The length of a rectangle is $1 \mathrm{yd}$ more than twice the width, and the area of the rectangle is $45 \mathrm{yd}^{2}$. Find the dimensions of the rectangle.
Answer
\(\boxed{\text{Final Answer: The dimensions of the rectangle are }\frac{9}{2}\text{ yards (width) and 10 yards (length).}}\)
Step 1 ：Given that the length (L) is 1 more than twice the width (W), and the area (A) is 45 square yards, we can write these as equations: \(L = 2W + 1\) and \(A = LW = 45\)
Step 2 ：Substitute the first equation into the second equation to solve for W: \(45 = (2W + 1) * W\)
Step 3 ：Solve the quadratic equation for W, and then find L using the first equation: \(W = \frac{9}{2}\) and \(L = 10\)
Step 4 ：\(\boxed{\text{Final Answer: The dimensions of the rectangle are }\frac{9}{2}\text{ yards (width) and 10 yards (length).}}\)
