Question

the sum of the perimeter of two squares is 40cm while the sum of the area is 58cm². Find length the side of the square.

Answer 1

$a,\ b-the\ length\ of\ sides\\\\(1)\qquad4a+4b=40\qquad\text{divide both sides by 4}\\\\a+b=10\to a=10-b\\\\(2)\qquad a^2+b^2=58\\\\\text{Substitute}\ (1)\ \text{to}\ (2):\\\\(10-b)^2+b^2=58\qquad\text{Use}\ (x-y)^2=x^2-2xy+y^2\\\\10^2-2(10)(b)+b^2+b^2=58\\\\100-20b+2b^2=58\qquad\text{subtract 58 from both sides}\\\\42-20b+2b^2=0\\\\2b^2-20b+42=0\qquad\text{divide both sides by 2}\\\\b^2-10b+21=0\\\\b^2-7b-3b+21=0\\\\b(b-7)-3(b-7)=0\\\\(b-7)(b-3)=0\iff b-7=0\ \vee\ b-3=0\\\\\boxed{b=7\ \vee\ b=3}$

$\text{Put the values of b to }\ (1):\\\\for\ b=7\to a=10-7=3\\\\for\ b=3\to a=10-3=7\\\\Answer:\ \boxed{7cm\ and\ 3cm.}$