Right Triangle

$First solution$ -The sides must be in a ratio of $3,4,5$ the next Pythagorean triple is $5,12,13$ which adds up to give $30$ and area gives $30$ .So the hypotenuse must be $13cm$ $Second - solution$

$a + b + h = 30$

$a b = 60$

$a ^2 + b ^2 = h^ 2$

Since $a + b + h = 30$ as follows

$a + b = 300 - h$

Square both sides

$(a + b)^2 = (60 - h)^2$

Expand both sides

$a^2 + b^2 + 2 a b$ = $30^2 + h^2 - 60h$

Combine the equation $a^2 + b^2 = h^2$ with the above equation to obtain

$2 a b = 30^2 - 60 h$

a *b is known to be equal to 60, hence the above equation becomes

$120 = 60^2 - 60 h$

Solve for h to obtain

$h = 13 cm$

Here a+b+c=30....(1) and 1/2(a * b)=30 or a * b=60.....(2) From (1) we get a+b+c=30 or a+b=30-c or a^2+b^2+2ab=900-60c+c^2 [squaring both sides] or c^2+2*60=900-60c+c^2 [using pythagorean theorem and from (2)] or c^2-c^2+60c=900-120 or 60c=780 or c=780/60 so c=13,so hypotenuse is ,c=13

We infer $ab=60$ , $a+b+h=30$ and $a^2+b^2=h^2$

Now taking, the $3^{rd}$ equation

$a^2+b^2+2ab=h^2+2ab$

$(a+b)^2=h^2+120$ ..........[We change RHS $ab=60$ ]

$(30-h)^2=h^2+120$ ......[first equation]

$h^2-60h+900=h^2+120$ ....Furthur is..