Question: User: 13. Find the area of the trapezoid. Graph: a blue trapezoid is in the center-bottom of the coordinate plane, with top base horizontal from $(-5, -3)$ to $(4, -3)$, bottom base horizontal from $(-7, -7)$ to $(6, -7)$, left side slanting down-right from $(-5, -3)$ to $(-7, -7)$, and right side slanting down-right from $(4, -3)$ to $(6, -7)$; position_hint = center.

1. **State the problem:** We need to find the area of a trapezoid with vertices at $(-5, -3)$, $(4, -3)$, $(-7, -7)$, and $(6, -7)$. 2. **Recall the formula for the area of a trapezoid:** $$\text{Area} = \frac{1}{2} (b_1 + b_2) h$$ where $b_1$ and $b_2$ are the lengths of the two parallel bases, and $h$ is the height (the perpendicular distance between the bases). 3. **Find the lengths of the bases:** - Top base from $(-5, -3)$ to $(4, -3)$: $$b_1 = 4 - (-5) = 9$$ - Bottom base from $(-7, -7)$ to $(6, -7)$: $$b_2 = 6 - (-7) = 13$$ 4. **Find the height:** The height is the vertical distance between the lines $y = -3$ and $y = -7$: $$h = |-3 - (-7)| = 4$$ 5. **Calculate the area:** $$\text{Area} = \frac{1}{2} (9 + 13) \times 4$$ $$= \frac{1}{2} \times 22 \times 4$$ Show cancellation: $$= \cancel{\frac{1}{2}} \times 22 \times \cancel{4} \times 2$$ $$= 22 \times 2 = 44$$ 6. **Final answer:** The area of the trapezoid is $44$ square units.