Let $ABC$ be a triangle with base $AB$. Let $D$ be the midpoint of $AB$ and $P$ be the midpoint of $CD$.
Extend $AB$ in both direction. Assuming $A$ to be on the left of $B$, let $X$ be a point on $BA$ extended further left such that $XA = AD$. Similarly, let $Y$ be a point on $AB$ extended further right such that $BY = BD$. Let $PX$ cut $AC$ at $Q$ and $PY$ cut $BC$ at $R$.
Let the sides of $ABC$ be $AC = 13$, $BC = 14$, and $AB = 15$.
What is the area of the pentagon $PQABR$?
Area $CPQ=PQD=a$
Area $XQA=AQD=b$
Area $CRP=PRD=c$
Area $RDB=BRY=d$
The reason being that each such pair of triangles has the same base length and height.
Triangle $ABC$ has area $\Delta=2a+b+d+2c$
Triangle $XPY$ also has area $\Delta$ since the base is doubled and the height is halved (compared to $ABC$).
Therefore we also have the equation $2b+a+c+2d=\Delta$
Adding these equations gives $$3(a+b+c+d)=2\Delta$$
Hence the required area of the pentagon is $a+b+c+d=\frac 23 \Delta$
It remains just to calculate the area $\Delta$ of the triangle $ABC$ using Heron's Formula, giving $\Delta=84$, so the final answer is $56$