For problem 5.6, parts a-d are all pretty much the same. Here I will do part d) since a hair more complex. The steps the same for the others.
Since AB = CD, we can substitute the following
AB is x + 6(x+2)
CD is 2(x+10)
To get
x + 6(x+2) = 2(x+10)
So to find the length of AB and CD, I need to figure out x. So solve the equation.
x + 6(x+2) = 2(x+10)
x + 6x + 12 = 2x + 20 – I just multiplied everything out
x + 6x – 2x = 20 – 12 > I moved all the x terms to the left and all the constants (numbers) to the right
5x = 8 > did the math
x = 8/5 = 1.6
Now I have x, but the problem asked for the lengths.
So now we can find the parts
AB = x + 6(x+2) = 1.6 + 6(1.6 + 2) = 1.6 + 6(3.6) = 23.2
AE = x = 1.6
EB = 6(x+2) = 21.6
CD = AB = 23.2
To do the others – follow the same process
