There seems to be a mistake in my solution to a problem, which I cannot find.
The problem:
There be the points $A(-2, -2, 0)$, $B(2, 6, 0)$, $C(8, 4, 0)$ and $D(0, 0, 9)$. Calculate the volume of the pyramid, whose base area is the triangle $ABC$ and whose apex is $D$.
Seems simple enough.
I calculated the three ground sides, which are $AB = \sqrt{80}$, $BC = \sqrt{52}$ and $CA = \sqrt{136}$
Then I got the height $h$ of $AB$ by finding the cosinus of the angle enclosed by $AB$ and $CA$, then using $\arccos$ to get the angle in radians $\Big(\dfrac{41\sqrt{170}}{680}\Big)$, whose sinus I multiplied with the side $CA$. Result: approximately $8.2523$.
The height of the base gets multiplied with the side it is perpendicular to (which in this case is $AB$), divided through $2$, which becomes the base area: approximately $36.9055$.
Lastly, the height of the pyramid is $9$, which can be determined by the z-coordinate of point $D$. Thus, according to the formula $V = \dfrac{A * h}{3}$, the volume is approximately $110.7165$.
But well, my solution does not match with the one presented in my book. So, where did I go wrong?