Question

the sum of the areas of two squares is 640 m^2. If the difference between their perimeters is 64m,then find
sides of the square.
​

Answer 1

Answer:

one side of first square=8

one side of second square=24

Step-by-step explanation:

Let the side of one square be x

Perimeter of this square = 4x

Given, Difference of perimeter of 2 squares = 64 m

Thus, Perimeter of the other square = (64 + 4x) m  

And, each side of this second square =  = (16 +x) m

According to the problem, sum of the areas of two squares is 640

According to the problem, sum of the areas of two squares is 640 m^{2}

\begin{array}{l}{x^{2}+(16+x)^{2}=640} \\ {x^{2}+256+32 x+x^{2}=640} \\ {2 x^{2}+256+32 x-640=0} \\ {2 x^{2}+32 x-384=0} \\ {x^{2}+16 x-192=0} \\ {x^{2}+16 x-192=0} \\ {x(x+24)-8(x+24)=0} \\ {(x+24)(x-8)=0} \\ {x=-24, \ 8}\end{array}

Since side of a square cannot be negative, each side of the square = 8 m.

And, each side of second square = (16+8) m = 24 m