SSC CHSL 13 Apr 2021 Shift 3 Paper

Given, an isosceles triangle of sum of the two sides $=16\mathit{c}\mathit{m}$ Let the two equal sides be $\mathit{a}\mathit{c}\mathit{m}$ and the third be $\mathit{b}\mathit{c}\mathit{m}$.
We known that, the sum of two sides of a triangle is always greater than the third side.
$\therefore$ Possible measurements, when $\mathit{a}=8$
Then $\mathit{b}=1,2,3,.............,15$
[total $=15$ triangles $]$ and when,
$\mathit{a}=6,7,9,10,11,12,13,14,15$
Then $\mathit{b}=10,9,7,6,5,4,3,2,1$ [total 9 triangles]
$\therefore$ Total possible triangles $=15+9=24$