11. The rectangle has a perimeter of 86 yards. Which is its area? (A) 210 sqyd (II) 420 sqyd (C) 560 sqyd (D) 840 sqyd

Let the length be \( x \) and the width be \( y \). We are given that the perimeter of the rectangle is \[ 2(x+y)=86. \] Dividing both sides by 2, we obtain \[ x+y=43. \] The area of the rectangle is \[ A = xy. \] Among the answer choices, we test whether there is a pair of whole numbers \( x \) and \( y \) with sum 43 that yields one of the given areas. Suppose the area is \(420\) square yards. Then we have \[ xy=420 \quad \text{and} \quad x+y=43. \] Express \( y \) as \[ y=43-x, \] and substitute into the area equation: \[ x(43-x)=420. \] This gives \[ 43x - x^2 = 420, \] or, rearranging, \[ x^2 - 43x + 420 = 0. \] To solve this quadratic, we look for factors of \(420\) that add to \(43\). Notice that \[ 15 \times 28 = 420 \quad \text{and} \quad 15+28=43. \] Thus the quadratic factors as \[ (x-15)(x-28)=0, \] so the solutions are \[ x=15 \quad \text{or} \quad x=28. \] If \( x=15 \) then \( y=43-15=28 \), and if \( x=28 \) then \( y=43-28=15 \). In either case, the area is \[ A = 15 \times 28 = 420 \, \text{sqyd}. \] Therefore, the area of the rectangle is \[ \boxed{420 \, \text{sqyd}}. \]