1. **State the problem:** We have an isosceles triangle ABC with AB = AC = 13 cm and BC = 10 cm. Points D and E lie on AC and AB respectively such that AD = AE = 6.5 cm, and ED is parallel to BC. We need to find the area of quadrilateral BCDE. 2. **Key properties and formula:** Since ED is parallel to BC and D, E lie on AC and AB respectively, triangle ADE is similar to triangle ABC by AA similarity (corresponding angles are equal). The area of a trapezium (quadrilateral BCDE) can be found by subtracting the area of triangle ADE from the area of triangle ABC: $$\text{Area}_{BCDE} = \text{Area}_{ABC} - \text{Area}_{ADE}$$ 3. **Calculate the area of triangle ABC:** Since AB = AC, triangle ABC is isosceles with base BC = 10 cm and equal sides 13 cm. Find the height $h$ from A to BC using Pythagoras theorem: $$h = \sqrt{13^2 - \left(\frac{10}{2}\right)^2} = \sqrt{169 - 25} = \sqrt{144} = 12$$ Area of ABC: $$\frac{1}{2} \times BC \times h = \frac{1}{2} \times 10 \times 12 = 60$$ 4. **Find the scale factor of similarity:** Since AD = AE = 6.5 cm and AB = AC = 13 cm, $$\text{scale factor} = \frac{AD}{AC} = \frac{6.5}{13} = \frac{1}{2}$$ 5. **Calculate the area of triangle ADE:** Area scales by the square of the similarity ratio: $$\text{Area}_{ADE} = \left(\frac{1}{2}\right)^2 \times \text{Area}_{ABC} = \frac{1}{4} \times 60 = 15$$ 6. **Calculate the area of quadrilateral BCDE:** $$\text{Area}_{BCDE} = 60 - 15 = 45$$ **Final answer:** $$\boxed{45 \text{ cm}^2}$$