To find the area of the triangle with sides measuring 11 cm, 8 cm, and 9 cm, we can use Heron's formula since we have all three sides.
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Calculate the semi-perimeter (s): \[ s = \frac{a + b + c}{2} = \frac{11 + 8 + 9}{2} = \frac{28}{2} = 14 , \text{cm} \]
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Calculate the area (A) using Heron's formula: \[ A = \sqrt{s(s-a)(s-b)(s-c)} \] where \(a = 11\), \(b = 8\), and \(c = 9\).
Calculate each term: \[ s - a = 14 - 11 = 3 \] \[ s - b = 14 - 8 = 6 \] \[ s - c = 14 - 9 = 5 \]
Now plug these into the formula: \[ A = \sqrt{14 \times 3 \times 6 \times 5} \]
First, calculate \(14 \times 3 = 42\). Then, calculate \(42 \times 6 = 252\). Finally, calculate \(252 \times 5 = 1260\).
Now, take the square root: \[ A = \sqrt{1260} \]
To simplify \(\sqrt{1260}\): \[ 1260 = 4 \times 315 = 4 \times 3 \times 105 = 4 \times 3 \times 3 \times 35 = 4 \times 3^2 \times 5 \times 7 \]
Thus, the square root is: \[ \sqrt{1260} = 2 \times 3 \sqrt{70} = 6\sqrt{70} \]
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Approximate the value (if necessary): Since \(\sqrt{70} \approx 8.37\), \[ 6 \sqrt{70} \approx 6 \times 8.37 \approx 50.22 , \text{cm}^2 \]
The area of the triangle is approximately:
\[ \boxed{50.22} \text{ cm}^2 \]
In exact form, the area is \(6\sqrt{70} \text{ cm}^2\).