1. The problem asks to find the perimeter and area of a rhombus with side length 10.7 ft and one diagonal 8 ft. 2. The perimeter $P$ of a rhombus is given by the formula: $$P = 4 \times \text{side}$$ Since all sides are equal, multiply the side length by 4. 3. The area $A$ of a rhombus can be found using the formula: $$A = \frac{1}{2} \times d_1 \times d_2$$ where $d_1$ and $d_2$ are the lengths of the diagonals. 4. We know one diagonal $d_1 = 8$ ft, but $d_2$ is unknown. We can find $d_2$ using the Pythagorean theorem because the diagonals of a rhombus are perpendicular and bisect each other. 5. Half of $d_1$ is $\frac{8}{2} = 4$ ft. Let half of $d_2$ be $x$. 6. Using the right triangle formed by half the diagonals and the side: $$10.7^2 = 4^2 + x^2$$ $$114.49 = 16 + x^2$$ $$x^2 = 114.49 - 16 = 98.49$$ $$x = \sqrt{98.49} = 9.92$$ 7. So the full diagonal $d_2 = 2 \times 9.92 = 19.84$ ft. 8. Calculate perimeter: $$P = 4 \times 10.7 = 42.8$$ ft 9. Calculate area: $$A = \frac{1}{2} \times 8 \times 19.84 = 79.36$$ sq ft 10. Final answers: Perimeter $P = 42.8$ ft Area $A = 79.36$ sq ft