SOLUTION: A rectangle has an area of 120 sq. cm.. Its length and width are whole numbers. What are the possibilities for the two numbers? Which possibility gives the smallest perimeter?

Question 255051: A rectangle has an area of 120 sq. cm.. Its length and width are whole numbers. What are the possibilities for the two numbers? Which possibility gives the smallest perimeter?
Found 3 solutions by palanisamy, Greenfinch, CharlesG2:
Answer by palanisamy(496)   (Show Source): You can put this solution on YOUR website!
A rectangle has an area of 120 sq. cm.. Its length and width are whole numbers.
The different possibilities are
length width
120 m 1 m
60 m 2 m
40 m 3 m
30 m 4 m
24 m 5 m
20 m 6 m
15 m 8 m
12 m 10 m
The smallest perimeter = 2(12+10) = 2*22 = 44m
Answer by Greenfinch(383)   (Show Source): You can put this solution on YOUR website!
1,120: 2,60: 3,40: 4,30: 5,24: 6,20: 8,15: 10;12 all multiply to 120.
The smallest perimeter is the one closest to a square, which is 10,12 giving a perimeter of 48 cm.
Answer by CharlesG2(834)   (Show Source): You can put this solution on YOUR website!
A rectangle has an area of 120 sq. cm.. Its length and width are whole numbers. What are the possibilities for the two numbers? Which possibility gives the smallest perimeter?
find the factors of 120:
120=1*120
120=2*60
120=3*40
120=4*30
120=5*24
120=6*20
120=8*15
120=10*12
perimeter of a rectangle = 2*length + 2 * width = 2 * (length + width)
perimeters of above factorizations:
2*(1+120)=2*121=142
2*(2+60)=2*62=124
2*(3+40)=2*43=86
2*(4+30)=2*34=68
2*(5+24)=2*29=58
2*(6+20)=2*26=52
2*(8+15)=2*23=46
2*(10+12)=2*22=44
the largest factors, 10 and 12 give smallest perimeter which is 44

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