TITLE: How to find the perimeter of two trapezoid figures inscribed in a rectangle?
QUESTION [0 upvotes]: The problem is as follows:

The figure from below represents a rectangle $ABCD$. Assume the area
of the right triangle $BQM$ is $120 \ \text{in}^2$ (in=inches).
Assume it also holds the following equality:
$$\csc \phi - \cot \phi =\frac{1}{5}$$

With all that information find the perimeter of the orange shaded
region.

The alternatives given in my book are as follows:
$\begin{array}{ll}
1.&\textrm{1.65 in}\\
2.&\textrm{1.72 in}\\
3.&\textrm{1.88 in}\\
4.&\textrm{1.56 in}\\
\end{array}$
The first part of this problem didn't seem difficult.
All what I did was as follows:
$$\csc \phi - \cot \phi =\frac{1}{5} \Rightarrow \csc \phi = \cot \phi + \frac{1}{5}$$
$$\csc \phi = \left(\cot \phi + \frac{1}{5}\right)^2=\cot^2 \phi+\frac{2}{5}\cot\phi+\frac{1}{25}$$
Then using: $1+\cot^2\phi=\csc^2\phi$
$$1+\cot^2\phi=\cot^2+\frac{2}{5}\cot\phi+\frac{1}{25}$$
$$\cot\phi=\frac{12}{5}$$
But the thing is that with the given information I can find $CN$ and $MA$ and $BQ$.
As the area is: $\frac{a^2\tan\phi}{2}=120$, $BQ=a=24$
But the thing is how to get $BC$? Is it just me or something is strange here?
Can someone help me here, please?

REPLY [1 votes]: Drop the perpendicular from $M$ onto $CD$ and call the foot $F$. Then $\triangle MFN$ and $\triangle BQM$ are similar and $FN+NC=BM=26$; since $NC=\frac{BQ}4=6$, $FN=20$ and hence $BC=MF=48$.
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