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? please 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
