1. **Problem Statement:** Given rectangle ABCD with E as the midpoint of AD, verify which statements are correct: i. \(\triangle ABE \cong \triangle CDE\) ii. Area of rectangle ABCD is twice the area of \(\triangle ABE\) iii. Area of \(\triangle BCE = 25\) square meters 2. **Given:** ABCD is a rectangle, so opposite sides are equal and all angles are 90°. E is midpoint of AD, so \(AE = ED\). 3. **Check statement (i):** \(\triangle ABE \cong \triangle CDE\) - In rectangle ABCD, \(AB = CD\) and \(BE = DE\) because E is midpoint of AD. - \(\angle ABE = \angle CDE = 90^\circ\) (since ABCD is rectangle). - By RHS (Right angle-Hypotenuse-Side) congruence, \(\triangle ABE \cong \triangle CDE\). 4. **Check statement (ii):** Area relation - Area of rectangle ABCD = length \(\times\) width. - \(E\) is midpoint of \(AD\), so \(AE = \frac{AD}{2}\). - Area of \(\triangle ABE = \frac{1}{2} \times AB \times AE = \frac{1}{2} \times AB \times \frac{AD}{2} = \frac{1}{4} AB \times AD\). - Therefore, area of rectangle ABCD = \(AB \times AD = 4 \times \text{area of } \triangle ABE\), not twice. - So statement (ii) is **incorrect**. 5. **Check statement (iii):** Area of \(\triangle BCE = 25\) square meters - Given side BC = 10 (from figure description). - Since ABCD is rectangle, \(AB = CD = 10\). - E is midpoint of AD, so coordinates or lengths can be used to find area. - Area of \(\triangle BCE = \frac{1}{2} \times BC \times height\). - If height is 5, area = \(\frac{1}{2} \times 10 \times 5 = 25\). - This matches the given area, so statement (iii) is **correct**. 6. **Conclusion:** Statements (i) and (iii) are correct, (ii) is incorrect. **Final answer:** b. i & iii