Distance to the corner

Labelling horizontal and vertical distances using letters

Drawing on vertical and horizontal distances from the well to the corners and labelling them $a,b,c$ and $d$ gives the image below.

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From the blue rectangle, $a^2+b^2=x^2$.

From the green rectangle, $b^2+c^2=100$.

From the red rectangle, $a^2+d^2=121$. 

From the yellow rectangle, $c^2+d^2=25$.

Adding the equations given by the green and red rectangles gives $$\begin{array}{rl}{b}^2+c^2+a^2+d^2=& 100+121\\ \Rightarrow (a^2+b^2)+(c^2+d^2)=& 221\end{array}$$ And now substituting $a^2+b^2=x^2$ and $c^2+d^2=25$ gives $$\begin{array}{rl}& x^2+25=221\\ \Rightarrow & x^2=196\\ \Rightarrow & x=14\end{array}$$

Using a diagram

In the diagram below, the courtyard has been split into 4 rectangles whose corners are at the well. The squares drawn on are for the diagrammatic representation of Pythagoras' theorem, so that the area of shaded square A added to the area of shaded square B is equal to $x^2$, because $x$ is the hypotenuse of the triangle in the blue rectangle.

Image

 

Without using the blue rectangle, how else can we get A+B?

The square of the hypotenuse in the red rectangle is A+C, and the square of the green rectangle is B+D. Adding these together gives A+B+C+D, which is too much by C+D. But the square of the hypotenuse in the yellow rectangle is C+D.

So A+B=11$^2$+10$^2-$5$^2$=196.

So $x^2$ = 196, so $x$ = 14.

Using vectors and the scalar product

In the diagram below, the vectors from the well to the four corners of the rectangle are labelled $\bf{a,b,c}$ and $\bf{x}$.

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From the diagram below, it can be seen that the vector $\bf{x}$ is equal to $\bf{a}+\bf{c}-\bf{b}$, and that the sides of the rectangle can be written as $\bf{b}-\bf{a}$ and $\bf{c}-\bf{b}$.

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The sides of the rectangle are perpendicular, so $$\begin{array}{rl}\left(\mathbf{b}\mathbf{-}\mathbf{a}\right).\left(\mathbf{c}\mathbf{-}\mathbf{b}\right)& =0\\ \Rightarrow \mathbf{b}\mathbf{.}\mathbf{c}\mathbf{-}\mathbf{b}\mathbf{.}\mathbf{b}\mathbf{-}\mathbf{a}\mathbf{.}\mathbf{c}\mathbf{+}\mathbf{a}\mathbf{.}\mathbf{b}& =0\\ \Rightarrow \mathbf{b}\mathbf{.}\mathbf{c}\mathbf{-}\mathbf{b}\mathbf{.}\mathbf{b}\mathbf{+}\mathbf{a}\mathbf{.}\mathbf{b}\mathbf{=}\mathbf{a}\mathbf{.}\mathbf{c}\end{array}$$ And we are looking for $\sqrt{\bf{x.x}}$, where $$\begin{array}{rl}\mathbf{x}\mathbf{.}\mathbf{x}& =\left(\mathbf{a}\mathbf{+}\mathbf{c}\mathbf{-}\mathbf{b}\right).\left(\mathbf{a}\mathbf{+}\mathbf{c}\mathbf{-}\mathbf{b}\right)\\ & =\mathbf{a}\mathbf{.}\mathbf{a}\mathbf{+}\mathbf{b}\mathbf{.}\mathbf{b}\mathbf{+}\mathbf{c}\mathbf{.}\mathbf{c}\mathbf{+}\mathbf{2}\mathbf{a}\mathbf{.}\mathbf{c}\mathbf{-}\mathbf{2}\mathbf{a}\mathbf{.}\mathbf{b}\mathbf{-}\mathbf{2}\mathbf{c}\mathbf{.}\mathbf{b}\end{array}$$So, substituting $\bf{b.c-b.b+a.b}=\bf{a.c}$, $$\begin{array}{rl}\mathbf{x}\mathbf{.}\mathbf{x}& =\mathbf{a}\mathbf{.}\mathbf{a}\mathbf{+}\mathbf{b}\mathbf{.}\mathbf{b}\mathbf{+}\mathbf{c}\mathbf{.}\mathbf{c}\mathbf{+}\mathbf{2}\left(\mathbf{b}\mathbf{.}\mathbf{c}\mathbf{-}\mathbf{b}\mathbf{.}\mathbf{b}\mathbf{+}\mathbf{a}\mathbf{.}\mathbf{b}\right)\mathbf{-}\mathbf{2}\mathbf{a}\mathbf{.}\mathbf{b}\mathbf{-}\mathbf{2}\mathbf{c}\mathbf{.}\mathbf{b}\\ & =\mathbf{a}\mathbf{.}\mathbf{a}\mathbf{+}\mathbf{b}\mathbf{.}\mathbf{b}\mathbf{+}\mathbf{c}\mathbf{.}\mathbf{c}\mathbf{-}\mathbf{2}\mathbf{b}\mathbf{.}\mathbf{b}\\ & =\mathbf{a}\mathbf{.}\mathbf{a}\mathbf{+}\mathbf{c}\mathbf{.}\mathbf{c}\mathbf{-}\mathbf{b}\mathbf{.}\mathbf{b}\end{array}$$

But $\bf{a.a}$$={10}^2=100,\$\$\mathbf{b}\mathbf{.}\mathbf{b}$$=5^2=25,$ $\bf{c.c}$$={11}^2=121\$,so\$\mathbf{x}\mathbf{.}\mathbf{x}$$=100+121-25=196=14^2$.

So the remaining distance is $14$.