Answer:
The dimension of the lot is 201.28 ft by 148.28 ft
Step-by-step explanation:
Given;
diagonal of the parking lot, d = 250 ft
let the length of the parking lot = L
the width, W = L - 53
The diagonal of the lot, length of the lot, and width of the lot form a right triangle.
Apply Pythagoras theorem to determine the length, L
L² + W² = 250²
L² + (L - 53)² = 250²
L² + L² - 106L + 2809 = 62,500
2L² - 106L + 2809 - 62,500 = 0
2L² - 106L - 59,691 = 0
this forms quadratic equation; a = 2, b = -106 and c = -59,691
[tex]L = \frac{- b \ \ +/- \ \ \sqrt{b^2 - 4ac} }{2a} \\\\L = \frac{- (-106) \ \ +/- \ \ \sqrt{(-106)^2 - 4(-59691 \times 2)} }{2(2)}\\\\L = \frac{106 \ \ +/- \ \ \sqrt{488764} }{4} \\\\L = \frac{106 + 699.12}{4} \\\\L = 201.28 \ ft[/tex]
The width, W = 201.28 - 53
W = 148.28 ft
