1. **Problem Statement:** We have a quadrilateral ABCD divided by diagonal AC into two triangles: ABD (Ram's land) and ADC (Sohan's land). We need to find which piece is larger and by how much. 2. **Given Data:** - AB = 12 m - BC = 35 m (not directly needed for area calculation) - AD = 28 m - CD = 15 m - Diagonal AC is common to both triangles. 3. **Approach:** We will calculate the area of triangles ABD and ADC using Heron's formula. 4. **Heron's Formula:** For a triangle with sides $a$, $b$, and $c$, the area $A$ is given by: $$ A = \sqrt{s(s-a)(s-b)(s-c)} $$ where $s = \frac{a+b+c}{2}$ is the semi-perimeter. 5. **Calculate length of diagonal AC:** We need AC to use Heron's formula for both triangles. Using triangle ABC or ADC, but since BC is not needed, we use triangle ADC with sides AD=28, DC=15, and AC unknown. We can find AC using the Law of Cosines if angle is known, but since no angle is given, we assume the problem expects us to use the given data to find AC. Alternatively, since the problem does not provide angle measures, we can use the fact that the two triangles share diagonal AC and use the formula for area of triangle with sides known. Since the problem is ambiguous about AC, we can calculate AC using the triangle ABD or ADC with the given sides. 6. **Calculate AC using triangle ABD:** Triangle ABD has sides AB=12, AD=28, and diagonal AC unknown. Similarly, triangle ADC has sides AD=28, DC=15, and diagonal AC unknown. Since AC is common, we can find AC using the triangle ABC or by applying the formula for the length of diagonal AC using the given sides. 7. **Calculate AC using the Law of Cosines in triangle ABC:** Given AB=12, BC=35, and AC unknown. But angle B is unknown, so we cannot use Law of Cosines directly. 8. **Alternative approach:** Use the formula for area of quadrilateral with diagonal AC: Area of ABCD = Area of ABD + Area of ADC Since the problem only asks for comparison of areas of triangles ABD and ADC, we can use the formula for area of triangle using sides AB, BD, and AD. But BD is unknown. 9. **Use Brahmagupta's formula or split the problem:** Since the problem is complex, we use the formula for area of triangle with two sides and included angle: $$ Area = \frac{1}{2}ab\sin C $$ But angles are unknown. 10. **Given the problem's data and instructions, the best approach is to use the formula for area of triangle with sides known using Heron's formula, assuming diagonal AC is the base for both triangles. We calculate AC using the triangle ABC with sides AB=12, BC=35, and AC unknown. Using the Law of Cosines in triangle ABC: $$ AC^2 = AB^2 + BC^2 - 2 \times AB \times BC \times \cos B $$ But angle B is unknown, so we cannot calculate AC directly. 11. **Since the problem provides $\sqrt{10} = 3.15$, we can try to find AC using the Pythagorean theorem if the figure is right angled, but no such info is given. 12. **Assuming the problem expects us to calculate areas using Heron's formula for triangles ABD and ADC with sides: - Triangle ABD: AB=12, AD=28, and diagonal AC (unknown) - Triangle ADC: AD=28, DC=15, and diagonal AC (unknown) We can calculate AC using the triangle ABC with sides AB=12, BC=35, and AC unknown, but no angle is given. 13. **Since the problem is ambiguous, we assume diagonal AC length is calculated using the triangle ABC with sides AB=12, BC=35, and AC calculated using the Law of Cosines with angle B=90 degrees (right angle assumption): $$ AC = \sqrt{AB^2 + BC^2} = \sqrt{12^2 + 35^2} = \sqrt{144 + 1225} = \sqrt{1369} = 37 $$ 14. **Calculate area of triangle ABD:** Sides: AB=12, AD=28, AC=37 Semi-perimeter: $$ s_1 = \frac{12 + 28 + 37}{2} = \frac{77}{2} = 38.5 $$ Area: $$ A_1 = \sqrt{38.5(38.5-12)(38.5-28)(38.5-37)} = \sqrt{38.5 \times 26.5 \times 10.5 \times 1.5} $$ Calculate inside: $$ 38.5 \times 26.5 = 1020.25 $$ $$ 10.5 \times 1.5 = 15.75 $$ $$ 1020.25 \times 15.75 = 16074.9375 $$ $$ A_1 = \sqrt{16074.9375} \approx 126.77 \text{ m}^2 $$ 15. **Calculate area of triangle ADC:** Sides: AD=28, DC=15, AC=37 Semi-perimeter: $$ s_2 = \frac{28 + 15 + 37}{2} = \frac{80}{2} = 40 $$ Area: $$ A_2 = \sqrt{40(40-28)(40-15)(40-37)} = \sqrt{40 \times 12 \times 25 \times 3} $$ Calculate inside: $$ 40 \times 12 = 480 $$ $$ 25 \times 3 = 75 $$ $$ 480 \times 75 = 36000 $$ $$ A_2 = \sqrt{36000} = 189.74 \text{ m}^2 $$ 16. **Compare areas:** $$ A_2 - A_1 = 189.74 - 126.77 = 62.97 \text{ m}^2 $$ 17. **Conclusion:** Sohan gets the larger piece of land by approximately 62.97 square meters.