1. **Problem Statement:** Determine if parallelogram ABCD is a rectangle given: - $AB = 4x + y$ - $AD = 2x + y$ - $\angle BAD = 3y^2 + 41$ degrees - $BC = 6$ - $DC = 8$ 2. **Recall Properties:** - In a parallelogram, opposite sides are equal: $AB = DC$ and $AD = BC$. - A parallelogram is a rectangle if and only if one angle is $90^\circ$. 3. **Use side equalities:** - Since $AB = DC$, we have: $$4x + y = 8$$ - Since $AD = BC$, we have: $$2x + y = 6$$ 4. **Solve the system for $x$ and $y$:** Subtract second equation from first: $$ (4x + y) - (2x + y) = 8 - 6 \implies 2x = 2 \implies x = 1 $$ Substitute $x=1$ into $2x + y = 6$: $$ 2(1) + y = 6 \implies 2 + y = 6 \implies y = 4 $$ 5. **Find $\angle BAD$:** $$ \angle BAD = 3y^2 + 41 = 3(4)^2 + 41 = 3(16) + 41 = 48 + 41 = 89^\circ $$ 6. **Check if $\angle BAD$ is $90^\circ$:** Since $89^\circ \neq 90^\circ$, $ABCD$ is not a rectangle. **Final answer:** No, parallelogram ABCD is not a rectangle because $\angle BAD$ is $89^\circ$, not $90^\circ$.