KCC's Quizzes AQQ284 about a ladder on a wall

Thank you for your request! Let's recheck the calculations step by step to ensure accuracy and evaluate confidence in the final result.

### Step 1: Ratios Between Similar Triangles
We know the smaller triangle at the box has:
- Vertical side (height) = 1 meter,
- Horizontal side (base) = 1 meter,
- Hypotenuse = \( \sqrt{2} \) meters.

The larger triangle formed by the ladder has:
- Hypotenuse = 10 meters,
- Base = \( x \),
- Height above the box = \( h \).

The ratio between the hypotenuse of the smaller triangle and the hypotenuse of the larger triangle is:
\[
\frac{\sqrt{2}}{10}.
\]

### Step 2: Height Calculation
Using the similarity of triangles:
\[
\frac{1}{h} = \frac{\sqrt{2}}{10}.
\]
Rearranging to solve for \( h \):
\[
h = \frac{10}{\sqrt{2}}.
\]

Simplify \( h \):
\[
h = 5\sqrt{2} \, \text{meters}.
\]

### Step 3: Total Height
Adding the box’s height (1 meter):
\[
H = h + 1 = 5\sqrt{2} + 1 \approx 7.071 + 1 = 8.071 \, \text{meters}.
\]

### Final Answer
After rechecking, the total height of the ladder's upper end above the floor is approximately **8.07 meters**.

### Confidence Level
I am **very confident** in the result because:
1. The problem relies entirely on geometric principles, which have been applied correctly.
2. Each step uses accurate calculations based on triangle similarity and the Pythagorean theorem.
3. No assumptions or errors were introduced during the recheck.

If you'd like to explore it further or visualize the geometry, I’m here to assist! 

It might still be not the the correct answer as it seems Co-Pilot takes the small triangle's hight as 1, but that's the height of the vertical side, not the Hypotenuse. I can't get Copilot to take this into consideration, even after 4 or 5 iterations and prompting the AI to take this into consideration and use the congruent relationship...

I actually found a website that solved the equations for me :)

(One) Solution seems to be 

0.5+sqrt(101)/2+sqrt(0.5*(49-sqrt(101)))

or

9.9379937m

Best regards,

Andre

If x=H, y =the floor distance, a=the top angle, we have tg(a)=y/x=1/(x-1) 

we have  (Pythagore): x^2+y^2=100.

x^2+x^2/(x-1)^2=100 

x^4 - 2x^3 - 98x^2 + 200x - 100=0

therefore (by successive approximations): x=H=9,937993..... m

Three possible (real positive) solutions:  9.99382,  1.1058  and 0.9045.   The two last ones are about a ladder almost horizontal.

How:  By proportional triangles (triangles B*H  and triangle 1*(H-1)) :  B/1  = H/(H-1)    where B is the base of the largest triangle.  Giving B= H/(H-1).  B^2 +H2 thus come to H^2/( H^2 -2H +1)  + H^2  = 10^2. 
We derive:  H^2 + H^4 -2H^3 +H = 100H^2 -200H +100,  or  H^4 -2H^3 -99H^2 +201H - 100 = 0.  There are four real roots, one is negative (-10.004... ).

Furthermore, 0.9045 is not acceptable either since H must be greater than 1.

For the symmetry of the problem you'll have 2 solutions. In fact you can "swap" the floor and the wall. The two solutions for the height, say it h, are

h1=9.938m

h2=1.112m

Let's call b the base of the bigger rectangular triangle generated by the ladder. 

h=x+1

b=y+1

 Where the values "1" come from the known red box dimensions.

By using Pitagora on the bigger triangle, you have

(x+1)^2 + (y+1)^2 =100

By using the proprierties of similar triangles, you can write: y=1/x, so

(x+1)^2 + (1/x + 1)^2=100

You get:

x^4+2x^3-98x^2+2x+1=0

By solving numerically you discard the two negative solutions and keep the two positive ones:

x1=8.938

x2=0.112

Then you simply use h=x+1 to get the final answers to the problem. 

h1=9.938m

h2=1.112m

The figure contains three similar right triangles.  The hypotenuse and base of the smallest one are marked as y and x respectively.

From that:

10/y = (1+x)/x = h/1

From the smallest triangle:

y**2 = 1+x**2

The two equations above give this equality for x:

[10x/(x+1)]**2 = 1+x**2

Rather than dealing with a fourth order polynomial, run this in LTspice:

FIND INTERSECTION

B1 n1 0 V=(10*time/(time+1))**2

B2 n2 0 V=1+time**2

.tran 150m

.end

The intersection is at t=112ms or x=0.112

H = 1.112/0.112 = 9.92857

Try to avoid 4th order poly...but its still complicated. 

Sum of two triangle hypotenuse =10

1/cos(theta) +1/sin(theta)=10...Theta angle between ground and ladder

The symetry is to be expected and the two intercepts avoid 4th order soln...

Solns for angle 6.3838degrees and 90-6.3838 degrees

Height up wall = 1.111879m and 9.937994m  The first soln is valid but not much use.

See Mind Your Decisions Channel for a special solution approach using geometry.