Triangle, perimeter = area?

$Perimeter=a+b+c=60.~~~\therefore~S=30. \\ Area=\sqrt{S(S-a)(S-b)(S-c)}=60.\\ \therefore~60^2=30(30-a)(30-b)(30-c).\\ But~\dfrac{60^2}{30}=2^3*3*5.~~~\\ a,b,c~are~integers, \\ ~~~\\ (1)~one~of~them,~say~a~is~divisible~by~15\\ \therefore~8=(2-\dfrac a {15})(30-b)(30-c).\\ Since~a<30,~only~a=15~is~possible.~~So~~b+c=60-a=45.\\ So ~WLOG~say~b~is~odd~and~c~is~even~then~~(30-b)(30-c)=8~must ~have~b=29~and~c=22.\\ But~then~a+b+c=15+29+22\neq60. ~No~solution.\\ ~~~\\ (2)~Or~~~WLOG~say~a~divisible~by~5~~~and~~~b~by~3.\\ \implies~~~\color{#3D99F6}{ 8=(6-\dfrac a 5)(10-\dfrac b 3)(30-c)}.\\ So~probable~ (6-\dfrac a 5)=4,~2,~or~1;~~~~~ (10-\dfrac b 3)=1,~2,~4,~or~8.~~~~~ (30-c)=1,~2,~4,~or~8...............(3).\\ ~~~~\\ a,~b,~c<30.\\ So~probables~~~~a=25,20,15,10,5.\\ But~by~(3)~corresponding~~(6-\dfrac a 5)~~only=1,2,~or~4~are~probable.\\ So~\color{#20A900}{a=25, ~20,~or~10}~only~are~probable.\\ Again~probable~~~~b = 3,6,9,12,15,18,21,24,27.\\ But~by~(3)~corresponding~~(10-\dfrac b 3)~~only=1,2,4,~or~8~are~probale.\\ So~\color{#20A900}{b=6,~18,~24,~27}~only~are~probable.\\ Let~us~start~trials~with~\Large~~If~a=25,~~ (6-\dfrac a 5)=1.\\ \therefore~(1)(10-\dfrac b 3)(30-c)=8.\\ ~~~~~If~b=6,~~ 8=(1) (10-\dfrac 6 3)(30-c)=8*(30-c).~~So~(30-c)=1~~and~~~c=29.\\ So~a+b+c=25+6+29=60 ~~as~required.\\ So~the~longest ~side~=\Large~\color{#D61F06}{29}.$

$From (3) ~we~may~proceed ~as~follows~to~ go~ through~all~probabilities.\\ \begin{array}{c} \\ \hline (6-\dfrac a 5)& a &~&(10-\dfrac b 3)& b &~&(30-c)& c &~&a+b+c&~&Solution?&\\ \hline 1 &\color{#20A900}{25}&~& 1 & \color{#20A900}{27} &~& 8 & \color{#20A900}{22} &~& 74 &~& NO &\\ & &~& 2 & \color{#20A900}{24} &~& 4 & \color{#20A900}{26} &~& 75 &~& NO &\\ & &~& 4 & \color{#20A900}{18} &~& 2 & \color{#20A900}{28} &~& 71 &~& NO &\\ & \color{#D61F06}{25} &~& 8 & \color{#D61F06}{6} &~& 1 & \color{#D61F06}{29 } &~& \color{#D61F06}{60}&~&\color{#D61F06}{Yes}&\\ 2 &\color{#20A900}{20}&~& 1 & \color{#20A900}{27} &~& 4 & \color{#20A900}{26} &~& 73 &~& NO &\\ & &~& 2 & \color{#20A900}{24} &~& 2 & \color{#20A900}{28 } &~& 72 &~& NO &\\ & &~& 4 & \color{#20A900}{18} &~& 1 & \color{#20A900}{29} &~& 67 &~& NO &\\ 4 &\color{#20A900}{10} &~& 1 & \color{#20A900}{27} &~& 2 & \color{#20A900}{28} &~& 65 &~& NO &\\ & &~& 2 & \color{#20A900}{24} &~& 1 & \color{#20A900}{29} &~& 63 &~& NO &\\ \hline \end{array}$

Let a, b and c be the length of the sides of the triangle.

Using Heron's formula for finding area

$S=\frac{a+b+c}{2}$

$\sqrt{S(S-a)(S-b)(S-c)}=60~units^2$

Since $a+b+c=60~units---(1)$

$S=30~units$

$\sqrt{30(30-a)(30-b)(30-c)}=60---(2)$

From $(2)$

$30(30-a)(30-b)(30-c)=3600~units^2$

$(30-a)(30-b)(30-c)=120~units^2$

This means that $(a,b,c < 30)$ and $((30-a),(30-b),(30-c)<30)$

There are only 9 possible values of $30-a, 30-b,30-c<30$

From $(1)$

$(30-a)+(30-b)+(30-c)=30~units$

From the 9 possibilities, only $(1, 5, 24)$ is in agreement with $(30-a)+(30-b)+(30-c)=30~units$

Therefore the longest side is $30-1=29~units$