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A rectangular lot has an area of 240 m^2. What is the width of the lot if it requires 64 m of fencing material to enclose it?

Sagot :

A=240 m²
P=64 m

Let x-length and y-width
xy= 240m²
(2x+2y=64m)/2
x+y=32m

Find the pair of factors of 240 that has a sum of 32.
xy = 240m² - (1&240 = 241) (2&120 = 122) (3&80 = 83) (4&60 = 64) (5&48 = 53) (6&40 = 46) (8&30 = 38) (10&24 = 34) (12&20 = 32)

Then x(length) and y(width) could be 20m and 12m.