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a school is constructing a rectangular play area against an exterior wa…

Question

a school is constructing a rectangular play area against an exterior wall of the school building. it uses 120 feet of fencing material to enclose three sides of the play area. width is the side perpendicular to the building. complete the table by giving the length and area for each width. (the width (feet): 10, 30, 40, w; length (feet); area (square feet))

Explanation:

Step1: Set up length - width relationship

Since we are fencing three sides and the total length of fencing is 120 feet, and the width is \(w\), the length \(l = 120 - 2w\).

Step2: Calculate length and area for \(w = 10\)

Length \(l=120 - 2\times10=100\) feet. Area \(A = l\times w=100\times10 = 1000\) square feet.

Step3: Calculate length and area for \(w = 30\)

Length \(l=120 - 2\times30 = 60\) feet. Area \(A=l\times w=60\times30=1800\) square feet.

Step4: Calculate length and area for \(w = 40\)

Length \(l=120 - 2\times40=40\) feet. Area \(A = l\times w=40\times40 = 1600\) square feet.

width (feet)length (feet)area (square feet)
30601800
40401600

Answer:

width (feet)length (feet)area (square feet)
30601800
40401600