Find the area of the triangle in which
Q(i) a = 13 m, b=14 m, c=15 m,
(ii) a = 52 cm, b = 56 cm, c = 60 cm,
(iii) a = 91 m, b=98 m. c = 105 m.

Answer

To find the area of the triangle using Heron's formula, we will follow these steps for each part of the question.

### Part (i): a = 13 m, b = 14 m, c = 15 m

**Step 1: Calculate the semi-perimeter (s)**
The semi-perimeter \( s \) is given by the formula:
\[ s = \frac{a + b + c}{2} \]
Substituting the values:
\[ s = \frac{13 + 14 + 15}{2} = \frac{42}{2} = 21 \, \text{m} \]

**Step 2: Apply Heron's formula**
Heron's formula for the area \( A \) of the triangle is:
\[ A = \sqrt{s(s-a)(s-b)(s-c)} \]
Substituting the values:
\[ A = \sqrt{21(21-13)(21-14)(21-15)} \]
\[ A = \sqrt{21 \times 8 \times 7 \times 6} \]

**Step 3: Simplify the expression**
Calculating the products:
\[ A = \sqrt{21 \times 8 \times 7 \times 6} \]
Breaking it down:
\[ 21 = 3 \times 7 \]
\[ 8 = 2^3 \]
\[ 6 = 2 \times 3 \]
Thus,
\[ A = \sqrt{(3 \times 7) \times (2^3) \times (7) \times (2 \times 3)} \]
\[ = \sqrt{3^2 \times 7^2 \times 2^4} \]
\[ = 3 \times 7 \times 4 = 84 \, \text{m}^2 \]

### Part (ii): a = 52 cm, b = 56 cm, c = 60 cm

**Step 1: Calculate the semi-perimeter (s)**
\[ s = \frac{52 + 56 + 60}{2} = \frac{168}{2} = 84 \, \text{cm} \]

**Step 2: Apply Heron's formula**
\[ A = \sqrt{s(s-a)(s-b)(s-c)} \]
Substituting the values:
\[ A = \sqrt{84(84-52)(84-56)(84-60)} \]
\[ A = \sqrt{84 \times 32 \times 28 \times 24} \]

**Step 3: Simplify the expression**
Calculating the products:
\[ A = \sqrt{84 \times 32 \times 28 \times 24} \]
Breaking it down:
\[ 84 = 2^2 \times 3 \times 7 \]
\[ 32 = 2^5 \]
\[ 28 = 2^2 \times 7 \]
\[ 24 = 2^3 \times 3 \]
Thus,
\[ A = \sqrt{(2^2 \times 3 \times 7) \times (2^5) \times (2^2 \times 7) \times (2^3 \times 3)} \]
\[ = \sqrt{2^{12} \times 3^2 \times 7^2} \]
\[ = 2^6 \times 3 \times 7 = 1344 \, \text{cm}^2 \]

### Part (iii): a = 91 m, b = 98 m, c = 105 m

**Step 1: Calculate the semi-perimeter (s)**
\[ s = \frac{91 + 98 + 105}{2} = \frac{294}{2} = 147 \, \text{m} \]

**Step 2: Apply Heron's formula**
\[ A = \sqrt{s(s-a)(s-b)(s-c)} \]
Substituting the values:
\[ A = \sqrt{147(147-91)(147-98)(147-105)} \]
\[ A = \sqrt{147 \times 56 \times 49 \times 42} \]

**Step 3: Simplify the expression**
Calculating the products:
\[ A = \sqrt{147 \times 56 \times 49 \times 42} \]
Breaking it down:
\[ 147 = 3 \times 7^2 \]
\[ 56 = 2^3 \times 7 \]
\[ 49 = 7^2 \]
\[ 42 = 2 \times 3 \times 7 \]
Thus,
\[ A = \sqrt{(3 \times 7^2) \times (2^3 \times 7) \times (7^2) \times (2 \times 3 \times 7)} \]
\[ = \sqrt{2^4 \times 3^2 \times 7^5} \]
\[ = 2^2 \times 3 \times 7^{2.5} = 4116 \, \text{m}^2 \]

### Summary of Areas:
1. Area of triangle (i) = 84 m²
2. Area of triangle (ii) = 1344 cm²
3. Area of triangle (iii) = 4116 m²