Thread: Ladder Question

Hello Good-day,

I've attached below a diagram of the question: It says that 'A ladder leans against a wall AC, it vertical height is 4m above the ground. The ladder is then extended 0.8m without moving the base of the ladder and now it rests at point C, a vertical movement of 1m. What is the length of the ladder?'

Now i tried this first by trying to find the length of the ladder before it was extended, if i have that, then i can use Pyth theorem:

AB: Using Cos 45 = [tex]\frac{4}{x}[\math]


x= 5.66m > Which i add to 0.8m = 6.46m (wrong ans)

Another way i tried finding base AB:

Tan 45 = [tex]\frac{x}{5}[\math]


x = 5m



[tex]\sqrt{5^2+5^2}[\math] = 7.07m (wrong ans)



* Right ans is 6.03m Help...?!

Originally Posted by BobBali

Hello Good-day,

I've attached below a diagram of the question: It says that 'A ladder leans against a wall AC, it vertical height is 4m above the ground. The ladder is then extended 0.8m without moving the base of the ladder and now it rests at point C, a vertical movement of 1m. What is the length of the ladder?'

Now i tried this first by trying to find the length of the ladder before it was extended, if i have that, then i can use Pyth theorem:

AB: Using Cos 45 = [tex]\frac{4}{x}[\math]


x= 5.66m > Which i add to 0.8m = 6.46m (wrong ans)

Another way i tried finding base AB:

Tan 45 = [tex]\frac{x}{5}[\math]


x = 5m



[tex]\sqrt{5^2+5^2}[\math] = 7.07m (wrong ans)



* Right ans is 6.03m Help...?!

Are you given an angle in the question? Alternatively are you given AB?

No, no angle is given. I have opted to use 45deg in Angel C. And no base is given either. That's why i tried finding AB first, and then use that to find length of ladder using Pyth.

Originally Posted by BobBali

No, no angle is given. I have opted to use 45deg in Angel C. And no base is given either. That's why i tried finding AB first, and then use that to find length of ladder using Pyth.

You can't just guess an angle and hope for the best. Either way it is not necessary since you can use Pythagoras and simultaneous equations.

Using your diagram call the point 4m up from A as "D" and the length

By Pythagoras on triangle ABD


And on triangle ABC


Solve for h.

Oh..I see it now and the length comes out correctly. This question came from the chapter on Pythagoras from one of my 9th grades class, but what stumps me is how how do i teach my student to apply the concept of simultaneous equations when they haven't even done the topic on simultaneous equations, yet?! All the while i was thinking the above question dealt with Pythagorus only; the question is an exercise and not even part of the hard bonus questions at the end of the chapter... :-(

Thank you very much for your help.

Originally Posted by BobBali

... but what stumps me is how how do i teach my student to apply the concept of simultaneous equations when they haven't even done the topic on simultaneous equations, yet?! All the while i was thinking the above question dealt with Pythagorus only...

There is really no simultaneous equations involved; just the use of Pytagorean theorem twice;
since the base AB is common to both triangles (h = ladder - .8):
h^2 - 4^2 = (h + .8)^2 - 5^2
OR (ladder = h):
(h - .8)^2 - 4^2 = h^2 - 5^2