Let a,band c a,bandc are three sides of a right triangle where cc is the hypotenuse , This means c>a and c>bc>aandc>b.
As per given condition of the problem a,band c a,bandc 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.
