[Guest]

Look at the diagram. There are two walls that are not vertical, but are parallel. There are two ladders placed in it that intersect each other at point X. One side of each ladder is touching one wall and the other touches the intersection of other wall and floor. The top of bigger ladder is vertically 5m above the top of smaller one. The top of smaller one is vertically 4m from the point of intersection of the two ladders. Now what is the height of the point of intersection X from the ground(d=?). I guess everything is clear in the diagram.

Edited April 17, 2008 by imran

[Guest]

Imagine a horizontal line across the figure through X. That creates 2 pairs of similar triangles (easy to prove using vertical angles and alternate interior angles). The ratio that's created by the similar triangles is d/9=4/d. Therefore d^2=36 and d=6.

I'm assuming the fact that the lines are not vertical is negligible.

[Guest]

Imagine a horizontal line across the figure through X. That creates 2 pairs of similar triangles (easy to prove using vertical angles and alternate interior angles). The ratio that's created by the similar triangles is d/9=4/d. Therefore d^2=36 and d=6.

I'm assuming the fact that the lines are not vertical is negligible.

The lines are not vertical and can be significantly tilted.

Edited April 17, 2008 by imran

[Guest]

The lines are not vertical and can be significantly tilted.

Even if the lines are not vertical, the above answer is still right

Following the Thales Theorem we get the following equations were m1 and m2 are the lengths of the 2 parts of ladder one and l1 and l2 are the lengths of the 2 parts of ladder two:

m1/m2 = 9/d

l1/l2 = d/4

but knowing that the two walls are parallel makes the two triangles formed by the ladders and the walls similar triangles and thus, l1/l2 = m1/m2

We then get d/4 = 9/d and so d = 6

[Guest]

Even if the lines are not vertical, the above answer is still right

Following the Thales Theorem we get the following equations were m1 and m2 are the lengths of the 2 parts of ladder one and l1 and l2 are the lengths of the 2 parts of ladder two:

m1/m2 = 9/d

l1/l2 = d/4

but knowing that the two walls are parallel makes the two triangles formed by the ladders and the walls similar triangles and thus, l1/l2 = m1/m2

We then get d/4 = 9/d and so d = 6

Yes I knew that result will be same but it needed some reasoning which you provide.

Good job both ChuckJerry and mojail.

[Guest]

Indeed the answer remains the same even if the lines are not vertical.

From parallel lines property, Angle(fab)=Angle(fde) = alpha.

Triangles dxc and dfb are similar ==> df/dx=(d+9)/d

On the other hand we have xf=df-dx. Combining with the above equation, we get xf=dx(9/d).

Triangles axf and gxd are also similar. ==> af/dg=xf/xd.

==> (d+9)sin(alpha)/(d+4)sin(alpha) = xd(9/d)/xd

Solving for d, we get d=6

[Guest]

Indeed the answer remains the same even if the lines are not vertical.

From parallel lines property, Angle(fab)=Angle(fde) = alpha.

Triangles dxc and dfb are similar ==> df/dx=(d+9)/d

On the other hand we have xf=df-dx. Combining with the above equation, we get xf=dx(9/d).

Triangles axf and gxd are also similar. ==> af/dg=xf/xd.

==> (d+9)sin(alpha)/(d+4)sin(alpha) = xd(9/d)/xd

Solving for d, we get d=6

You get the right answer but your explanation is not clear. Also you are using two d's which make it confusing. Maybe you can clarify it.

[Guest]

You get the right answer but your explanation is not clear. Also you are using two d's which make it confusing. Maybe you can clarify it.

From parallel lines property, Angle(A) in triangle FAD=Angle(D) in triangle (GDE) = alpha.

Triangles DXC and DFB are similar (Angle-Angle similarity) ==> DF/DX=(d+9)/d

On the other hand we have XF=DF-DX. Combining with the above equation, we get XF=DX*(9/d) ............(i)

Triangles AXF and GXD are also similar (Angle-Angle similarity). ==> AF/GD=XF/XD. ..................(ii)

==> (d+9)sin(alpha)/(d+4)sin(alpha) = DX*(9/d)/XD [combining (i) and (ii)]

==> (d+9)/(d+4) = 9/d ....... the other terms cancel each other.

Solving for d, we get d=6

QED.

[Guest]

From parallel lines property, Angle(A) in triangle FAD=Angle(D) in triangle (GDE) = alpha.

Triangles DXC and DFB are similar (Angle-Angle similarity) ==> DF/DX=(d+9)/d

On the other hand we have XF=DF-DX. Combining with the above equation, we get XF=DX*(9/d) ............(i)

Triangles AXF and GXD are also similar (Angle-Angle similarity). ==> AF/GD=XF/XD. ..................(ii)

==> (d+9)sin(alpha)/(d+4)sin(alpha) = DX*(9/d)/XD [combining (i) and (ii)]

==> (d+9)/(d+4) = 9/d ....... the other terms cancel each other.

Solving for d, we get d=6

QED.

Now clear.

You have a mistake(which in this case does not effect result ) . The length of AF should be (d+9)/sin(alpha) not (d+9)sin(alpha). Similarly it should be

(d+4)/sin(alpha) instead of (d+4)sin(alpha). But since it appears in both numerator and denominator so cancels out.

Good job though.

[Guest]

Now clear.

You have a mistake(which in this case does not effect result ) . The length of AF should be (d+9)/sin(alpha) not (d+9)sin(alpha). Similarly it should be

(d+4)/sin(alpha) instead of (d+4)sin(alpha). But since it appears in both numerator and denominator so cancels out.

Good job though.

That is true. In the diagram GD = (d+4)/sin(alpha) ... [not d+9].

Thanx for the correction.