A triangle has sides of length 12, 15, and 18. 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:

• $$a = 12$$
• $$b = 15$$
• $$c = 18$$

First, calculate the semi-perimeter:

\[ s = \frac{12+15+18}{2} = \frac{45}{2} = 22.5 \]

Now, apply Heron's formula:

\[ \text{Area} = \sqrt{22.5(22.5-12)(22.5-15)(22.5-18)} \]

\[ \text{Area} = \sqrt{22.5(10.5)(7.5)(4.5)} \]

\[ \text{Area} = \sqrt{7973.4375} \]

\[ \text{Area} \approx 89.3 \]

Alternatively, we can use fractions:

\[ s = \frac{45}{2} \]

\[ \text{Area} = \sqrt{\frac{45}{2}(\frac{45}{2}-\frac{24}{2})(\frac{45}{2}-\frac{30}{2})(\frac{45}{2}-\frac{36}{2})} \]

\[ \text{Area} = \sqrt{\frac{45}{2}(\frac{21}{2})(\frac{15}{2})(\frac{9}{2})} \]

\[ \text{Area} = \sqrt{\frac{45 \times 21 \times 15 \times 9}{16}} \]

\[ \text{Area} = \frac{1}{4} \sqrt{79734.375} \]

\[ \text{Area} = \frac{1}{4} \sqrt{\frac{1275750}{16}} \]

\[ \text{Area} = \frac{1}{4} \sqrt{\frac{1275750}{16}} = \frac{1}{4} \frac{\sqrt{1275750}}{4} = \frac{\sqrt{255150}}{16} \]

\[ \text{Area} = \frac{1}{4} \sqrt{22.5 \times 10.5 \times 7.5 \times 4.5} = \frac{1}{4} \sqrt{\frac{45}{2} \times \frac{21}{2} \times \frac{15}{2} \times \frac{9}{2}} = \frac{1}{4} \sqrt{\frac{127575}{16}} = \frac{1}{16} \sqrt{127575} \]

\[ \text{Area} = \frac{1}{16} \sqrt{25 \times 5103} = \frac{5}{16} \sqrt{5103} \]

\[ \text{Area} = \frac{5}{16} \sqrt{9 \times 567} = \frac{5 \times 3}{16} \sqrt{567} = \frac{15}{16} \sqrt{81 \times 7} = \frac{15 \times 9}{16} \sqrt{7} = \frac{135}{16}\sqrt{7} \]

Let's re-calculate using simpler numbers. The semi-perimeter is $$s = (12+15+18)/2 = 45/2$$.

\[ \text{Area} = \sqrt{\frac{45}{2}(\frac{45}{2}-12)(\frac{45}{2}-15)(\frac{45}{2}-18)} \]

\[ \text{Area} = \sqrt{\frac{45}{2}(\frac{21}{2})(\frac{15}{2})(\frac{9}{2})} \]

\[ \text{Area} = \sqrt{\frac{45 \times 21 \times 15 \times 9}{16}} \]

\[ \text{Area} = \frac{1}{4} \sqrt{(9 \times 5) \times (3 \times 7) \times (3 \times 5) \times 9} \]

\[ \text{Area} = \frac{1}{4} \sqrt{9 \times 9 \times 3 \times 3 \times 5 \times 5 \times 7} \]

\[ \text{Area} = \frac{1}{4} (9 \times 3 \times 5 \sqrt{7}) \]

\[ \text{Area} = \frac{135}{4} \sqrt{7} \]

Ответ: $$\frac{135}{4}\sqrt{7}$$