Chapter 12: Vectors and the Geometry of Space - Section 12.4 - The Cross Product - Exercises 12.4 - Page 719: 42

$$2$$

We have $\vec{AB}= 4\hat{i}+4\hat{j} \\ \vec{AC}= 3\hat{i}+2\hat{j}$ The area of the triangle is: $$A=\dfrac{1}{2}|\vec{AB}\times\vec{AC}|$$ Now, $$\vec{AB}\times\vec{AC}=\begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\4&4&0\\3&2&0\end{vmatrix} \\=(8-12)\hat{k} \\= -4\hat{k}$$ and $|\vec{AB}\times\vec{AC}|=\sqrt {16}= 4$ Now, $Area= 2$