Integral perimeter and Hypotenuse problem

 The IOQM seems to have some nice problems, though the solutions given by the many of the Indian book authors/coaching classes are sometimes wrong and/or lacking for the harder problems.

Here is a problem where one of the books got the wrong answer, even though the problem isn't hard.

The altitude dropped on the hypotenuse of a right triangle is the integer $12$. Given that the hypotenuse and the perimeter of the triangle are integers too, what is the smallest possible length of the hypotenuse?

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If $c$ is the hypotenuse (and $a,b$ are the legs), then we have that $12 c = ab$ (by equating the area of the triangle). $a+b+c$ and $c$ are both integers.

We also have that $(a+b)^2$ is a perfect square.

$$(a+b)^2 = a^2 + b^2 + 2ab = c^2 + 24c$$

By imagining a semicircle of diameter $c$, we can easily see that $c \ge 24$, and note that for any $c \ge 24$, a right triangle exists.

Now $c^2 + 24c$ has to a perfect square, and $c=25$ is the smallest that works.

The book had the answer as $24$ which is wrong.