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Algebra- Rectangle Homework Problem?
A rectangle is 4cm longer than it is wide. If the length and width are both decreased by 2cm, the area is decreased by 24 cm sq. What are the dimensions of the original rectangle?
2 Answers
- SqdancefanLv 79 years agoFavorite Answer
w(w+4) = (w-2)(w+2)+24
w^2 + 4w = w^2 + 20
4w = 20
w = 5
The original rectangle is 5 cm wide and 9 cm long.
- boundsLv 44 years ago
permit w equivalent the kind of ft interior the width. The length is longer than the width. it is 4 situations longer ( 4w ) -- and then even 2 ft better than that ( 4w + 2 ) So the kind of ft interior the size is ( 4w + 2 ) the fringe is the sum of all 4 of those aspects w + ( 4w + 2 ) + w + ( 4w + 2 ) = 80 4 a million) assemble like words w + 4w + w + 4w + 2 + 2 = 80 4 2) combine like words 10w + 4 = 80 4 3) Subtract 4 from the two aspects 10w = 80 4) Divide the two aspects by using 10 w = 8 One section is 8 ft so the different section is ( 4*8 + 2 ) ft, that's 34 ft. answer: The rectangle is 8 by using 34 ft.
