The diagonal of a rectangular field is 60 meters more than the shorter side.

Solution:

Let the length of smaller side of rectangle be $=x$ metres then larger side be $=x+30$ metres and their diagonal be $=x+60$ metres

Then, as we know that Pythagoras theorem

$x^{2}+(x+30)^{2}=(x+60)^{2}$

$x^{2}+(x+30)^{2}=(x+60)^{2}$

$x^{2}+x^{2}+60 x+900=x^{2}+120 x+3600$

$2 x^{2}+60 x+900-x^{2}-120 x-3600=0$

$x^{2}-60 x-2700=0$

$x^{2}-90 x+30 x-2700=0$

$x(x-90)+30(x-90)=0$

$(x-90)(x+30)=0$

$(x-90)=0$

$x=90$

or

$(x+30)=0$

$x=-30$

But, the side of rectangle can never be negative.

Therefore, when $x=90$ then

$x+30=90+30$

$=120$

Hence, length of smaller side of rectangle be $=90$ metres and larger side be $=120$ metres