1. **Problem statement:** We have an acute triangle with sides $x$, $2x$, and $15$ cm, where $15$ cm is the longest side. We need to find the smallest whole-number value of $x$ such that the triangle is acute. 2. **Recall the property of an acute triangle:** For a triangle with sides $a \leq b \leq c$, it is acute if and only if $$a^2 + b^2 > c^2.$$ Here, $c=15$ is the longest side. 3. **Identify sides:** The sides are $x$, $2x$, and $15$. Since $15$ is the longest side, we have $15 > 2x$ and $15 > x$. 4. **Apply the acute triangle condition:** $$x^2 + (2x)^2 > 15^2$$ $$x^2 + 4x^2 > 225$$ $$5x^2 > 225$$ 5. **Solve for $x$:** $$x^2 > \frac{225}{5}$$ $$x^2 > 45$$ $$x > \sqrt{45}$$ 6. **Calculate $\sqrt{45}$:** $$\sqrt{45} = \sqrt{9 \times 5} = 3\sqrt{5} \approx 6.708$$ 7. **Check triangle inequality:** For a triangle, the sum of any two sides must be greater than the third side. - $x + 2x > 15 \Rightarrow 3x > 15 \Rightarrow x > 5$ - $x + 15 > 2x \Rightarrow 15 > x$ - $2x + 15 > x \Rightarrow 15 > -x$ (always true) So $x$ must be greater than 5 and less than 15. 8. **Find smallest whole number $x$ satisfying both conditions:** - From acuteness: $x > 6.708$ - From triangle inequality: $x > 5$ The smallest whole number greater than $6.708$ is $7$. 9. **Check if $x=7$ satisfies all conditions:** - $x=7$ - $2x=14$ - $15$ longest side - Check acuteness: $$7^2 + 14^2 = 49 + 196 = 245 > 225$$ (true) - Check triangle inequality: $$7 + 14 = 21 > 15$$ (true) 10. **Check $x=6$ to confirm it fails acuteness:** $$6^2 + 12^2 = 36 + 144 = 180 < 225$$ (not acute) **Final answer:** The smallest whole-number value of $x$ for which the triangle is acute is **7**.