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# hi please answer this question?

a rectangle is 6 times as long as it is wide, the perimater is 120 cm. find the dimension of rectangle found to the nearest tenth if necessary?

### 5 Answers

- 1 decade agoFavorite Answer
Length = L

Wide = W

perimeter = (2L) + (2W)

length = 6W

120 cm = (2.6W) + (2W)

120 cm = 12W + 2W

120 cm = 14W

W (wide) = 120/14 = 8,6

L = 6W

L= 6. (8.57)

L= 51,4

- Jun AgrudaLv 71 decade ago
Width (x):

2(x + 6x) = 120

x + 6x = 60

7x = 60

x = 60/7 or 8 4/7

Length:

= 6(60/7)

= 360/7 or 51 3/7

Answer: Approximated to the nearest tenth dimension: 8.6 by 51.4 cm

Proof (perimeter: 120 cm):

= 2(8.6 cm) + 2(51.4 cm)

= 17.2 cm + 102.8 cm

= 120 cm

- Anonymous1 decade ago
Let's consider both the width and length to be 'x' and consider the perimeter of a rectangle, P = 2l + 2w, then:

P = 2(6x + x)

120 = 12x + 2x

120 = 14x

120/14 = 14x/14

x = 120/14 or 8.6

- 1 decade ago
let breadth be x which means length is 6x

perimeter, 2(6x+x)=120

14x=120

x=120/14

x= 8.57=9(approx)

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- Anonymous1 decade ago
Perimeter = 2(length+Width)

p =2(w+l)

l=6w

p=2(w+6w)

120=2(7w)

60= 7W

W=8.571

L= 51.426