Find the product $S1.S2$ in the figure below

Solution 1:

$S_{MXTR} = 2 \cdot S_{\triangle ARM} =8$

Note that $NJTX$ and $RTSQ$ are rhombus and hence $S_{MXTR} = \sqrt{S_1 \cdot S_2}$. Here is another way to look at it -

By midpoint theorem, $MQ \parallel BD$ and $AR = RT$. Similarly, $MN \parallel AC$

Also, can you see $\triangle ARQ \cong \triangle DSQ$ and $\triangle BXN \cong \triangle CJN$?

So, $NJTX$ and $RTSQ$ are rhombus with lengths $RM$ and $AR$ respectively. If $\angle ARM = \theta$,

$S_1 = RM^2 \sin\theta, S_2 = AR^2 \sin\theta$

$S_1 \cdot S_2 = (2 \cdot \frac 1 2 \cdot AR \cdot RM \cdot \sin\theta)^2 = (2 \cdot S_{\triangle ARM})^2 = 64$