1. **State the problem:** We have an isosceles triangle with a base of 40 ft and a height of 13 ft. There is a center vertical support beam at the altitude, and two additional vertical support beams placed equidistant from the center beam. We need to find the distance $x$ between the center beam and each of the side beams. 2. **Understand the setup:** The altitude divides the base into two equal segments of 20 ft each (since base is 40 ft). The vertical support beams are placed parallel to the altitude, with the side beams each at distance $x$ from the center beam. 3. **Use the given information:** The top vertex to the altitude beam is marked as 9.5 ft. This likely represents the length of the segment from the top vertex down to the point where the side beams intersect the triangle sides. 4. **Apply the Pythagorean theorem:** Consider the right triangle formed by the altitude, half the base, and the side of the triangle. The altitude is 13 ft, half the base is 20 ft, so the side length $s$ is: $$s = \sqrt{13^2 + 20^2} = \sqrt{169 + 400} = \sqrt{569} \approx 23.85 \text{ ft}$$ 5. **Find the horizontal distance $x$:** The vertical support beams are placed so that the vertical distance from the top vertex to the side beams is 9.5 ft. The vertical distance from the top vertex to the altitude beam is 13 ft (the height). The difference is $13 - 9.5 = 3.5$ ft down from the top vertex. 6. **Calculate the horizontal distance at that height:** The triangle sides slope from the top vertex to the base. The horizontal distance from the center line to the side beam at height 9.5 ft is proportional to the ratio of vertical distances: $$x = 20 \times \frac{13 - 9.5}{13} = 20 \times \frac{3.5}{13} = \frac{70}{13} \approx 5.38 \text{ ft}$$ 7. **Final answer:** The distance $x$ between the center beam and each side beam is approximately 5.4 ft to the nearest tenth. **Answer:** $x \approx 5.4$ ft