Sum of the areas of two squares is $640\ m^2$. If the difference of their perimeter is 64 m, find the sides of the two squares.

Given:

Sum of the areas of two squares is $640\ m^2$.

The difference of their perimeters$=64\ m$.

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,

$4y-4x=64$

$4(y-x)=64$

$y-x=\frac{64}{4}=16$

$y=x+16\ m$----(1)

Area of a square of length $s=s^2$

Area of the larger square $=y^2\ m^2$.

Area of the smaller square$=x^2\ m^2$.

According to the question,

$x^2+y^2=640$

$x^2+(x+16)^2=640$   (From equation 1)

$x^2+x^2+32x+256=640$

$2x^2+32x+256-640=0$

$2x^2+32x-384=0$

$2(x^2+16x-192)=0$

$x^2+16x-192=0$

Solving for $x$ by factorization method, we get,

$x^2+24x-8x-192=0$

$x(x+24)-8(x+24)=0$

$(x+24)(x-8)=0$

$x+24=0$ or $x-8=0$

$x=-24$ or $x=8$

Length cannot be negative. Therefore, the value of $x=8$.

$y=x+16=8+16=24\ m$

The sides of the two squares are $8\ m$ and $24\ m$.

Tutorialspoint

Simply Easy Learning

Updated on: 10-Oct-2022

46 Views

Kickstart Your Career

Get certified by completing the course

Get Started