Question

Area of $\triangle ABC$ is

• A. 8$\sqrt{2}$
• B. 4$\sqrt{2}$
• C. 2$\sqrt{2}$
• D. 6$\sqrt{2}$

Answer 1

Answer: (B)

4$\sqrt{2}$

Answer 2

The vertices of the triangle are $A(0, 0)$, $B(2, 2\sqrt{2})$, and $C(2, -2\sqrt{2})$. The base BC is a vertical segment with length $|2\sqrt{2} - (-2\sqrt{2})| = 4\sqrt{2}$. The height of the triangle from vertex A to the base BC is the perpendicular distance from A(0,0) to the line $x=2$, which is 2. The area of $\triangle ABC = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times (4\sqrt{2}) \times 2 = 4\sqrt{2}$.