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.