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 $16\ cm$.

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Updated on: 10-Oct-2022

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