1. **State the problem:** We have a quadrilateral ABCD with AB parallel to DC, AB = 9, BC = 10, AD = 4, and angle B = 63°. We need to find the perimeter of ABCD. 2. **Recall the properties:** Since AB \parallel DC, ABCD is a trapezoid. The perimeter is the sum of all sides: $P = AB + BC + CD + DA$. 3. **Known sides:** $AB = 9$, $BC = 10$, $DA = 4$. We need to find $CD$. 4. **Find length CD:** Since AB \parallel DC and angle B = 63°, we can use trigonometry in triangle BCD. Drop a perpendicular from C to line AD to find CD. 5. **Calculate CD:** Angle at B is 63°, so angle between BC and AB is 63°. Since BC is horizontal, vertical side CD can be found using sine: $$CD = BC \times \sin(63^\circ) = 10 \times \sin(63^\circ)$$ 6. **Calculate numerical value:** $$\sin(63^\circ) \approx 0.8910$$ $$CD = 10 \times 0.8910 = 8.91$$ 7. **Calculate perimeter:** $$P = AB + BC + CD + DA = 9 + 10 + 8.91 + 4 = 31.91$$ **Final answer:** The perimeter of ABCD is approximately $31.91$ units.