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–³ ACD. This means that triangles â–³ ABE and â–³ ACD are similar with a length scale factor of 2. This also means that the perimeter of â–³ ACD is twice that of â–³ ABE.
P_(â–³ ACD) = 2 P_(â–³ ABE)
Therefore, we can find the perimeter of â–³ ACD by first finding the perimeter of â–³ ABE. Since we are given all the side lengths of â–³ ABE, we can easily find its perimeter by adding them up.
P_(â–³ ABE) = 3 + 7 + 5 = 13
Next, let's substitute the found value of P_(â–³ ABE) into the equation relating it to the value of P_(â–³ 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–³ ABE and â–³ 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–³ ACD)= 4 A_(â–³ ABE)
Since we are given that A_(â–³ ABE) is 6.5units^2, we can substitute this value into equation above and solve the other area.
