This is a geometric problem that requires finding the value of $x$. The figure is a triangle $\triangle ABC$, with a perpendicular drawn from point $B$ to the base $AC$, meeting at point $D$.

Here’s what we know:

• $AB = x$
• $BC = 8$
• $AD = 3$
• $DC = 4$
• $BD$ is perpendicular to $AC$

We can solve this using the Pythagorean Theorem in both right triangles $\triangle ABD$ and $\triangle BDC$:

Step 1: Apply the Pythagorean theorem to $\triangle ABD$

$AB^2 = AD^2 + BD^2$ $x^2 = 3^2 + BD^2$ $x^2 = 9 + BD^2 \quad \text{(1)}$

Step 2: Apply the Pythagorean theorem to $\triangle BDC$

$BC^2 = BD^2 + DC^2$ $8^2 = BD^2 + 4^2$ $64 = BD^2 + 16$ $BD^2 = 64 - 16 = 48 \quad \text{(2)}$

Step 3: Substitute $BD^2 = 48$ into equation (1)

$x^2 = 9 + 48$ $x^2 = 57$ $x = \sqrt{57} \approx 7.55$

Thus, the value of $x$ is approximately 7.55.

Would you like further details on any part of this solution?

Here are 5 related questions:

1. How would the problem change if $BD$ were not perpendicular to $AC$?
2. How is the Pythagorean theorem used in general for right triangles?
3. Can this problem be solved using trigonometry instead of the Pythagorean theorem?
4. What are the properties of an altitude in a triangle?
5. How would this solution differ if the lengths of $AD$ and $DC$ were swapped?

Tip: Always check if a triangle is a right triangle before applying the Pythagorean theorem, as it is only valid for right triangles.