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 a^{2} : b^{2}.
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
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