# The area of the square formed on the diagonal of a rectangle as its side is 108 1/3 % more than the area of the rectangle. If the perimeter of the rectangle is 28 units, find the difference between the sides of the rectangle

#### Comments and Answers (3)

**Vivek** * *

how did the answer came

**Andrew** * *

please anyone explain this answer

**Soumen Sarkar Khulna ** * ~ Khulna*

let length of rectangle = x and breth =y therefore 2(x+y)=28, x+y=14,

side of the square=diagonal of the rectangle=sruare root of x^2+y^2 therefore area of the square=x^2+y^2, according to the question x^2+y^2=xy(1+108.1/3%)

=>12x^2+12y^2=25xy

=>12x^2+24xy+12y^2=49xy

=> 12(x^2+2xy+y^2)=49xy

=>12(x+y)^2=49xy

=>12*14^2=49xy

=>xy=48,

now (x-y)^2=(x+y)^2-4xy

=> (x-y)^2=(14)^2-4*48

=>(x-y)^2=4

=>(x-y)=2 Ana.

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