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The height of a triangle is 2 feet less than 5 times its base. If the base of the triangle is [tex]x[/tex] feet, and the area of the triangle is 12 square feet, which equation models this situation?

A. [tex]5x^2 - 2x - 12 = 0[/tex]
B. [tex]5x^2 - 2x - 24 = 0[/tex]
C. [tex]25x^2 - 10x - 24 = 0[/tex]
D. [tex]5x^2 - 2x - 6 = 0[/tex]


Sagot :

To solve this problem, we need to create an equation that models the relationship between the base and height of the triangle and then relate it to the given area. Let's go through the problem step-by-step:

1. Understand the problem:
- The height of the triangle is [tex]\(2\)[/tex] less than [tex]\(5\)[/tex] times its base.
- The base of the triangle is [tex]\(x\)[/tex] feet.
- The area of the triangle is [tex]\(12\)[/tex] square feet.

2. Express the height in terms of [tex]\(x\)[/tex]:
The height [tex]\(h\)[/tex] can be written as:
[tex]\[ h = 5x - 2 \][/tex]

3. Write the formula for the area of a triangle:
The area [tex]\(A\)[/tex] of a triangle is given by:
[tex]\[ A = \frac{1}{2} \times \text{base} \times \text{height} \][/tex]
Substitute the base [tex]\(x\)[/tex] and the height [tex]\(5x - 2\)[/tex] into the formula:
[tex]\[ 12 = \frac{1}{2} \times x \times (5x - 2) \][/tex]

4. Simplify the equation:
Multiply both sides by [tex]\(2\)[/tex] to clear the fraction:
[tex]\[ 24 = x \times (5x - 2) \][/tex]
Distribute [tex]\(x\)[/tex] on the right side:
[tex]\[ 24 = 5x^2 - 2x \][/tex]

5. Rearrange the equation into standard quadratic form:
Subtract [tex]\(24\)[/tex] from both sides to set the equation to [tex]\(0\)[/tex]:
[tex]\[ 5x^2 - 2x - 24 = 0 \][/tex]

6. Identify the correct equation from the options:
Let's compare our resulting equation with the given options:
- [tex]\(5x^2 - 2x - 12 = 0\)[/tex]
- [tex]\(5x^2 - 2x - 24 = 0\)[/tex]
- [tex]\(25x^2 - 10x - 24 = 0\)[/tex]
- [tex]\(5x^2 - 2x - 6 = 0\)[/tex]

The equation we derived matches option B:
[tex]\[ 5x^2 - 2x - 24 = 0 \][/tex]

So, the correct equation that models this situation is:
[tex]\[ \boxed{5x^2 - 2x - 24 = 0} \][/tex]