grade 9 math
posted by olav .
Trapezoid ABCD, top AD and bottom BC are parallel. AD is the smaller of the two parallel lines. Left AB is 45, bottom BC is 60, right CD is 30, top AD is unknown.
Trapezoid EFGH is similar to ABCD, top EH and bottom FG are parallel. EH is the smaller of the two parallel lines. Top EH is 10, other sides are unknown.
Area of ABCD is 9 times larger than EFGH.
What is the perimeter of EFGH ?
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 ?