Question

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.

Answer 1

Let the side of the first square be 'a'm and that of the second be ${}^{\prime }{A}^{\prime }$ m.

Area of the first square $=a^2$ sq m.

Area of the second square $=A^2$ sq m.

Their perimeters would be $4a$ and $4A$ respectively.

Given $4A-4a=24$

$A-a=6$ --(1)

${A^2+a}^2=468$ --(2)

From (1), $A=a+6$

Substituting for $A$ in (2), we get

${{(a+6)}^2+a}^2=468a^2+12a+36{+a}^2=4682a^2+12a+36=468a^2+6a+18=234a^2+6a-216=0a^2+18a-12a-216=0a(a+18)-12(a+18)=0(a-12)(a+18)=0a=12,-18$

So, the side of the first square is $12$ m. and the side of the second square is $18$ m.