1. The problem asks to find the value of $c$ such that $4a^2 + 44a + c$ is a perfect square trinomial. 2. A perfect square trinomial has the form $\left(\sqrt{A}a + B\right)^2 = A a^2 + 2B\sqrt{A} a + B^2$. 3. Here, $A = 4$, so $\sqrt{A} = 2$. The middle term is $44a$, which equals $2B \times 2a = 4Ba$. 4. Set $4B = 44$ to find $B$: $$4B = 44 \implies B = \frac{44}{4} = 11$$ 5. The constant term $c$ must be $B^2 = 11^2 = 121$. 6. Therefore, the value of $c$ is $121$ to make the trinomial a perfect square. Final answer: $121$