1. **Stating the problem:** We have a right triangle with legs 32 and 2, and inside it a smaller right triangle with legs labeled $a$ and $b$. We need to analyze the given values for $a$ and $b$ in each case. 2. **Formula and rules:** In a right triangle, the Pythagorean theorem applies: $$a^2 + b^2 = c^2$$ where $c$ is the hypotenuse. 3. **Check each case:** **a.** $a=8$, $b=8\sqrt{17}$ Calculate $a^2 + b^2$: $$8^2 + (8\sqrt{17})^2 = 64 + 64 \times 17 = 64 + 1088 = 1152$$ **b.** $a=64$, $b=68$ Calculate $a^2 + b^2$: $$64^2 + 68^2 = 4096 + 4624 = 8720$$ **c.** $a=8$, $b=2\sqrt{17}$ Calculate $a^2 + b^2$: $$8^2 + (2\sqrt{17})^2 = 64 + 4 \times 17 = 64 + 68 = 132$$ **d.** $a=18$, $b=2\sqrt{17}$ Calculate $a^2 + b^2$: $$18^2 + (2\sqrt{17})^2 = 324 + 68 = 392$$ 4. **Interpretation:** Since the original triangle legs are 32 and 2, the hypotenuse is: $$c = \sqrt{32^2 + 2^2} = \sqrt{1024 + 4} = \sqrt{1028}$$ Approximate $\sqrt{1028} \approx 32.06$. 5. **Compare sums to $c^2$:** - For (a), $a^2 + b^2 = 1152$ which is greater than $1028$, so this cannot be a right triangle inside the original. - For (b), $a^2 + b^2 = 8720$ which is much greater than $1028$, so no. - For (c), $a^2 + b^2 = 132$ which is less than $1028$, so possible but smaller. - For (d), $a^2 + b^2 = 392$ which is less than $1028$, also possible but smaller. **Final answer:** Only cases (c) and (d) have $a$ and $b$ values consistent with smaller right triangles inside the original triangle with legs 32 and 2.