here's the problem:
A ladder leans against a wall with its base 10 ft from it. When the ladder is pulled 3 ft further from the wall, the upper end slides down 7 ft. How long is the ladder?
Here's what I've done so far:
First, I did a drawing with the initial conditions. Assuming the right triangle I got:
$a^2+10^2=h^2$; $h$ is the length if the ladder.
From the other conditions, I created a new triangle adding 3 ft to the base, and substracting 7 ft from the other side of the triangle. I obtained:
$(a-7)^2+13^2=h^2$
Then I equaled the equations to get $a$ and got:
$a=\frac{59}{7}$
I went ahead and replaced it to find $h$, but the answer was not in the options. The possible outcomes are: 10 ft, 11 ft, 12 ft, 13 ft, 14 ft.
I don't see what am I doing wrong. I tried the classic approach with the differentials, but I got more confused, because that path is used when the rate in which the ladder falls is being ask, but this is not the case.
I would appreciate any help, thank you for reading.
$\endgroup$ 11 Answer
$\begingroup$Your calculations look fine. When you substitute $a=\frac{59}{7}$ into the original equation, you get
$$\left(\frac{59}{7}\right)^2+100=h^2$$
$$h=\sqrt{\frac{3481}{49}+\frac{4900}{49}}=\sqrt{\frac{8381}{49}} \approx 13.08 \ \text{ft}$$
While it's not exactly $13$, the best option to select would be $13$ ft.
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