State 'T' for true and 'F' for false.
(I) The lengths of the three sides of a triangular field are 40 m, 24 m and 32 m respectively. The area of the triangle is 384 `m^(2)`.
(II) The area of a quadrilateral ABCD in which AB=3 cm, BC=4 cm, CD=4 cm, DA = 5 cm and AC = 5 cm is 18 `cm^(2)`.
(III) An advertisement board is in the form of an isosceles triangle with its sides equal to 12 m, 10 m and 10 m. The cost of painting it at Rs 2.25 per `m^(2)` is Rs 112.
(IV) Heron's formula cannot be used to calculate area of quadrilaterals.

Options

A. `{:(I,II,III,IV),(T,F,F,T):}`
B. `{:(I,II,III,IV),(F,T,F,F):}`
C. `{:(I,II,III,IV),(T,F,T,F):}`
D. `{:(I,II,III,IV),(T,F,F,F):}`

Answer

To solve the given statements, we will analyze each one step by step using Heron's formula and other relevant geometric principles.

### Statement (I):
The lengths of the three sides of a triangular field are 40 m, 24 m, and 32 m respectively. The area of the triangle is 384 m².

**Step 1:** Calculate the semi-perimeter (s) of the triangle.
\[ s = \frac{A + B + C}{2} = \frac{40 + 24 + 32}{2} = \frac{96}{2} = 48 \text{ m} \]

**Step 2:** Use Heron's formula to find the area (A) of the triangle.
\[ A = \sqrt{s(s - A)(s - B)(s - C)} = \sqrt{48(48 - 40)(48 - 24)(48 - 32)} \]
\[ = \sqrt{48 \times 8 \times 24 \times 16} \]

**Step 3:** Calculate the values inside the square root.
\[ = \sqrt{48 \times 8 \times 24 \times 16} = \sqrt{48 \times 8 \times 384} = \sqrt{147456} \]

**Step 4:** Simplify the square root.
\[ = 384 \text{ m}^2 \]

**Conclusion for Statement (I):** The statement is **True (T)**.

---

### Statement (II):
The area of a quadrilateral ABCD in which AB=3 cm, BC=4 cm, CD=4 cm, DA=5 cm, and AC=5 cm is 18 cm².

**Step 1:** Divide the quadrilateral into two triangles by diagonal AC.
- Triangle ABC with sides AB = 3 cm, BC = 4 cm, AC = 5 cm.
- Triangle ACD with sides AD = 5 cm, CD = 4 cm, AC = 5 cm.

**Step 2:** Calculate the area of triangle ABC using Heron's formula.
\[ s_{ABC} = \frac{3 + 4 + 5}{2} = 6 \text{ cm} \]
\[ A_{ABC} = \sqrt{s_{ABC}(s_{ABC} - 3)(s_{ABC} - 4)(s_{ABC} - 5)} = \sqrt{6(6 - 3)(6 - 4)(6 - 5)} = \sqrt{6 \times 3 \times 2 \times 1} = \sqrt{36} = 6 \text{ cm}^2 \]

**Step 3:** Calculate the area of triangle ACD using Heron's formula.
\[ s_{ACD} = \frac{5 + 4 + 5}{2} = 7 \text{ cm} \]
\[ A_{ACD} = \sqrt{s_{ACD}(s_{ACD} - 5)(s_{ACD} - 4)(s_{ACD} - 5)} = \sqrt{7(7 - 5)(7 - 4)(7 - 5)} = \sqrt{7 \times 2 \times 3 \times 2} = \sqrt{84} \approx 9.165 \text{ cm}^2 \]

**Step 4:** Total area of quadrilateral ABCD.
\[ A_{ABCD} = A_{ABC} + A_{ACD} = 6 + 9.165 \approx 15.165 \text{ cm}^2 \]

**Conclusion for Statement (II):** The statement is **False (F)**.

---

### Statement (III):
An advertisement board is in the form of an isosceles triangle with its sides equal to 12 m, 10 m, and 10 m. The cost of painting it at Rs 2.25 per m² is Rs 112.

**Step 1:** Calculate the semi-perimeter (s) of the triangle.
\[ s = \frac{10 + 10 + 12}{2} = 16 \text{ m} \]

**Step 2:** Use Heron's formula to find the area (A) of the triangle.
\[ A = \sqrt{s(s - 10)(s - 10)(s - 12)} = \sqrt{16(16 - 10)(16 - 10)(16 - 12)} = \sqrt{16 \times 6 \times 6 \times 4} \]
\[ = \sqrt{384} \approx 19.6 \text{ m}^2 \]

**Step 3:** Calculate the cost of painting.
\[ \text{Cost} = \text{Area} \times \text{Cost per m}^2 = 19.6 \times 2.25 \approx 44.1 \text{ Rs} \]

**Conclusion for Statement (III):** The statement is **False (F)**.

---

### Statement (IV):
Heron's formula cannot be used to calculate the area of quadrilaterals.

**Conclusion for Statement (IV):** This statement is **False (F)**. Heron's formula can be used to find the area of quadrilaterals by dividing them into triangles.

---

### Final Answers:
1. (I) T
2. (II) F
3. (III) F
4. (IV) F

---