Find area of the triangle ABC, given the coordinates of vertices in plane 
$A, B$ and $C$ are the points $(7,3), (-4,1)$ and $(-3,-2)$ respectively. Find the area of the triangle $ABC$.

I've worked out the lengths of each side of the triangle which are $AB=5\sqrt5$, $BC=\sqrt10$ and $AC=5\sqrt5$.
I know that the formula for the area of a triangle is $\frac12hb$ but when I checked the solutions the answer to the area of this triangle is $17\frac12$.
I do not understand how this answer is achieved.

Follow-through
As you have said before, the side lengths of $\triangle ABC$ is $AB=AC=5\sqrt{5}$, $BC=\sqrt{10}$, using Heron's formula, we can compute the answer.
Heron's formula states that given side lengths $a,b,c$ of $\triangle ABC$, the area is given $$\sqrt{s(s-a)(s-b)(s-c)}\tag{1}$$
Where $s$ Is the semi perimeter. ($s=\frac {a+b+c}{2}$).
So in your case, we have $$a=5\sqrt{5},b=5\sqrt{5},c=\sqrt{10}\tag{2}$$
The semi perimeter is $$\frac {10\sqrt{5}+\sqrt{10}}{2}\tag{3}$$ and plugging in the values, we have $$\sqrt{\frac {10\sqrt{5}+\sqrt{10}}{2}\left(\frac {10\sqrt{5}+\sqrt{10}}{2}-5\sqrt{5}\right)\left(\frac {10\sqrt{5}+\sqrt{10}}{2}-5\sqrt{5}\right)\left(\frac {10\sqrt{5}+\sqrt{10}}{2}-\sqrt{10}\right)}=\boxed{17.5}\tag{4}$$
