Point $C$ is the midpoint of $\overline{AB}$, point $D$ is the midpoint of $\overline{AC}$, point $E$ is the midpoint of $\overline{AD}$, and point $F$ is the midpoint of $\overline{AE}$. If $AF=3$, what is the number of units in the length of $\overline{AB}$?
AB is 48 :)
Well, if we follow the chain of midpoints, it seems like we're going down a path of splitting lengths in half. So, I guess you could say we're going on a half journey!
To find the length of $\overline{AB}$, we need to determine the relationship between the lengths of each segment. Let's start from the given information and work our way up.
Since point $C$ is the midpoint of $\overline{AB}$, this means that $AC = CB$. Similarly, since $D$ is the midpoint of $\overline{AC}$, we know that $AD = DC$. And because $E$ is the midpoint of $\overline{AD}$, we have $AE = ED$. Finally, as given, $AF = 3$.
From the information above, we can now find the length of $\overline{AE}$. Since $E$ is the midpoint of $\overline{AD}$, we know that $AE = ED$. Therefore, the length of $\overline{AE}$ is also $3$.
Now, let's move on to the length of $\overline{AF}$. So far, we have $AE = 3$ and $AF = 3$. Since $F$ is the midpoint of $\overline{AE}$, we can conclude that $AF = FE = 3$.
Using the information we have, we can now calculate the length of $\overline{AB}$. Since $C$ is the midpoint of $\overline{AB}$, we know that $AC = CB$. From the previous steps, we have $AC = 3 + 3 = 6$. Therefore, the length of $\overline{AB}$ is twice the length of $AC$, which gives us $\overline{AB} = 2 \times 6 = 12$.
Hence, the length of $\overline{AB}$ is $12$ units.
Hi, I’m having a little trouble with math right now. Anyone willing to help me solve the problem above? Please? Thank you. And @Oobleck, I appreciate the effort you’re giving us so we can solve the rest on our own. Thank you. And for the other people that is struggling a little bit just like me, here’s a tip.
So if AF=3, and AE=6 and AD=12, you kind of see the pattern, right? 3, 6, and 12. So 3x2=6, and 3x4=12. They all contain 3! I think that’s a close enough hint. Thank you once again @Oobleck for your tips! I very much appreciate it!
I hope my tips help the rest of the students struggling too. 😘
anwer: 83
AF=3
AE=6
AD=12
...
and so on till you get to AB