The minorities

Let the height and base width of the big parallelogram $ABCD$ be $h$ and $b$ respectively. Then the area $[ABCD] = hb = 12 + 36 + 48 + 24 = 120$ .

Since $AMSQ$ and $BNSM$ have the same height, $\dfrac {QS}{SN} =$ $\dfrac {[AMSQ]}{[BNSM]} =$ $\dfrac {12}{36} =$ $\dfrac 13$ $\implies QS = \dfrac 14b$ . Similarly, $\dfrac {QT}{TN} =$ $\dfrac {[DQTP]}{[CPTN]} =$ $\dfrac {24}{48} =$ $\dfrac 12$ $\implies QT = \dfrac 13b$ .

Now, we note that the area colored blue:

$\begin{aligned} [BTDS] & = \frac 12 (ST) h \\ & = \frac 12 (QT-QS)h \\ & = \frac 12 \left(\frac b3 - \frac b4\right) h \\ & = \frac 12 \cdot \frac 1{12} \cdot \color{#3D99F6}bh & \small \color{#3D99F6} \text{Note that }bh = 120 \\ & = \frac 12 \cdot \frac 1{12} \cdot 120 \\ & = \boxed{5} \end{aligned}$