Question

The sides of a right-angled triangle are integers. The length of one of the sides is $12.$ The largest possible radius of the incircle of such a triangle is

• A. $2$
• B. $3$
• C. $4$
• D. $5$

Answer 1

The correct option is B $3$


For a incircle of maximum radius, we need a triangle with largest sides.
Therefore, $12$ is the shortest side.
Let the other two sides are $x$ and $y$ units.
$\therefore x^2-y^2=144$
$(x+y)(x-y)=144$
Since, sides are integers, therefore the possible values of $(x+y)$, $(x-y)$, $x$ and $y$ are:
$\begin{array}{cccc}(x+y)& (x-y)& x& y\\ 72& 2& 37& 35\\ 36& 4& 20& 16\\ 24& 6& 15& 9\\ 18& 8& 13& 5\end{array}$

Radius of incircle $r=\frac{△}{s}$

If sides are $37,35,12$,
$r=\frac{35\times 12}{37+35+12}=5$

If sides are $20,16,12$,
$r=\frac{16\times 12}{20+16+12}=4$

Therefore, maximum inradius is $4$ units.