I was preparing for my end-sem exams and I ran into this problem that asks me to figure out the area of the triangle given its mid-points. Either I could've computed the vertices of the triangle which is very tedious or use this fact for $\Delta ABC$ with mid-points of sides $AB, BC, CA$ as $D, E, F$ respectively:
$$\text{ar}(\Delta DEF)=\dfrac{1}{4}\text{ar}(\Delta ABC)$$
I use the latter for $D(-1/5, 5/2), E(7, 3), F(7/2, 7/2)$ and build up the following expression:
$$\text{ar}(\Delta ABC)=4\cdot \underbrace{\dfrac{1}{2}\left|\begin{matrix}7/2 && 7/2 && 1\\ 7 &&3 && 1\\ -1/5 && 5/2 && 1\end{matrix}\right|}_{\text{ar}(\Delta DEF)}=\left|\begin{matrix}7 && 7 && 2\\ 7 &&3 && 1\\ -1/5 && 5/2 && 1\end{matrix}\right|=10.7 \text{ sq. units}$$
This does not agree with the answer in my textbook which says it to be $11$ sq. units. Is there a mistake in my reasoning? Thanks