To solve for the height of the triangle, we need to follow a systematic approach. We know the following:
- The formula for the area of a triangle is given by [tex]\( \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} \)[/tex].
- The height ([tex]\(h\)[/tex]) of the triangle is [tex]\(6c\)[/tex] meters.
- The base ([tex]\(b\)[/tex]) of the triangle is [tex]\(c-1\)[/tex] meters.
- The area ([tex]\(A\)[/tex]) of the triangle is [tex]\(18\)[/tex] square meters.
Given these pieces of information, we set up the equation for the area:
[tex]\[ \frac{1}{2} \times (c - 1) \times (6c) = 18 \][/tex]
We now simplify and solve for [tex]\(c\)[/tex]:
1. Multiply both sides of the equation by [tex]\(2\)[/tex] to eliminate the fraction:
[tex]\[ (c - 1) \times (6c) = 36 \][/tex]
2. Distribute [tex]\(6c\)[/tex] over [tex]\((c - 1)\)[/tex]:
[tex]\[ 6c^2 - 6c = 36 \][/tex]
3. Move [tex]\(36\)[/tex] to the left side of the equation to set it equal to zero:
[tex]\[ 6c^2 - 6c - 36 = 0 \][/tex]
4. Divide the entire equation by [tex]\(6\)[/tex] to simplify:
[tex]\[ c^2 - c - 6 = 0 \][/tex]
5. Factor the quadratic equation [tex]\(c^2 - c - 6 = 0\)[/tex]:
[tex]\[ (c - 3)(c + 2) = 0 \][/tex]
6. Set each factor equal to zero:
[tex]\[ c - 3 = 0 \quad \text{or} \quad c + 2 = 0 \][/tex]
7. Solve for [tex]\(c\)[/tex]:
[tex]\[ c = 3 \quad \text{or} \quad c = -2 \][/tex]
Since [tex]\(c\)[/tex] represents a dimension in the problem (base and height), it must be a positive value:
[tex]\[ c = 3 \][/tex]
Now, we substitute [tex]\(c\)[/tex] back into the expression for the height [tex]\(6c\)[/tex] to find the height of the triangle:
[tex]\[ \text{height} = 6c = 6 \times 3 = 18 \text{ meters} \][/tex]
Therefore, the correct equation and the correct measure of the height of the triangle are:
[tex]\[ 0.5 (c - 1)(6c) = 18; \text{ height} = 18 \text{ meters} \][/tex]
