I would use coordinate geometry here. Begin by defining the vertices of rectangle \(\displaystyle ABCD\):
\(\displaystyle A=(0,0)\)
\(\displaystyle B=(0,h)\)
\(\displaystyle C=(b,h)\)
\(\displaystyle D=(b,0)\)
Now, from the information given, we have:
\(\displaystyle \displaystyle E=\left(0,\frac{2}{3}h\right)\)
\(\displaystyle \displaystyle Z=\left(\frac{1}{2}b,h\right)\)
\(\displaystyle \displaystyle M=\left(\frac{1}{2}b,\frac{1}{3}h\right)\)
\(\displaystyle \displaystyle N=\left(\frac{3}{4}b,\frac{1}{2}h\right)\)
The formula for the area of a triangle whose vertices are 3 points in the plane is:
\(\displaystyle \displaystyle A=\frac{1}{2}\left|(x_3-x_1)(y_2-y_1)-(x_2-x_1)(y_3-y_1) \right|\)
Given that the area of \(\displaystyle \triangle EMN\) is 5 units squared, we may state:
\(\displaystyle \displaystyle 5=\frac{1}{2}\left|\left(\frac{3}{4}b-0\right)\left(\frac{1}{3}h-\frac{2}{3}h\right)- \left(\frac{1}{2}b-0\right)\left(\frac{1}{2}h-\frac{2}{3}h\right) \right|\)
From this, we obtain:
\(\displaystyle bh=60\)
Now, find the area of the quadrilateral \(\displaystyle EBZN\) by breaking it into two triangles, use the above formula for their areas in terms of the product \(\displaystyle bh\), and then add the areas of the 3 triangles to get the requested area.
