What will be area of shaded region, in two rectangles, where 2 vertex are common?

Question

In a figure, there are two rectangles $ABCD$ and $DEBG$, each of lengh $7\ cm$ and width $3\ cm$. The area of shaded region. in $cm^2$ is approximately ?

Options given : $12$, $10$, $8$, $4$

Answer 1

Here, notice that $\Delta DEK$ and $\Delta BCK$ are congruent. If we find $|DK|$, we are done. By Pythagoras Theorem, $(7-a)^2 = a^2+9 \implies a = \frac{20}{7}$. Therefore, $|DK| = 7-\frac{20}{7} = \frac{29}{7}$ and shaded area is $3 \cdot \frac{29}{7} = \frac{87}{7} \approx12$.

Answer 2

Let $DG\cap AB=\{K\}$.

Thus, since $\Delta ABD\cong\Delta GDB$, we have $\measuredangle ABD=\measuredangle GDB$, which says $DK=KB$.

Let $KB=x$.

Thus, $$3^2+(7-x)^2=x^2,$$ which gives $$x=\frac{29}{7}$$ and the needed area it's $$\frac{29}{7}\cdot3=\frac{87}{7}.$$