# Find MN

#### MarkFL

Staff member
Here is my solution using integral calculus:

Without loss of generality, I choose to let the trapezoid to be mapped to the first quadrant region bounded by the coordinate axes, the line $x=1$ and the line $y=-2\sqrt{2}x+8\sqrt{2}$.

Next, choose a value $x=c$ such that:

$$\displaystyle \int_0^c-2\sqrt{2}x+8\sqrt{2}\,dx=\int_c^1-2\sqrt{2}x+8\sqrt{2}\,dx$$

Applying the FTOC, we find:

$$\displaystyle -\sqrt{2}c^2+8\sqrt{2}c=-\sqrt{2}+8\sqrt{2}-\left(-\sqrt{2}c^2+8\sqrt{2}c \right)$$

After simplifying, we have:

$$\displaystyle 2c^2-16c+7=0$$

Taking the root such that $$\displaystyle 0\le c\le1$$ we obtain:

$$\displaystyle c=4-\frac{5}{\sqrt{2}}$$

Hence:

$$\displaystyle \overline{MN}=y\left(4-\frac{5}{\sqrt{2}} \right)=-2\sqrt{2}\left(4-\frac{5}{\sqrt{2}} \right)+8\sqrt{2}=10$$