To solve for the base and height of the triangle, we'll use the formula for the area of a triangle:

$\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}$

Given:

• The area of the triangle $\text{Area} = 600 \, \text{in}^2$
• The base is 8 inches more than the height. If we let the height be $h$ inches, then the base will be $h + 8$ inches.

Using the area formula:

$600 = \frac{1}{2} \times (h + 8) \times h$

Simplify and solve for $h$:

$600 = \frac{1}{2} \times (h^2 + 8h)$

Multiply both sides by 2 to eliminate the fraction:

$1200 = h^2 + 8h$

Now, rearrange the equation into a standard quadratic form:

$h^2 + 8h - 1200 = 0$

This quadratic equation can be solved using the quadratic formula:

$h = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Here, $a = 1$, $b = 8$, and $c = -1200$. Substituting these values into the formula:

$h = \frac{-8 \pm \sqrt{8^2 - 4(1)(-1200)}}{2(1)}$

$h = \frac{-8 \pm \sqrt{64 + 4800}}{2}$

$h = \frac{-8 \pm \sqrt{4864}}{2}$

$h = \frac{-8 \pm 69.7}{2}$

Now, calculate the two possible values for $h$:

1. $h = \frac{-8 + 69.7}{2} = \frac{61.7}{2} = 30.85$
2. $h = \frac{-8 - 69.7}{2} = \frac{-77.7}{2} = -38.85$

Since height cannot be negative, the valid height is:

$h \approx 30.85 \, \text{inches}$

Now, calculate the base:

$\text{base} = h + 8 \approx 30.85 + 8 = 38.85 \, \text{inches}$

So, the base is approximately $38.9$ inches, and the height is approximately $30.9$ inches.

Would you like further details or have any questions? Here are five related questions:

1. How do you derive the quadratic formula?
2. What is the geometric significance of a triangle's area?
3. How would the problem change if the area were different?
4. What are the properties of a triangle with a specific base and height ratio?
5. How would you verify the solution using an alternative method?

Tip: When solving quadratic equations, always check for both positive and negative roots, and ensure the solution makes sense in the context of the problem.