1. **Problem statement:** We have a rectangular yard ABCD with sides AB = 5m and AD = 12m. Point B is 5cm lower than A, and point D is 8cm lower than A. We need to find how much lower point C is compared to point A, rounding to the nearest centimeter. 2. **Set up coordinate system and points:** Let A be at the origin with height 0cm. Then: - A = (0,0,0) - B = (5,0,-5) since B is 5m along x-axis and 5cm lower (negative z direction) - D = (0,12,-8) since D is 12m along y-axis and 8cm lower 3. **Find coordinates of C:** Since ABCD is a rectangle, C is at (5,12,z_C). We need to find $z_C$. 4. **Assumption:** The surface is a plane passing through points A, B, and D. The height at C lies on this plane. 5. **Find the plane equation:** Points: $A=(0,0,0)$ $B=(5,0,-5)$ $D=(0,12,-8)$ Vectors: $\vec{AB} = (5,0,-5)$ $\vec{AD} = (0,12,-8)$ Normal vector $\vec{n} = \vec{AB} \times \vec{AD}$: $$\vec{n} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 5 & 0 & -5 \\ 0 & 12 & -8 \end{vmatrix} = (0 \times -8 - (-5) \times 12)\mathbf{i} - (5 \times -8 - (-5) \times 0)\mathbf{j} + (5 \times 12 - 0 \times 0)\mathbf{k}$$ $$= (0 + 60)\mathbf{i} - (-40 - 0)\mathbf{j} + (60)\mathbf{k} = (60,40,60)$$ 6. **Plane equation:** Using point A (0,0,0), plane equation is: $$60x + 40y + 60z = 0$$ 7. **Find $z_C$ at point C (5,12,z_C):** $$60 \times 5 + 40 \times 12 + 60 z_C = 0$$ $$300 + 480 + 60 z_C = 0$$ $$780 + 60 z_C = 0$$ $$60 z_C = -780$$ $$z_C = -13$$ 8. **Interpretation:** $z_C = -13$ means point C is 13cm lower than point A. **Final answer:** Point C is 13cm lower than point A. **Answer choice:** A 13cm.