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.

Image

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 $$\begin{align}b^2+c^2+a^2+d^2=&100+121\\ \Rightarrow \left(a^2+b^2\right)+\left(c^2+d^2\right)=&221\end{align}$$ And now substituting $a^2+b^2=x^2$ and $c^2+d^2=25$ gives $$\begin{align}&x^2+25=221\\ \Rightarrow&x^2=196\\ \Rightarrow &x=14\end{align}$$

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 $\bf{c}-\bf{b}$.

Image

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

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

So the remaining distance is $14$.