1. **State the problem:** We need to find the projection of the shortest side of a triangle with sides 6, 7, and 8 onto the longest side. 2. **Identify the sides:** The shortest side is 6, and the longest side is 8. 3. **Use the Law of Cosines to find the angle opposite the shortest side:** $$c^2 = a^2 + b^2 - 2ab\cos(C)$$ Here, let $c=6$ (shortest side), $a=7$, and $b=8$ (longest side). 4. **Calculate $\cos(C)$:** $$6^2 = 7^2 + 8^2 - 2 \times 7 \times 8 \times \cos(C)$$ $$36 = 49 + 64 - 112 \cos(C)$$ $$36 = 113 - 112 \cos(C)$$ $$112 \cos(C) = 113 - 36 = 77$$ $$\cos(C) = \frac{77}{112}$$ 5. **Find the projection of the shortest side (length 6) onto the longest side (length 8):** Projection = $6 \times \cos(C) = 6 \times \frac{77}{112} = \frac{462}{112} = \frac{231}{56} \approx 4.125$. **Final answer:** The projection of the shortest side onto the longest side is approximately $4.125$ units.