We notice that the sum of 2 consecutive triangular numbers is always a square number : \(\dfrac{\left(k-1\right)k}{2}+\dfrac{k\left(k+1\right)}{2}=\dfrac{k.2k}{2}=k^2\)

So, we need to find the largest triangular number less than 100

\(\dfrac{n\left(n+1\right)}{2}< 100\Leftrightarrow n^2+n< 200\Leftrightarrow n^2+2n.\dfrac{1}{2}+\dfrac{1}{4}< 200\dfrac{1}{4}\)

\(\Leftrightarrow\left(n+\dfrac{1}{2}\right)^2< 200\dfrac{1}{4}\Rightarrow n< 14\)

So, n = 13 and the answer is : \(\dfrac{13.14}{2}=91\)

(One of them is 91 and the other is \(\dfrac{12.13}{2}=78\))