1. **Problem statement:** The perimeter of a rectangle is 16 cm and its area is 12 cm\textsuperscript{2}. Calculate the length and width of the rectangle. 2. **Formulas and rules:** - Perimeter of a rectangle: $$P = 2(l + w)$$ where $l$ is length and $w$ is width. - Area of a rectangle: $$A = l \times w$$ 3. **Set up equations:** From the perimeter: $$2(l + w) = 16 \implies l + w = 8$$ From the area: $$l \times w = 12$$ 4. **Express one variable:** From $l + w = 8$, we get $$l = 8 - w$$ 5. **Substitute into area equation:** $$l \times w = 12 \implies (8 - w)w = 12$$ 6. **Expand and form quadratic:** $$8w - w^2 = 12$$ Rearranged: $$-w^2 + 8w - 12 = 0$$ Multiply both sides by $-1$ to simplify: $$\cancel{-}w^2 + \cancel{8}w - \cancel{12} = 0 \implies w^2 - 8w + 12 = 0$$ 7. **Solve quadratic equation:** Use quadratic formula $$w = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ where $a=1$, $b=-8$, $c=12$. Calculate discriminant: $$\Delta = (-8)^2 - 4 \times 1 \times 12 = 64 - 48 = 16$$ 8. **Find roots:** $$w = \frac{8 \pm \sqrt{16}}{2} = \frac{8 \pm 4}{2}$$ So, $$w_1 = \frac{8 + 4}{2} = 6$$ $$w_2 = \frac{8 - 4}{2} = 2$$ 9. **Find corresponding lengths:** If $w=6$, then $l = 8 - 6 = 2$. If $w=2$, then $l = 8 - 2 = 6$. 10. **Answer:** The length and width are 6 cm and 2 cm (order does not matter).