1. **State the problem:** We are given values $a = \sqrt{81}$, $b = -\frac{3}{4}$, and $c = \pi$. We need to determine which set represents the irrational numbers among the options: $\{c\}$, $\{a, c\}$, $\{a\}$, and $\{a, b\}$. 2. **Evaluate each variable:** - $a = \sqrt{81} = 9$ because $81$ is a perfect square. - $b = -\frac{3}{4}$ is a rational number since it is a ratio of two integers. - $c = \pi$ is a well-known irrational number. 3. **Recall definitions:** - Rational numbers can be expressed as a fraction of two integers. - Irrational numbers cannot be expressed as such fractions and have non-repeating, non-terminating decimal expansions. 4. **Analyze each set:** - $\{c\}$ contains only $\pi$, which is irrational. - $\{a, c\} = \{9, \pi\}$ contains $9$ (rational) and $\pi$ (irrational). - $\{a\} = \{9\}$ contains only a rational number. - $\{a, b\} = \{9, -\frac{3}{4}\}$ contains only rational numbers. 5. **Conclusion:** The set that contains only irrational numbers is $\{c\}$. **Final answer:** $\boxed{\{c\}}$