In the standard $(x,y)$ coordinate plane below, $\triangle{ABC}$ is bounded by $\overline{AB}$, $\overline{AC}$, and the $y$-axis. Which of the following values is closest to the area, in square coordinate units, of $\triangle{ABC}$ ?

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The base of the triangle is the distance from $B$ to $C$, which is 14.

The height of the triangle is the line from the base to point $A$, perpendicular to $\overline{BC}$. The distance of this line is 5.

$$A = \frac{1}{2}bh$$ $$A = \frac{1}{2}(14)(5)$$ $$A = \boxed{35}$$