  triangle ABC ~ triangle DEF

 

 

 

Source: ABEKA Plane Geometry 2006

Reasons for proof, exercise #6

​2.  each angle of an equilateral triangle is 60 degrees

​3.  substitution

​4.  definition of equilateral triangles

​5.  division

​6.  definition ~ polygons

Source: ABEKA Plane Geometry 2006

Exercise #7

Prove that any two squares are similar.

​Given:  squares ABCD, EFGH

​Prove:  Square ABCD ~ Square EFGH

*the squares do not necessarily have to be the same size.

​Statements for proof:

​2.  AB = BC = CD = DA, EF = FG = GH = HE

3.  AB:EF = BC:FG = CD:GH = DA:HE

4.  Square AC and square EG are rectangles

5.  <A = right angle

     <E = right angle

6.  Draw AC, BD, EG, FH

7.  AC = BD, EG = FH

8.  triangle ABC is congruent to triangle BCD is congruent to triangle CDA is congruent to triangle DAB

triangle EFG is congruent to triangle FGH is congruent to triangle GHE is congruent to triangle HEF

9.  <A = <B = <C = <D = <E = <F = <G = <H

10.  <A = <E, <B = <F, <C = <G, <D = <H

11.  Square AC ~ square EG 

 

Source: ABEKA Plane Geometry 2006

Reasons for proof, exercise #7

​2.  all sides of a square are equal

​3.  division

​4.  definition of a square

​5.  definition of a rectangle

​6.  two points determine a straight line

​7.  the diagonals of a rectangle are equal

​8.  S.S.S.

​9.  C.P.C.T.E.

​10.  substitution

​11.  definition of similar polygons

Source: ABEKA Plane Geometry 2006

Exercise #8

Prove that any two equilangular triangles are similar.

​Given:  triangle ABC and triangle DEF are both equilangular

​Prove:  triangle ABC ~ triangle DEF

Source: ABEKA Plane Geometry 2006

Reasons for proof, exercise #8

1.  given

​2.  an equilangular triangle is also equilateral

​3.  any 2 equilateral triangles are similar

Source: ABEKA Plane Geometry 2006