Find height of a wall if at the beginning it exceeds $10$ meters and then $8$ meters

Question

When the foot of a staircase is $5$ meters from the base of a wall, it protrudes $10$ meters above the wall; and if it is $9$ meters from the base, it stands $8$ meters.

Find the height of the wall.

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Using the Pythagorean theorem for the two situations I have

$$\begin{cases}(10-x)^2=5^2+y^2\\(8-x)^2=9^2+y^2,\end{cases}$$ where $x$ is the hypotenuse and $y$ is the height of the wall.

Solving that system of equations I have that $x=23$ and $y=\pm12$, but since a height is always positive, the solution is $\boxed{12~\text{meters}}$.

Is it correct?

I am not sure of the $x$ value because first it has a value but then it has another, so maybe we have to use $x_1$ and $x_2$, but then we have $3$ equations with $2$ variables, so it is not possible.

Thanks!

Answer 1

Let $z$ be the length of the ladders, then

$$(z-10)^2=5^2+y^2$$

$$(z-8)^2=9^2+y^2$$

subtracting the two equations, we have

$$-2(2z-18) = (5-9)(5+9)$$

$$4(z-9) = (9-5)(5+9)=4(14)$$

$$z=9+14=23$$

$$(23-10)^2-5^2=y^2$$

$$y^2=13^2-5^2=12^2$$

Hence $y=12$.

Note my definition of $z$ is not the hypothenus but the length of ladder.

Answer 2

The hypotenuse is $x - 10$ and $x - 8$ where $x$ is the total length of the staircase with overhang and is the same for both.

\begin{cases}(x-10)^2=5^2+y^2\\(x-8)^2=9^2+y^2\end{cases}

$(x-8)^2 - (x-10)^2 = 56$

$(x^2 - 16x + 64)-(x^2 - 20x +100) = 56$

$4x -36 = 56$

$x = 23$

$y = \sqrt{13^2 - 5^2} = 12$

Also $y = \sqrt{15^2 - 9^2} = 12$