b The area scale factor between similar figures is the square of the length scale factor between them.
A
a 48 units
B
b 216cm^2
Practice makes perfect
a Consider the given diagram.
We are told that BE is the midsegment of â–³ ACD. This means that â–³ ABE and â–³ ACD are similar triangles with the length scale factor of 2. From this, we know that the perimeter of â–³ ACD is twice that of â–³ ABE.
P_(â–³ ACD) = 2 P_(â–³ ABE)
We can find the perimeter of â–³ ACD if we first find 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) = 7 + 11 + 6 = 24
Next, let's substitute the value of P_(â–³ ABE) into the equation relating it to the value of P_(â–³ ACD).
b In Part A we determined that â–³ ABE and â–³ ACD are similar with a length scale factor of 2. Recall that the area scale factor between similar figures is the square of the linear scale factor between them.
area scale factor = ( length scale factor )^2Since â–³ ABE and â–³ ACD are similar with a length scale factor of 2, this means that the area of the larger triangle is 2^2 = 4 times larger than the area of the smaller triangle.
A_(â–³ ACD)= 4 A_(â–³ ABE)
We are given that the area of A_(â–³ ABE) is 54 square centimeters. Let's substitute this value into the equation and solve the larger area.
