Triangles $ABC$ and $DEF$ are identical with $EF=10, EO=2, CO=6.$
Find the shaded area.
I have tried my best to solve the area-finding question mentioned below, but nothing is working out. So anyone interested in helping me out with this problem.
Q: As shown in the figure, two identical triangles are partially overlapped. $BC\parallel EF$, points $A$, $D$, $C$, and $F$ are on the same straight line; $OC=6\text{cm}$, $OE=2\text{cm}$, $EF=10\text{cm}$.
First off, $\angle{DAB}=\angle{CDE}$ as the triangles are identical. This means that ${AB}\parallel{DE}$. So, $\angle{DOC}=90°$. So, the triangles $DOC$ and $DEF$ are similar by $AA$. If we let $DO=x$, then using similarity, $\frac{x}{x+2}=0.6$. Solving this, we get $x=3$. Now, the shaded area is given by $area(ABC)-area(DOC)=25-9=16$.
Edit: Though you haven't specified it, I've assumed the triangles are right triangles because they're drawn to look that way.
Edit 2: Just saw your comment.
Look at this figure
You must have the same slope for DF and DC.
This condition can be written, with the notations of this figure :
$$\frac{DE}{EF}=\frac{DO}{OC} \ \iff \ \frac{x+2}{10}=\frac{x}{6}$$
a first degree equation giving $x:= DO = 3$.
As a consequence :
$$ \text{area}(DEF)=\text{area}(ABC)= \frac12(3+2)\times 10=25$$ $$\text{ and area}(DOC)=\frac12 6 \times 3=9$$
Conclusion : the area of trapezoid $ABOD$ is
$$\text{area}(ABC) - \text{area}(DOC) = 25 - 9 =16$$
