A wire with length of 52 cm is cut into two unequal lengths. Each par is bent to form a square. If the sum of the area of the two squares is 97 sq. cm., what is the area of the smaller square?

$A_1 + A_2 = 97$

$\left( \dfrac{x}{4} \right)^2 + \left( \dfrac{52 - x}{4} \right)^2 = 97$
 

 

$\dfrac{x^2}{16} + \dfrac{2704 - 104x + x^2}{16} = 97$

$x^2 + (2704 - 104x + x^2) = 1552$

$2x^2 - 104x + 1152 = 0$

$x^2 - 52x + 576 = 0$

$(x - 16)(x - 36) = 0$

$x = 16 ~ \text{and} ~ 36$
 

$\text{Area of smaller square} = \left( \dfrac{16}{4} \right)^2$

$\text{Area of smaller square} = 16 ~ \text{cm}^2$