A triangle has sides of length 7, 9, and 11. Use Heron's formula to find the area.

Heron's formula for the area of a triangle with sides $$a, b, c$$ is: Area $$= \sqrt{s(s-a)(s-b)(s-c)}$$, where $$s$$ is the semi-perimeter, $$s = \frac{a+b+c}{2}$$.

Given:

• $$a = 7$$
• $$b = 9$$
• $$c = 11$$

First, calculate the semi-perimeter:

\[ s = \frac{7+9+11}{2} = \frac{27}{2} = 13.5 \]

Now, apply Heron's formula:

\[ \text{Area} = \sqrt{13.5(13.5-7)(13.5-9)(13.5-11)} \]

\[ \text{Area} = \sqrt{13.5(6.5)(4.5)(2.5)} \]

\[ \text{Area} = \sqrt{987.1875} \]

\[ \text{Area} = \sqrt{\frac{27}{2}(\frac{13}{2})(\frac{9}{2})(\frac{5}{2})} \]

\[ \text{Area} = \sqrt{\frac{27 \times 13 \times 9 \times 5}{16}} \]

\[ \text{Area} = \frac{1}{4} \sqrt{(9 \times 3) \times 13 \times 9 \times 5} \]

\[ \text{Area} = \frac{1}{4} \sqrt{9 \times 9 \times 3 \times 13 \times 5} \]

\[ \text{Area} = \frac{1}{4} (9 \sqrt{195}) \]

\[ \text{Area} = \frac{9}{4} \sqrt{195} \]

Ответ: $$\frac{9}{4}\sqrt{195}$$