The hypotenuse: # color(blue)(AB) = 6.1# cm (assuming length to be in cm)

Let the shorter leg: #color(blue)(BC) =x # cm

Let the longer leg: #color(blue)(CA) =(x +4.9 ) # cm

As per Pythagoras Theorem :

#(AB)^2 = (BC)^2 + (CA)^2#

#(6.1)^2 = (x)^2 + (x+4.9)^2#

#37.21 = (x)^2 +color(green)( (x+4.9)^2#

Applying the below property to # color(green)( (x+4.9)^2# :

#color(blue)((a+b)^2 = a^2 +2ab +b^2#

#37.21 = (x)^2 +[color(green)( x^2 + 2 xx x xx4.9 + 24.01]] #

#37.21 = (x)^2 +[color(green)( x^2 + 9.8x + 24.01]] #

#37.21 = 2x^2 + 9.8x + 24.01#

#13.2 = 2x^2 + 9.8x #

# 2x^2 + 9.8x -13.2 =0 #

Multiplying the entire equation by #10# to remove the decimal

# 20x^2 + 98x -132 =0 #

Dividing the entire equation by #2# for simplicity

# 10x^2 + 49x -66=0 #

The equation is now of the form #color(blue)(ax^2+bx+c=0# where:

#a=10, b=49, c=-66#

The **Discriminant** is given by:

#Delta=b^2-4*a*c#

# = (49)^2-(4*(10)*(-66))#

# = 2401 +2640 = 5041#

The solutions are found using the formula

#x=(-b+-sqrtDelta)/(2*a)#

#x = ((-49)+-sqrt(5041))/(2*10) = (-49+-(71))/20#

#x = = (-49+(71))/20 = 22/20 = 1.1#

#x = = (-49-(71))/20# (not applicable since side cannot be negative)

So, the shorter side #color(blue)(x=1.1 cm #

The longer side #= color(blue)(x +4.9 = 6cm #