Question

The width of a rectangular parking lot is 53 ft less than its length. Determine the dimensions of the parking lot if it measures 250 ft diagonally.

Answer 1

The dimension of the lot is 201.28 ft by 148.28 ft

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

`L = (- b \ \ +/- \ \ √(b^2 - 4ac) )/(2a) \\\\L = (- (-106) \ \ +/- \ \ √((-106)^2 - 4(-59691 * 2)) )/(2(2))\\\\L = (106 \ \ +/- \ \ √(488764) )/(4) \\\\L = (106 + 699.12)/(4) \\\\L = 201.28 \ ft`

The width, W = 201.28 - 53

W = 148.28 ft