FINDING SCALE FACTOR OF SIMILAR FIGURES AND FIND PERIMETER

If the sides of a rectangle are multiplied by k, a similar rectangle is obtained.

Finding the perimeter :

Perimeter of old shape = 2 (a + b)

Perimeter of new shape = 2(ka + kb)

= 2k(a + b)

Note :

If two shapes are similar, then their perimeter are proportional to the measures of the corresponding sides.

More simply, if two shapes are in the ratio a : b, then its perimeter will also be in the ratio a : b.

If two shapes are in the ratio a : b, then its area will be in the ratio a² : b². 

The figures in each pair are similar. Compare the first figure to the second. Give the ratio of the perimeters and the ratio of the areas.

Problem 1 :

Solution :

Ratio of corresponding sides = 2 : 4

Ratio of perimeter = a : b

= 2 : 4

a : b = 1 : 2

Ratio of area = a² : b²

= 1² : 2²

a² : b² = 1 : 4

Problem 2 :

Solution :

Ratio of corresponding sides = 15 : 25

Ratio of perimeter = a : b

= 15 : 25

a : b = 3 : 5

Ratio of area = a² : b²

= 3² : 5²

a² : b² = 9 : 25

Problem 3 :

Solution :

Ratio of corresponding sides = 8 : 6

Ratio of perimeter = a : b

= 8 : 6

a : b = 4 : 3

Ratio of area = a² : b²

= 4² : 3²

a² : b² = 16 : 9

Find the scale factor and the ratio of perimeters for each pair of similar figures.

Problem 4 :

Two regular octagons with areas 4 ft² and 16 ft²

Solution :

a² : b² = 4 : 16

a : b = √4 : √16

a : b = 2 : 4

a : b = 1 : 2

Problem 5 :

Two trapezoids with areas 49 cm² and 9 cm²

Solution :

a² : b² = 49 : 9

a : b = √49 : √9

a : b = 7 : 3

Problem 6 :

Two equilateral triangles with areas 16√3 ft² and √3 ft²

Solution :

a²: b² = 16√3: √3

a: b = 4: 1

Problem 7 :

Two circles with areas 2π cm² and 200π cm²

Solution :

a² : b² = 2π : 200π

a : b = 1 : 100

Problem 8 :

Figure A has a perimeter of 60 inches and one of the side lengths is 5 inches. Figure B has a perimeter of 84 inches. Find the corresponding missing side length.

Solution :

Perimeter of figure A = 60 inches

One side length = 5 inches

Perimeter of figure B = 84 inches

Let x be the missing side length.

Ratio of the corresponding sides is equal to ratio of perimeters

5 : x = 60 : 84

5/x = 60/84

84(5) = 60x

x = 84(5) / 60

x = 7

So, the missing side is 7 inches.

Problem 9 :

Figure A has an area of 4928 square feet and one of the side lengths is 88 feet. Figure B has an area of 77 square feet. Find the corresponding missing side length.

Solution :

Area of figure A = 4928 square feet

One side length = 88 feet

Area of figure B = 77 square feet

Let x be the missing side length.

Ratio of squares of corresponding sides is equal to ratio of areas of similar figures.

88² : x² = 4928 : 77

88² / x² = 4928 / 77

7744 / x² = 4928 / 77

 4928x² = 7744(77)

x² = 7744(77)/4928

x² = 121

x = 11 feet

So, the missing measure is 11 feet.

Problem 10 :

a) For the pair of similar triangles below, list the corresponding angles and the corresponding sides.

b) Determine all unknown side lengths to the nearest tenth of a unit and determine all unknown angles

Solution :

Triangles FJG and HJI

<FJG = HJI

<FGJ = <JHI

Using AA the triangles are similar.

FJ/JI = FG/HI = GJ/HJ

18.9/6.3 = FG/4.3

FG = 18.9(4.3) / 6.3

= 12.9

Since it is right triangle, we may use Pythagorean theorem, we get

18.9² = 12.9² + GJ²

357.21 = 166.41 + GJ²

GJ² = 357.21 - 166.41

GJ² = 190.8

GJ = √190.8

= 13.8

Scale factor = 18.9 : 6.3

= 18.9/6.3

= 189/63

= 3/1

= 3 : 1

Using the scale factor,

GJ : HJ = 3 : 1

13.8 : HJ = 3 : 1

13.8/HJ = 3/1

HJ = 13.8/3

= 4.6

a) 

<FJG = HJI

<FGJ = <JHI

FJ/JI = FG/HI = GJ/HJ

b) GJ = 13.8, HJ = 4.6 and FG = 12.9