Let #a,band c # are three sides of a right triangle where #c# is the hypotenuse , This means #c>a and c>b#.
As per given condition of the problem #a,band c # are all positive whole numbers ,
and
#1/2ab =a+b+c.....[1]#
Again by Pythagorean theorem
#c^2=a^2+b^2........[2]#
Combining [1] and [2] we get
#=>(1/2ab-a-b)^2=a^2+b^2#
#=>1/4a^2b^2+a^2+b^2-2*1/2a^2b-2*1/2ab^2+2ab=a^2+b^2#
#=>1/4a^2b^2-a^2b-ab^2+2ab=0#
#=>1/4ab-a-b+2=0.# as #ab!=0#
#=>1/2ab-2a-2b+4=0........[3]#
Combining {1} and {3] we get
#2a+2b -4=a+b+c.#
#=>c=a+b-4......[4]#
Combining [1] and [4] we get
#2a+2b-4=1/2ab#
#=>a+b=2+1/4ab......[5]#
LHS of this relation is an integer. To satisfy this both #a and b# should be even or any of #aand b# is a multiple of #4#. So minimum value of # a or b=4#. If minimum value of # b=4# then minimum value of #a# will be #3# and #c=5# But it does not satisfy equation [1]
For integer values of #a,band c# satisfying these conditions and equation {2] we get the following when #b=8#,the next integral multiple of #4#
#a=6,b=8and c=10#
For #b# taking other higher multiple of #4 # as 12,16,20..etc the relation (5) is not satisfied.
