WireA piece of wire of length 50 m is cut into two parts. Each part is then bent to form a square. It is found that the total area of the sq

okay

when you cut the peice of wire let the peices be x and 50-x  metes long     (that works because 50-x+x=50)

so

one of the squares will have side lengths of x/4 metres and the other will have sides of (50-x)/4 metres

So

The areas if the squares will be        $$\left(\frac{x}{4}\right)^2+\left(\frac{50-x}{4}\right)^2$$

so

$$\begin{array}{rll} \left(\frac{x}{4}\right)^2+\left(\frac{50-x}{4}\right)^2&=&100\\\\ \frac{x^2}{16}+\frac{2500-100x+x^2}{16}&=&100\\\\ \frac{2x^2-100x+2500}{16}&=&100\\\\ 2x^2-100x+2500&=&1600\\\\ 2x^2-100x+900&=&0\\\\ x^2-50x+450&=&0\\\\ x&=&\frac{50\pm\sqrt{2500-1800}}{2}\\\\ x&=&\frac{50\pm\sqrt{700}}{2}\\\\ x&=&\frac{50\pm10\sqrt{7}}{2}\\\\ x&=&25\pm5\sqrt{7}\\\\ \end{array}$$

now one of the peices will be    $$25+5\sqrt7$$    and the other will be   $$25-5\sqrt7$$

the difference between these will be       $$25+5\sqrt7-(25-5\sqrt7)=10\sqrt7\;m$$

the difference between the sides will be a quarter of this =   $$2.5\sqrt7\;\;metres$$