Step 1: Coordinate setup.
Place $A(0,0),\ B(w,0),\ C(w,h),\ D(0,h)$ so $\text{Area}_{ABCD}=wh=90$.
Then $P\!\left(\frac{w}{3},0\right),\ Q\!\left(\frac{2w}{3},0\right)$ (trisect $AB$) and $R\!\left(\frac{w}{2},h\right)$ (midpoint of $CD$). The diagonal $AC$ is $y=\frac{h}{w}x$.
Step 2: Intersections with $AC$.
Line $PR$ has slope $\displaystyle \frac{h}{\frac{w}{2}-\frac{w}{3}}=\frac{6h}{w}$, so
\[
PR:\ y=\frac{6h}{w}\Big(x-\frac{w}{3}\Big).
\]
Intersect with $AC$: $\displaystyle \frac{h}{w}x=\frac{6h}{w}\Big(x-\frac{w}{3}\Big)⇒ x=\frac{2w}{5},\ y=\frac{2h}{5}.$
Hence $M\!\left(\frac{2w}{5},\frac{2h}{5}\right)$.
Similarly, $QR$ has slope $\displaystyle \frac{h}{\frac{w}{2}-\frac{2w}{3}}=-\frac{6h}{w}$, so
\[
QR:\ y=-\frac{6h}{w}\Big(x-\frac{2w}{3}\Big).
\]
Intersect with $AC$: $\displaystyle \frac{h}{w}x=-\frac{6h}{w}\Big(x-\frac{2w}{3}\Big)⇒ x=\frac{4w}{7},\ y=\frac{4h}{7}.$
Hence $N\!\left(\frac{4w}{7},\frac{4h}{7}\right)$.
Step 3: Area of $PQNM$ (shoelac(E).
Order the vertices $P\!\left(\frac{w}{3},0\right), Q\!\left(\frac{2w}{3},0\right), N\!\left(\frac{4w}{7},\frac{4h}{7}\right), M\!\left(\frac{2w}{5},\frac{2h}{5}\right)$. Then
\[
[\!PQNM\!]=\frac12\Big|\sum x_i y_{i+1}-\sum y_i x_{i+1}\Big|
=\frac{13}{105}\,wh.
\]
Since $wh=90$,
\[
[\!PQNM\!] = \frac{13}{105}\times 90=\frac{78}{7}\approx 11.14.
\]
\[
\boxed{[\!PQNM\!]=\frac{78}{7}\text{ sq. units }\approx 11.14}\ ⇒\ \text{falls in }(11,\,11.5].
\]