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To solve the problem, we need to find the length of the base of a triangle, given the relationship between the base and the altitude, as well as the area of the triangle.

Let's denote:
- \( b \) as the length of the base.
- \( h \) as the length of the altitude.

The problem states:
1. The length of the base is 4 cm smaller than the length of its altitude, which translates to:
\[ b = h - 4 \]

2. The area of the triangle is 96 cm². The formula for the area of a triangle is:
\[ \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} \]
Plugging in the values, we have:
\[ 96 = \frac{1}{2} \times b \times h \]

Now, substituting \( b = h - 4 \) into the area formula:
\[ 96 = \frac{1}{2} \times (h - 4) \times h \]

Multiplying both sides by 2 to eliminate the fraction:
\[ 192 = (h - 4) \times h \]

Expanding the equation:
\[ 192 = h^2 - 4h \]

Rearranging to form a quadratic equation:
\[ h^2 - 4h - 192 = 0 \]

We can solve this quadratic equation using the quadratic formula:
\[ h = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
where \( a = 1 \), \( b = -4 \), and \( c = -192 \).

Plugging in the values, we get:
\[ h = \frac{-(-4) \pm \sqrt{(-4)^2 - 4 \cdot 1 \cdot (-192)}}{2 \cdot 1} \]
\[ h = \frac{4 \pm \sqrt{16 + 768}}{2} \]
\[ h = \frac{4 \pm \sqrt{784}}{2} \]
\[ h = \frac{4 \pm 28}{2} \]

This gives us two potential solutions for \( h \):
\[ h = \frac{4 + 28}{2} = 16 \]
\[ h = \frac{4 - 28}{2} = -12 \]

Since a negative value for the altitude is not physically meaningful, we discard \( h = -12 \) and accept \( h = 16 \).

Now, using \( h = 16 \), we find the base \( b \):
\[ b = h - 4 = 16 - 4 = 12 \]

Therefore, the length of the base is:
\[ \boxed{12 \, \text{cm}} \]

Would you like more details or have any questions about the steps?

Here are some related questions to expand your understanding:
1. How do you derive the quadratic formula?
2. Can you prove that the area formula for a triangle is \(\frac{1}{2} \times \text{base} \times \text{height}\)?
3. What are the steps to factorize a quadratic equation?
4. How can you determine if a quadratic equation has real or complex solutions?
5. What are some other methods to solve quadratic equations besides the quadratic formula?

**Tip:** Always check if your solutions are physically meaningful, especially in geometry problems where negative lengths do not make sense.

The length of the base of a triangle is 4 cm smaller than the length of its altitude. The area of the triangle is 96 cm². The length of the base is?

Find the base length of a triangle whose base is 4 cm smaller than its altitude and has an area of 96 cm². Learn how to set up a quadratic equation and apply the quadratic formula to solve this problem. Perfect for students in Grades 8-10.

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