Consider a figure made by cutting two corners from a rectangle, as in the diagram to the right. The lengths of the sides are $$\begin{array}{lll}
\mathrm{AB} = 11, & \mathrm{BC} = 4, & \mathrm{CD} = 2\sqrt{13} \\
\mathrm{DE} = 5, & \mathrm{EF} = 2\sqrt{5}, & \mathrm{FA} = 6
\end{array}$$ We are to find the area of this figure. First, extend the sides of the figure as in the diagram and denote the sides forming the right angles by $x, y, u$ and $v$. Then $$u = \mathbf{A} - y, \quad v = x + \mathbf{B}.$$ Substituting these expressions in the equation $u^2 + v^2 = \mathbf{CD}$ and also using the equation $x^2 + y^2 = \mathbf{EF}$, we obtain $$x = \mathbf{G}.\, y - \mathbf{H}.$$ Then, since $$\mathbf{I}y^2 - \mathbf{J} = \mathbf{J}y - \mathbf{K} = 0,$$ we obtain $y = \mathbf{L}$. From this we have $x = \mathbf{M}$, and further $u = \mathbf{N}$ and $v = \mathbf{O}$. Finally we conclude that the area of this figure is $\mathbf{PQ}$.
Consider a figure made by cutting two corners from a rectangle, as in the diagram to the right. The lengths of the sides are
$$\begin{array}{lll}
\mathrm{AB} = 11, & \mathrm{BC} = 4, & \mathrm{CD} = 2\sqrt{13} \\
\mathrm{DE} = 5, & \mathrm{EF} = 2\sqrt{5}, & \mathrm{FA} = 6
\end{array}$$
We are to find the area of this figure.
First, extend the sides of the figure as in the diagram and denote the sides forming the right angles by $x, y, u$ and $v$. Then
$$u = \mathbf{A} - y, \quad v = x + \mathbf{B}.$$
Substituting these expressions in the equation $u^2 + v^2 = \mathbf{CD}$ and also using the equation $x^2 + y^2 = \mathbf{EF}$, we obtain
$$x = \mathbf{G}.\, y - \mathbf{H}.$$
Then, since
$$\mathbf{I}y^2 - \mathbf{J} = \mathbf{J}y - \mathbf{K} = 0,$$
we obtain $y = \mathbf{L}$.
From this we have $x = \mathbf{M}$, and further $u = \mathbf{N}$ and $v = \mathbf{O}$. Finally we conclude that the area of this figure is $\mathbf{PQ}$.