The parking lot is located along the front and one side of the school
building.
Reasons:
The given dimensions in the compound are;
Length of the school and the old parking lot = 300 + 75
Width of the school and the old parking lot = 165 + 75
Length of the school, the old and new parking lot = 300 + 75 + x
Width of the school and the old and new parking lot = 165 + 75 + x
Area of the new parking lot the school wants = Double the old parking lot
In terms of areas, we have;
New compound - School = 2 × (old compound - school)
Which gives;
(300 + 75 + x) × (165 + 75 + x) - 300 × 165 = 2 × ((300 + 75) × (165 + 75) - 300 × 165)
2 × (375 × 240 - 49,500) = (375 + x) × (240 + x) - 49,500
Simplifying gives;
81,000 = x² + 615·x + 40,500
x² + 615·x + 40,500 - 81,000 = 0
x² + 615·x - 40,500 = 0
Factorizing with a graphing calculator gives;
(x - 60)·(x + 675) = 0
Therefore;
x = 60 or x = -675
Given that x is a natural number, we have, the correct value is; x = 60;
The distance with which the lot should be expanded, x = 60 feet.
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