The problem asks us to find the area of an arrow-shaped figure plotted on a coordinate plane. The vertices of the figure are given as (-2, 2), (5, 2), (5, 6), (12,0), (5, -6), (5, -2), and (-2, -2).

1. Problem Description

The problem asks us to find the area of an arrow-shaped figure plotted on a coordinate plane. The vertices of the figure are given as (-2, 2), (5, 2), (5, 6), (12,0), (5, -6), (5, -2), and (-2, -2).

2. Solution Steps

We can divide the arrow into a rectangle and a triangle.

The rectangle has vertices (-2, 2), (5, 2), (5, -2), and (-2, -2).

The length of the rectangle is $5 - (-2) = 7$.

The width of the rectangle is $2 - (-2) = 4$.

The area of the rectangle is $length \times width = 7 \times 4 = 28$.

The triangle has vertices (5, 6), (12, 0), and (5, -6).

The base of the triangle is the vertical distance between (5, 6) and (5, -6), which is $6 - (-6) = 12$.

The height of the triangle is the horizontal distance between (12, 0) and (5, 0), which is $12 - 5 = 7$.

The area of the triangle is $\frac{1}{2} \times base \times height = \frac{1}{2} \times 12 \times 7 = 6 \times 7 = 42$.

The area of the arrow is the sum of the areas of the rectangle and the triangle.

Area of arrow = Area of rectangle + Area of triangle = $28 + 42 = 70$.

3. Final Answer

70