Full solution

Q. A rectangular school yard has an area of

3,485

square meters and a perimeter of

252

meters. What are the dimensions of the school yard?

Define Variables: Let's call the length of the school yard $L$ and the width $W$ . The area $A$ is given by $A = L \times W$ , and the perimeter $P$ is given by $P = 2(L + W)$ .
Area and Perimeter Equations: We know the area $A$ is $3,485$ square meters, so $L \times W = 3,485$ .
Solve for Length and Width: The perimeter $P$ is $252$ meters, so $2(L + W) = 252$ . Simplifying this, we get $L + W = 126$ .
Quadratic Equation Solution: Now we have two equations: $L \times W = 3,485$ and $L + W = 126$ . Let's solve the second equation for $W$ : $W = 126 - L$ .
Calculate Discriminant: Substitute $W$ in the first equation: $L \times (126 - L) = 3,485$ . This simplifies to $L^2 - 126L + 3,485 = 0$ .
Find Square Root: We need to solve this quadratic equation. Let's use the quadratic formula: $L = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ , where $a = 1$ , $b = -126$ , and $c = 3,485$ .
Calculate Possible Lengths: Plugging the values into the quadratic formula, we get $L = \frac{126 \pm \sqrt{126^2 - 4 \cdot 1 \cdot 3,485}}{2}$ .
Calculate Possible Widths: Calculate the discriminant: $126^2 - 4 \times 3,485 = 15,876 - 13,940 = 1,936$ .
Final Valid Solutions: Now, find the square root of the discriminant: $\sqrt{1936} = 44$ .
Final Valid Solutions: Now, find the square root of the discriminant: $\sqrt{1936} = 44$ .So, the possible values for $L$ are $L = (126 \pm 44) / 2$ . This gives us two possible solutions for $L$ : $L = (126 + 44) / 2$ or $L = (126 - 44) / 2$ .
Final Valid Solutions: Now, find the square root of the discriminant: $\sqrt{1936} = 44$ .So, the possible values for $L$ are $L = (126 \pm 44) / 2$ . This gives us two possible solutions for $L$ : $L = (126 + 44) / 2$ or $L = (126 - 44) / 2$ .Calculate the two possible values for $L$ : $L = 170 / 2$ or $L = 82 / 2$ .
Final Valid Solutions: Now, find the square root of the discriminant: $\sqrt{1936} = 44$ .So, the possible values for $L$ are $L = (126 \pm 44) / 2$ . This gives us two possible solutions for $L$ : $L = (126 + 44) / 2$ or $L = (126 - 44) / 2$ .Calculate the two possible values for $L$ : $L = 170 / 2$ or $L = 82 / 2$ .Simplify the values for $L$ : $L$ $0$ or $L$ $1$ .
Final Valid Solutions: Now, find the square root of the discriminant: $\sqrt{1936} = 44$ .So, the possible values for $L$ are $L = (126 \pm 44) / 2$ . This gives us two possible solutions for $L$ : $L = (126 + 44) / 2$ or $L = (126 - 44) / 2$ .Calculate the two possible values for $L$ : $L = 170 / 2$ or $L = 82 / 2$ .Simplify the values for $L$ : $L$ $0$ or $L$ $1$ .If $L$ $0$ , then $L$ $3$ . If $L$ $1$ , then $L$ $5$ . Since a rectangle can have the length and width swapped, both solutions are valid.