Welcome to our community

Be a part of something great, join today!

Find MN

Albert

Well-known member
Jan 25, 2013
1,225
Find MN.JPG
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
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:

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

Albert

Well-known member
Jan 25, 2013
1,225
MN.JPG