Until I came to this forum I had never heard of Heron's formula so I am going to do it without using Heron's formula.
$$\\A=(1/2)*a*b*sinC\\\\
200=(1/2)*26*40*sinC\\\\
sinC=\frac{400}{26*40}\\\\
sinC=\frac{5}{13}\\\\
cosC=\pm\frac{12}{13} \qquad $draw a triangle and use the pythagorean theorem to help you get this$\\\\
$now use cosine rule$\\\\
c^2=26^2+40^2\pm 2*26*40*\frac{12}{13}\\\\
c^2=26^2+40^2\pm 2*2*40*12\\\\
c^2=356,\;\; or\;\;c^2=4196\\\\
c^2=4*89,\;\; or\;\;c^2=4*1049\\\\
c=2\sqrt{89}\approx 18.8680cm,\;\; or\;\;c=2\sqrt{1049}\approx 64.7765cm$$
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check - using Heron's formula
check $$2+\sqrt{89}$$ first
$$\\$s=the semi perimeter $\\\\
s= \frac{26+40+2\sqrt{89}}{2}=33+\sqrt{89}\\\\
A=\sqrt{s(s-a)(s-b)(s-c)}\\\\
A^2=s(s-a)(s-b)(s-c)\\\\
A^2=(33+\sqrt{89})(33+\sqrt{89}-2\sqrt89)(33+\sqrt{89}-26)(33+\sqrt{89}-40)\\\\
A^2=(33+\sqrt{89})(33-\sqrt{89})(7+\sqrt{89})(-7+\sqrt{89})\\\\
A^2=(33^2-89)(89-49)\\\\
A^2=1000*40\\\\
A^2=40000\\\\
A=200$$
OR
now check the other one using the approximation.
semiperimeter= $${\mathtt{0.5}}{\mathtt{\,\times\,}}\left({\mathtt{64.776\: \!5}}{\mathtt{\,\small\textbf+\,}}{\mathtt{40}}{\mathtt{\,\small\textbf+\,}}{\mathtt{26}}\right) = {\mathtt{65.388\: \!25}}$$
Area
$${\sqrt{{\mathtt{65.388\: \!25}}{\mathtt{\,\times\,}}\left({\mathtt{65.388\: \!25}}{\mathtt{\,-\,}}{\mathtt{40}}\right){\mathtt{\,\times\,}}\left({\mathtt{65.388\: \!25}}{\mathtt{\,-\,}}{\mathtt{26}}\right){\mathtt{\,\times\,}}\left({\mathtt{65.388\: \!25}}{\mathtt{\,-\,}}{\mathtt{64.776\: \!5}}\right)}} = {\mathtt{200.003\: \!028\: \!623\: \!087\: \!389\: \!9}}$$
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$$\\
\mathbf{\mbox{So there are 2 possible exact lengths for the 3rd side}}\\\\
\mathbf{2\sqrt{89}\;cm\;\;\;\; or\;\;\;\;\;2\sqrt{1049}\;cm}$$
