The areas of similar figures are proportional to the square of their corresponding sides
since the areas are in the ratio of 9:1
the sides are in the ratio of 3:1
so AD/EH = 3/1
AD/10 = 3/1
AD = 30
Perimeter of ABCD = 165
Perimeter EFGH/Perimeter ABCD = 1/3
Perimeter EFGH/165 = 1/3
perimeter EFGH = 55
I am still lost.
How come the areas of similar figures (trapezoids) are proportional to the squeare of their corresponding perimeters.
let me illustrate with an example
consider the right-angled triange with sides 3, 4, and 5
(easy to find areas of right-angles triangles)
a triangle similar is 12, 16, 20 ( I multiplied each side by 4)
area of smaller = (1/2)(3)(4) = 6
area of larger = (1/2)(12)(16) = 96
notice the ratio of their sides = 1:4
ratio of their areas = 6:96 = 1:16 = 1^2 : 4^2
which is the square of their sides.
now look at the perimeters
per of smaller = 3+4+5 = 12
per of larger = 12+16+20 = 48
notice the ratio of perimeters is 12:48 = 1:4
the same as the ratio of sides
Perimeter is a linear relationship
while area is a second degree relationship
Does that help?
Does this work for all trapezoids ?
Is this true for all similar enclosed figures ?
Can you help me with my other question posted minutes later than this one ?
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