1. **State the problem:** We need to find the area of quadrilateral ABCD with sides AB = 7 cm, BC = 15 cm, CD = 25 cm, diagonal AC = 20 cm, and a right angle at vertex C. 2. **Analyze the shape:** Since angle C is 90°, triangle BCD is a right triangle with legs BC = 15 cm and CD = 25 cm. 3. **Calculate area of triangle BCD:** Use the formula for the area of a right triangle: $$\text{Area}_{BCD} = \frac{1}{2} \times BC \times CD = \frac{1}{2} \times 15 \times 25 = 187.5 \text{ cm}^2$$ 4. **Calculate area of triangle ABC:** We know sides AB = 7 cm, AC = 20 cm, and BC = 15 cm. Use Heron's formula: - Compute semi-perimeter: $$s = \frac{AB + BC + AC}{2} = \frac{7 + 15 + 20}{2} = 21$$ - Area formula: $$\text{Area}_{ABC} = \sqrt{s(s - AB)(s - BC)(s - AC)} = \sqrt{21(21 - 7)(21 - 15)(21 - 20)} = \sqrt{21 \times 14 \times 6 \times 1}$$ - Simplify inside the root: $$\sqrt{1764} = 42 \text{ cm}^2$$ 5. **Calculate total area of quadrilateral ABCD:** $$\text{Area}_{ABCD} = \text{Area}_{ABC} + \text{Area}_{BCD} = 42 + 187.5 = 229.5 \text{ cm}^2$$ **Final answer:** The area of quadrilateral ABCD is $229.5$ cm$^2$.