Similarity Learning

Introduction

     Similarity is defined as a property of being similar. It is denoted as ||| lr. We are learning polygons such that the angles of one are equal to the corresponding angles of the other but the corresponding sides are proportional.

     In similarity learning ,for two polygons (any) to be similar, they must have the same shape but not need the same size.

 

Triangle problem of similarity learning:

 

Example 1:  

     The areas of two similarity triangles are 20 cm2 and 45 cm2.The sums of their perimeters are 25 cm. Find the perimeter of each triangle.

      we are learning representation of triangles and call the two triangles t1 and t2 and let the scale factor of the two similarity triangles be a : b.

 area of triangle formula as follow:

 area t1/ area t2=(a/b)2  (From theorem)

                                   20/45= (a/b) 2

                   Reduce the fraction,

                                        4/9= (a/b)2

       Take square roots of both sides.

                                        2/3=a/b

 a: b is the reduced form of the scale factor. 2: 3 is then the reduced form

 of the comparison of the perimeters.

                         Let 2x = perimeter of t1

                        and 3x = perimeter of t2

Then 2x + 3x = 25 (The sum of the perimeters is 25 cm)

                   5x=25

                      x=5

Therefore Perimeter t1=2(5) =10 cm

                   Perimeter t2=3(5) =15 cm 

Answer: P t1=10 cm

                P t2 =15 cm

Example 2:

In below ΔPQR˜ΔMNW. Find the perimeter of ΔMNW.

Perimeter of ΔPQR/Perimeter of ΔMNW=PQ/MN

 24/Perimeter of ΔMNW=6/9

6(Perimeter of ΔMNW) =216    Using cross product property.

Perimeter of ΔMNW = 36 inches.

                 Answer: 36 inches

 

 

Finding area of similarity learning:

 

Example problem in similarity learning :

Finding the of similarity right angled triangles whose scale factor is 3 : 4.

               Area ΔABC=1/2(9) (12)

                                   =54

              Area ΔPQR=1/2(16) (12)

                                  =96

the ratio of the areas as folllow:

Now we can compare the ratio of the areas of the above similarity triangles.

                     Area ΔABC/area ΔPQR=54/96

                                                              = 9/16

                                                              = (3/4)2

                                               Answer : (3/4)2