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the perimeterof a rectangle is 52 feet. the length of the rectangle is 10 feet longer than the width. find the dimension of the rectangle?

Sagot :

The width is 31
The length is 21

the missing is the width
             
                 52
              -  10       --- subtract the feet longer first in the perimeter
              ----------
                 42

Then divide 42 ÷ 2 = 21 
21+ 10 = 31
the answer is 31
                    
let x = the width
let x + 10 = the length

P = 2L + 2W

52 = 2(x+10) + 2x
52 = 2x + 20 + 2x
52 = 4x + 20
52-20=4x
32=4x
1/4(32) = (4x)1/4
8 = x

..therefore, width = x = 8ft
length = x +10 = 8+10 = 18ft

dimension : 18 by 8