Question #19822

Let `R` be the radius of the incircle.

Radius of incircle of a right triangle `R = (a*b)/((a+b+c)`, where a and b are the legs of the triangle and c is the hypotenuse.

To get the largest possible radius of the incircle, 12 should be the smallest side of the triangle.

As this is a multiple-choice (objective) question, let's start with trying Option `D (R=5)`

Let `a` be the smallest side =`12`
`=> R=(12b)/(12+b+c)=5`

`=> c=(7b-60)/5` ...............(1)

As the triangle is right-angled, `=> c^2=a^2+b^2`
`=> c^2=144+b^2` ....... (2)

Substituting (1) into (2), we get `b=35, c=37`

So the triangle is in the ratio of : `12:35:37`

Check : `12^2+35^2=1369=37^2`, (OK)

`R=(axxb)/(a+b+c)=(12xx35)/(12+35+37)=(420/84)=5`,

Hence, option `D` is the answer.

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Footnote : the solution below is a much better one. A big "thank you" to `dk`_`ch` for kindly providing such an excellent solution.

We have `c^2−b^2=144`, where `c` is the hypotenuse and `b` is another side. `c>b`. By the given condition both `c and b` are integers.

Now `(c+b)(c−b)=144=72⋅2`...[1]

Here both `(c+b)and(c−b)` are even integers as their product `144` is even one.

To get maximum integer value satisfying the given condition both `c and b` will have large integer value and the value of `c−b` will have minimum possible even integer value.

So the minimum integer value of `(c−b)` should be `2`, i.e., `(c−b)=2`.

So from relation [1]

we have `(c+b)=72 and (c−b)=2`

Thus we get `c=37 and b=35`.