1. **Stating the problem:** We have a trapezoid ABCD with BC = 14 cm and CD = 10 cm. Inside it, triangle BFC has an area of 50 square units. We need to find the area of the shaded region, which is triangle FCD. 2. **Understanding the figure and given data:** - BC = 14 cm (bottom base) - CD = 10 cm (right side) - Area of triangle BFC = 50 3. **Key observations:** - Points B, F, C form triangle BFC. - Points F, C, D form triangle FCD (the shaded area). - Since B, C, D are vertices of the trapezoid, and F lies inside, triangles BFC and FCD share side FC. 4. **Using the area formula for triangles:** The area of a triangle is given by $$\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}$$ 5. **Relating the two triangles:** Since triangles BFC and FCD share side FC, and points B and D lie on opposite sides of FC, the heights from B and D to FC are proportional to the lengths of BC and CD respectively. 6. **Calculating the ratio of areas:** The ratio of areas of triangles BFC and FCD is equal to the ratio of their heights from points B and D to line FC. Given BC = 14 and CD = 10, the ratio of areas is $$\frac{\text{Area of } BFC}{\text{Area of } FCD} = \frac{14}{10} = \frac{7}{5}$$ 7. **Finding the area of triangle FCD:** Let the area of triangle FCD be $x$. Then $$\frac{50}{x} = \frac{7}{5} \implies 50 \times 5 = 7x \implies 250 = 7x \implies x = \frac{250}{7} \approx 35.71$$ **Final answer:** The area of the shaded region (triangle FCD) is approximately **35.71** square units.