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(D) Let the sides of the right-angled triangle be $a, b, c$ where $c$ is the hypotenuse. The inradius $r$ of a right-angled triangle is given by $r =

Vedclass · Accepted Answer

$5$

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(D) Let the sides of the right-angled triangle be $a, b, c$ where $c$ is the hypotenuse. The inradius $r$ of a right-angled triangle is given by $r = \frac{a+b-c}{2}$.Given one side is $12$. Let $a = 12$.Then $r = \frac{12+b-c}{2}$ $\Rightarrow 2r = 12+b-c$ $\Rightarrow c-b = 12-2r$.Also,$a^2 = c^2-b^2 = (c-b)(c+b)$.$144 = (12-2r)(c+b) \Rightarrow c+b = \frac{144}{2(6-r)} = \frac{72}{6-r}$.Since $c+b$ and $c-b$ must be integers,$6-r$ must be a divisor of $72$.Also,$r = \frac{a+b-c}{2}$. For $r$ to be maximum,we test possible values.If $a=12$ is the hypotenuse,$12^2 = b^2+c^2$. Integer solutions for $b^2+c^2=144$ are not possible for non-zero sides.If $a=12$ is a leg,$r = \frac{12+b-c}{2}$.Using $r = \frac{ab}{a+b+c}$,for a right triangle with legs $12$ and $b$,hypotenuse $c = \sqrt{144+b^2}$.$r = \frac{12b}{12+b+\sqrt{144+b^2}}$.Testing integer sides $(12, 35, 37) \Rightarrow r = \frac{12+35-37}{2} = 5$.Testing $(12, 16, 20) \Rightarrow r = \frac{12+16-20}{2} = 4$.Testing $(12, 9, 15) \Rightarrow r = \frac{12+9-15}{2} = 3$.The largest possible radius is $5$.

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

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