In the figure below, be is parallel to cf , cf is parallel to dg and ab =5, bc =10, cd=2, and ae=8. whats the length of ag? tell me how you got your answer

Check out the attached image. It's essentially the same as the given figure but I've added on x and y. 

Given:
AB = 5
BC = 10
CD = 2
AE = 8
Unknowns:
EF = x
FG = y

The goal is to find the values for x and y. Once we know them we can find the length of AG

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Let's find x first. 
AB/BC = AE/EF
5/10 = 8/x
5x = 10*8 <<--- cross multiply
5x = 80
5x/5 = 80/5
x = 16
Or more simply: EF is twice as long as AE since BC is two times longer than AB

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Now use that x value to find y
AC/CD = FA/FG
(AB+BC)/CD = (AE+EF)/FG
(5+10)/2 = (8+x)/y
15/2 = (8+x)/y
15/2 = (8+16)/y <<--- plugged in x = 16
15/2 = 24/y
15y = 2*24 <<--- cross multiply
15y = 48
15y/15 = 48/15
y = 48/15
y = 16/5
y = 3.2

Therefore, AG is...
AG = AE+EF+FG
AG = 8+x+y
AG = 8+16+3.2
AG = 24+3.2
AG = 27.2

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choice C) 27.2

The answer is -pie over 2