1. **Stating the problem:** We have a trapezoid ABCD with sides AD = 4.0 cm, DC = 2.0 cm, and AB = 5.0 cm. We need to find: a. The length of BC b. The perimeter of the trapezoid c. The area of the trapezoid 2. **Understanding the trapezoid:** AD is perpendicular to AB, and DC is perpendicular to AD, so ABCD is a right trapezoid. 3. **Finding BC (a):** Since DC is parallel to AB and the height between these bases is AD = 4.0 cm, we can drop a perpendicular from C to AB, creating a right triangle with legs 4.0 cm (height) and the horizontal distance from B to the foot of the perpendicular. The length of DC is 2.0 cm, and AB is 5.0 cm, so the horizontal segment from A to the foot of the perpendicular from C is $5.0 - x$, where $x$ is the horizontal distance from B to the foot of the perpendicular. Because DC is parallel to AB, the horizontal distance from D to C is 2.0 cm, so the foot of the perpendicular from C lies 2.0 cm from D along AB. Therefore, the horizontal distance from B to the foot of the perpendicular is $5.0 - 2.0 = 3.0$ cm. Now, BC is the hypotenuse of a right triangle with legs 4.0 cm (vertical) and 3.0 cm (horizontal): $$BC = \sqrt{4.0^2 + 3.0^2} = \sqrt{16 + 9} = \sqrt{25} = 5.0\text{ cm}$$ 4. **Finding the perimeter (b):** The perimeter $P$ is the sum of all sides: $$P = AB + BC + CD + DA = 5.0 + 5.0 + 2.0 + 4.0 = 16.0\text{ cm}$$ 5. **Finding the area (c):** The area $A$ of a trapezoid is given by: $$A = \frac{(\text{sum of parallel sides}) \times \text{height}}{2}$$ Here, the parallel sides are AB = 5.0 cm and DC = 2.0 cm, and the height is AD = 4.0 cm. $$A = \frac{(5.0 + 2.0) \times 4.0}{2} = \frac{7.0 \times 4.0}{2} = \frac{28.0}{2} = 14.0\text{ cm}^2$$ **Final answers:** - BC = 5.0 cm - Perimeter = 16.0 cm - Area = 14.0 cm²