1. **State the problem:** We need to find the area of an isosceles triangle DEF where sides DF and DE are each 25 cm, and the base FE is 16 cm. 2. **Formula for the area of a triangle:** $$\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}$$ 3. **Find the height:** Since the triangle is isosceles, the height from vertex D to base FE bisects FE. - Half of base FE is $$\frac{16}{2} = 8$$ cm. - Use the Pythagorean theorem in right triangle formed by height (h), half base (8 cm), and side (25 cm): $$25^2 = h^2 + 8^2$$ 4. **Calculate height:** $$h^2 = 25^2 - 8^2 = 625 - 64 = 561$$ $$h = \sqrt{561} \approx 23.7 \text{ cm}$$ 5. **Calculate area:** $$\text{Area} = \frac{1}{2} \times 16 \times 23.7 = 8 \times 23.7 = 189.6 \text{ cm}^2$$ 6. **Final answer:** The area of the triangle is approximately **189.6 cm²** to 1 decimal place.