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Sum of the areas of two squares is $468\ m^2$. If the difference of their perimeters is 24 m, find the sides of the two squares.
Given:
Sum of the areas of two squares is $468\ m^2$.
The difference of their perimeters$=24\ m$.
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,
$4y-4x=24$
$4(y-x)=24$
$y-x=\frac{24}{4}=6$
$y=x+6\ m$----(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,
$x^2+y^2=468$
$x^2+(x+6)^2=468$ (From equation 1)
$x^2+x^2+12x+36=468$
$2x^2+12x+36-468=0$
$2x^2+12x-432=0$
$2(x^2+6x-216)=0$
$x^2+6x-216=0$
Solving for $x$ by factorization method, we get,
$x^2+18x-12x-216=0$
$x(x+18)-12(x+18)=0$
$(x+18)(x-12)=0$
$x+18=0$ or $x-12=0$
$x=-18$ or $x=12$
Length cannot be negative. Therefore, the value of $x=12$.
$y=x+6=12+6=18\ m$
The sides of the two squares are $12\ m$ and $18\ m$.
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