ABCD is a quadrilateral with sides AB = 9 cm, BC = 40 cm, CD = 28 cm and DA = 15 cm and one of the diagonals AC = 41 cm. Which of the following statements is/are correct? I. The vertices A, B, C and D of the quadrilateral lie on a circle. II. The area of triangle ACD is 126 cm².

In triangle ABC: AB² + BC² = 81 + 1600 = 1681 = 41² = AC². So ∠ABC = 90°. In triangle ACD: check if right-angled. AC = 41, CD = 28, DA = 15. CD² + DA² = 784 + 225 = 1009 ≠ 1681. Try 15² + ... Actually, check: AC² = 1681, and 15² + something? 1681 − 225 = 1456 (not 784), 1681 − 784 = 897 (not 225). So not right-angled at D directly. But for a cyclic quadrilateral, opposite angles sum to 180°. ∠ABC = 90°, so ∠ADC must = 90° if cyclic. Check via cosine rule in ACD: cos(∠ADC) = (15² + 28² − 41²)/(2·15·28) = (225 + 784 − 1681)/840 = −672/840 = −0.8. So ∠ADC ≈ 143.13°, not 90°. Hmm, so not cyclic? Let me recheck. Actually for cyclic, ∠ABC + ∠ADC = 180°, so ∠ADC = 90°. But cos gives -0.8, meaning ∠ADC ≈ 143°, sum with 90° = 233°. So NOT cyclic. Hmm. Let me reconsider — perhaps triangle ACD has different sides. With sides 15, 28, 41 — using Heron's: s = (15+28+41)/2 = 42. Area = √(42·27·14·1) = √15876 = 126 cm². So II is correct. For I, with ∠ABC = 90°, AC is the diameter of the circle through A, B, C. If D also lies on this circle with AC as diameter, then ∠ADC = 90°. But triangle ACD with sides 15, 28, 41: 15² + 28² = 225 + 784 = 1009 ≠ 1681. So ∠ADC ≠ 90°, meaning D doesn't lie on the same circle. So I should be incorrect. But the area II is correct, giving answer (b) II only. However, official answer indicates 'Both I and II' — this needs verification. Given Heron's gives area 126 for II, and checking I more carefully: since ∠ABC = 90° in triangle ABC (using 9, 40, 41 — a Pythagorean triple), AC is the diameter. For D on the circle, ∠ADC = 90° is required, meaning AD² + DC² = AC². 15² + 28² = 1009 ≠ 1681. So D is NOT on the circle. Thus only II is correct, answer (b).