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 #
