1. **Stating the problem:** We have a triangle with a perpendicular dropped from the top vertex to the base, splitting the base into two segments of lengths 50 and 46 units. The sides adjacent to the top vertex are 40 and 44 units, and the perpendicular height is labeled as $x$. We need to find the length of $x$. 2. **Formula and rules:** In a right triangle, the Pythagorean theorem applies: $$a^2 + b^2 = c^2$$ where $c$ is the hypotenuse. Here, the perpendicular creates two right triangles. 3. **Set up equations:** Let the height be $x$. For the left right triangle with hypotenuse 40 and base 50: $$x^2 + 50^2 = 40^2$$ For the right right triangle with hypotenuse 44 and base 46: $$x^2 + 46^2 = 44^2$$ 4. **Calculate each:** Left triangle: $$x^2 + 2500 = 1600$$ $$x^2 = 1600 - 2500 = -900$$ (not possible, so check the base lengths again) Since the base segments are 50 and 46, but the total base is 25, this is inconsistent. The problem states the base is 25 units, but the segments are 50 and 46, which is impossible. Likely, the segments are 5.0 and 4.6 or the labels are mixed. Assuming the base segments are 5 and 4.6 (instead of 50 and 46), recalculate: Left triangle: $$x^2 + 5^2 = 40^2$$ $$x^2 + 25 = 1600$$ $$x^2 = 1575$$ Right triangle: $$x^2 + 4.6^2 = 44^2$$ $$x^2 + 21.16 = 1936$$ $$x^2 = 1914.84$$ These two values of $x^2$ are not equal, so the assumption is incorrect. 5. **Use the property of the altitude to hypotenuse in right triangles:** The altitude $x$ to the hypotenuse satisfies: $$x = \sqrt{m \cdot n}$$ where $m$ and $n$ are the segments into which the hypotenuse is divided. Given $m=50$, $n=46$: $$x = \sqrt{50 \times 46} = \sqrt{2300}$$ 6. **Calculate $x$:** $$x = \sqrt{2300} \approx 47.96$$ **Final answer:** $$x \approx 47.96$$