b The area scale factor between similar figures is a square of the length scale factor between them.
A
a 26
B
b 26units^2
Practice makes perfect
a Consider the given diagram.
We are told that BE is a midsegment of â–ł ACD. This means that triangles â–ł ABE and â–ł ACD are similar with a length scale factor of 2. This also means that the perimeter of â–ł ACD is twice that of â–ł ABE.
P_(â–ł ACD) = 2 P_(â–ł ABE)
Therefore, we can find the perimeter of â–ł ACD by first finding the perimeter of â–ł ABE. Since we are given all the side lengths of â–ł ABE, we can easily find its perimeter by adding them up.
P_(â–ł ABE) = 3 + 7 + 5 = 13
Next, let's substitute the found value of P_(â–ł ABE) into the equation relating it to the value of P_(â–ł ACD).
b Let's recall that the area scale factor between similar figures is a square of their linear scale factor.
Area scale factor
=
( Length scale factor )^2In Part A we found that the triangles â–ł ABE and â–ł ACD are similar with a length scale factor of 2. This means that the area of the bigger one is 2^2 = 4 times larger than the first one.
A_(â–ł ACD)= 4 A_(â–ł ABE)
Since we are given that A_(â–ł ABE) is 6.5units^2, we can substitute this value into equation above and solve the other area.
