Question

Sum of the area of two squares is $468m^2$ if the difference of the the is perimeter is $25m$ find the sides of the two squares.

Answer 1

$a_2^1+a_2^2=468$

$4a_1-4a_2=25$

$(a_1-a_2)-\frac{25}{4}$ ________ (1)

$a_1^2+a_2^2-2a_1{a}_2=\frac{625}{16}$

$2a_1{a}_2=468-\frac{625}{16}=\frac{6863}{16}$

$(a_1+a_2)=a_1^2+a_2^2+2a_1{a}_2$

$=468+\frac{6863}{16}=\frac{14351}{16}$

$a_1+a_2=\frac{119.8}{4}$ ________ (2)

adding (1) & (2)

$2a_1=\frac{44.8}{4}=36.2$

$a_1=18.1$ $a_2=\frac{119.8}{4}-8.1$

$a_1=18.1$ $a_2=11.85$