Question

the perimeter of a rectangle is 60 meters. what are the possible lengths of any of the rectangle’s sides if its area exceeds 216 square meters?
○ (12, 18)
○ 12,18)
○ (-∞, 12) ∪ (18, ∞)
○ (0, 12) ∪ (18, 60)

Explanation:

Step1: Define variables and perimeter formula

Let the length of the rectangle be \( l \) and the width be \( w \). The perimeter of a rectangle is given by \( P = 2(l + w) \). We know \( P = 60 \), so \( 2(l + w)=60 \), which simplifies to \( l + w = 30 \), so \( w = 30 - l \).

Step2: Define area formula and inequality

The area of a rectangle is \( A = lw \). We want \( A>216 \). Substituting \( w = 30 - l \) into the area formula, we get \( l(30 - l)>216 \).

Step3: Rearrange the inequality

Expanding and rearranging the inequality: \( 30l - l^{2}>216 \) becomes \( l^{2}-30l + 216<0 \).

Step4: Solve the quadratic equation

First, solve the quadratic equation \( l^{2}-30l + 216 = 0 \). Factoring, we get \( (l - 12)(l - 18)=0 \). So the roots are \( l = 12 \) and \( l = 18 \).

Step5: Analyze the quadratic inequality

The quadratic function \( f(l)=l^{2}-30l + 216 \) is a parabola opening upwards (since the coefficient of \( l^{2} \) is positive). The inequality \( f(l)<0 \) holds between the roots. So \( 12 < l < 18 \). Since length and width are interchangeable, the possible lengths of any side (length or width) are in the interval \( (12, 18) \).

Answer:

\( (12, 18) \)