1. **State the problem:** Find the area of triangle $\triangle BCD$ with vertices $B(-8,-10)$, $C(10,-10)$, and $D(4,-4)$.\n\n2. **Formula for area of a triangle given coordinates:**\nThe area can be found using the formula:\n$$\text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right|$$\nwhere $(x_1,y_1)$, $(x_2,y_2)$, and $(x_3,y_3)$ are the coordinates of the vertices.\n\n3. **Assign points:**\nLet $B = (x_1,y_1) = (-8,-10)$, $C = (x_2,y_2) = (10,-10)$, and $D = (x_3,y_3) = (4,-4)$.\n\n4. **Substitute values into the formula:**\n$$\text{Area} = \frac{1}{2} | (-8)(-10 - (-4)) + 10(-4 - (-10)) + 4(-10 - (-10)) |$$\n\n5. **Simplify inside parentheses:**\n$$-10 - (-4) = -10 + 4 = -6$$\n$$-4 - (-10) = -4 + 10 = 6$$\n$$-10 - (-10) = -10 + 10 = 0$$\n\n6. **Calculate each term:**\n$$(-8)(-6) = 48$$\n$$10(6) = 60$$\n$$4(0) = 0$$\n\n7. **Sum the terms:**\n$$48 + 60 + 0 = 108$$\n\n8. **Calculate the area:**\n$$\text{Area} = \frac{1}{2} |108| = \frac{108}{2} = 54$$\n\n**Final answer:** The area of $\triangle BCD$ is $54$ square units.