Find the number of isosceles triangles with integral side lengths and having pe-rimeter 144 and only one side being largest.
My approach - If $a,a,b$ are sides such that $b>a$ then by triangle inequality $b<2a$. So $a<b<2a$, and by adding $2a$ on all sides, $3a<b+2a<4a$, so $3a<144<4a$.
So $36<a<48$ so $a$ has $11$ values from $37$ to $47$ and $11$ triangles are formed according to me but as per the book answer is $37$. Where am I wrong?
You are correct and the book is wrong. We can list all partitions of $144$ into 3 positive integers, 2 of them equal:
$$(1,1,142)\\(2,2,140)\\:\\(36,36,72)\\(37,37,70)\\:\\(47,47,50)\\(48,48,48)\\(49,49,46)\\:\\(71,71,2)$$
There are a total of $71$ partitions, and as you have noted, only the ones from $(37,37,70)$ to $(47,47,50)$ are valid: the first $36$ violate the triangle inequality, and the last $24$ have two longest sides. The book's author probably forgot to account for the triangle inequality.