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Sum of the areas of two squares is $400\ cm^2$. If the difference of their perimeters is 16 cm, find the sides of two squares.
Given:
Sum of the areas of two squares is $400\ cm^2$.
The difference of their perimeters$=16\ cm$.
To do:
We have to find the sides of the squares.
Solution:
Let the side of the smaller square be $x$ and the side of the larger square be $y$.
We know that,
Perimeter of a square of side of length $s=4s$.
Perimeter of larger square$=4y$.
Perimeter of smaller square$=4x$.
This implies,
$4y-4x=16$
$4(y-x)=16$
$y-x=\frac{16}{4}=4$
$y=x+4\ cm$----(1)
Area of a square of length $s=s^2$
Area of the larger square $=y^2\ cm^2$.
Area of the smaller square$=x^2\ cm^2$.
According to the question,
$x^2+y^2=400$
$x^2+(x+4)^2=400$ (From equation 1)
$x^2+x^2+8x+16=400$
$2x^2+8x+16-400=0$
$2x^2+8x-384=0$
$2(x^2+4x-192)=0$
$x^2+4x-192=0$
Solving for $x$ by factorization method, we get,
$x^2+16x-12x-192=0$
$x(x+16)-12(x+16)=0$
$(x+16)(x-12)=0$
$x+16=0$ or $x-12=0$
$x=-16$ or $x=12$
Length cannot be negative. Therefore, the value of $x=12$.
$y=x+4=12+4=16\ cm$
The sides of the two squares are $12\ cm$ and $16\ cm$.