Question

a man was at one corner of a 630 square - yard rectangular field. he then ran to the furthest corner, as shown in the diagram. if the length (l) of the field is 3 yards longer than three times the width (w), then how far did the man run? find both the exact solution and an approximate solution.
exact solution
he ran exactly (square) yards.
(simplify your answer. give an exact answer, using radicals as needed.)
approximate solution

Explanation:

Step1: Set up equations for the rectangle

We know that the area of a rectangle $A = L\times W$, and $A = 630$ square - yards, also $L=3W + 3$. Substitute $L$ into the area formula: $(3W + 3)W=630$. Expand to get $3W^{2}+3W - 630 = 0$. Divide through by 3: $W^{2}+W - 210 = 0$.

Step2: Solve the quadratic equation

Use the quadratic formula $W=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$ for the quadratic equation $ax^{2}+bx + c = 0$. Here, $a = 1$, $b = 1$, and $c=-210$. Then $W=\frac{-1\pm\sqrt{1^{2}-4\times1\times(-210)}}{2\times1}=\frac{-1\pm\sqrt{1 + 840}}{2}=\frac{-1\pm\sqrt{841}}{2}=\frac{-1\pm29}{2}$. We take the positive root since width cannot be negative, so $W=\frac{-1 + 29}{2}=14$ yards.

Step3: Find the length

Substitute $W = 14$ into $L=3W + 3$. Then $L=3\times14+3=42 + 3=45$ yards.

Step4: Calculate the distance the man ran

The man ran the length of the diagonal of the rectangle. By the Pythagorean theorem $d=\sqrt{L^{2}+W^{2}}$, where $d$ is the diagonal. Substitute $L = 45$ and $W = 14$: $d=\sqrt{45^{2}+14^{2}}=\sqrt{2025 + 196}=\sqrt{2221}$ yards.

Answer:

Exact Solution: $\sqrt{2221}$ yards
Approximate Solution: $\sqrt{2221}\approx47.13$ yards