To solve this problem, we need to calculate the area of the triangular region and determine how many cans of paint are needed to cover it, given that each can covers 70 square meters.
### Step-by-Step Solution:
1. Identify the sides of the triangle:
- Given sides: [tex]\(a = 63\)[/tex] meters, [tex]\(b = 66\)[/tex] meters, [tex]\(c = 73\)[/tex] meters.
2. Calculate the semi-perimeter of the triangle ([tex]\(s\)[/tex]):
[tex]\[
s = \frac{a + b + c}{2}
\][/tex]
Plugging in the given values:
[tex]\[
s = \frac{63 + 66 + 73}{2} = \frac{202}{2} = 101
\][/tex]
3. Use Heron's formula to calculate the area ([tex]\(A\)[/tex]) of the triangle:
[tex]\[
A = \sqrt{s(s-a)(s-b)(s-c)}
\][/tex]
Plugging in the values:
[tex]\[
A = \sqrt{101(101-63)(101-66)(101-73)}
\][/tex]
Simplifying inside the square root:
[tex]\[
A = \sqrt{101 \times 38 \times 35 \times 28}
\][/tex]
Evaluating this expression yields:
[tex]\[
A \approx 1939.39 \text{ square meters}
\][/tex]
4. Determine how many cans of paint are required:
- Each can of paint covers 70 square meters.
- To find the number of cans needed, divide the area by the coverage of each can and round up to the nearest whole number because we cannot use a fraction of a can.
[tex]\[
\text{Number of cans needed} = \left\lceil \frac{A}{\text{paint coverage}} \right\rceil = \left\lceil \frac{1939.39}{70} \right\rceil
\][/tex]
Calculating the value inside the ceiling function:
[tex]\[
\frac{1939.39}{70} \approx 27.70
\][/tex]
Rounding up to the nearest whole number:
[tex]\[
\left\lceil 27.70 \right\rceil = 28
\][/tex]
Therefore, the number of cans needed is [tex]\(28\)[/tex]. Thus, the correct answer is:
[tex]\[
\boxed{28 \text{ cans}}
\][/tex]
