In △ABC with angles A ,B and C and sides a, b and c opposite the respective angles, a=16, b=9 and c=11 measured in meters. Applying Heron's formula, the area of the triangle is found to be?

To find the area of triangle \( \triangle ABC \) using Heron's formula, we can follow these steps:

### Step 1: Calculate the semi-perimeter (s)
The semi-perimeter \( s \) is calculated as:
\[
s = \frac{a + b + c}{2}
\]
Substituting the given values \( a = 16 \) meters, \( b = 9 \) meters, and \( c = 11 \) meters:
\[
s = \frac{16 + 9 + 11}{2} = \frac{36}{2} = 18 \text{ meters}
\]

### Step 2: Apply Heron's formula
Heron's formula for the area \( A \) of a triangle is given by:
\[
A = \sqrt{s(s-a)(s-b)(s-c)}
\]
Substituting the values:
\[
A = \sqrt{18(18-16)(18-9)(18-11)}
\]
\[
A = \sqrt{18(2)(9)(7)}
\]
\[
A = \sqrt{18 \times 2 \times 9 \times 7}
\]

### Step 3: Simplify the expression
First, calculate the product inside the square root:
\[
18 \times 2 = 36
\]
\[
36 \times 9 = 324
\]
\[
324 \times 7 = 2268
\]

Now, take the square root of 2268:
\[
A = \sqrt{2268}
\]
\[
A \approx 47.63 \text{ square meters}
\]

### Final Answer:
The area of the triangle \( \triangle ABC \) is approximately \( 47.63 \) square meters.

Would you like more details or have any questions?

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**Related Questions:**
1. How do you derive Heron's formula?
2. What is the significance of the semi-perimeter in Heron's formula?
3. Can Heron's formula be used for any triangle?
4. How does Heron's formula compare to the basic area formula \( \frac{1}{2} \times \text{base} \times \text{height} \)?
5. How would the area change if one of the sides is doubled?

**Tip:** Always verify your calculations by checking if the square root and other intermediate steps are correct.

Discover how to calculate the area of triangle ABC with sides measuring 16 meters, 9 meters, and 11 meters using Heron's Formula. Follow a detailed step-by-step solution and learn about the application of Heron's Formula in geometry. Suitable for Grades 9-12.

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