A triangle has sides of length 10, 15, and 20. 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:

• $$a = 10$$
• $$b = 15$$
• $$c = 20$$

First, calculate the semi-perimeter:

\[ s = \frac{10+15+20}{2} = \frac{45}{2} = 22.5 \]

Now, apply Heron's formula:

\[ \text{Area} = \sqrt{22.5(22.5-10)(22.5-15)(22.5-20)} \]

\[ \text{Area} = \sqrt{22.5(12.5)(7.5)(2.5)} \]

\[ \text{Area} = \sqrt{5273.4375} \]

\[ \text{Area} = \sqrt{\frac{45}{2} \times \frac{25}{2} \times \frac{15}{2} \times \frac{5}{2}} = \sqrt{\frac{45 \times 25 \times 15 \times 5}{16}} \]

\[ \text{Area} = \frac{1}{4} \sqrt{(9 \times 5) \times 25 \times (3 \times 5) \times 5} \]

\[ \text{Area} = \frac{1}{4} \sqrt{9 \times 25 \times 25 \times 3 \times 5} \]

\[ \text{Area} = \frac{1}{4} \times 3 \times 5 \times 5 \sqrt{15} \]

\[ \text{Area} = \frac{75}{4} \sqrt{15} \]

Ответ: $$\frac{75}{4}\sqrt{15}$$