# Question 19822

Dec 30, 2016

$D . 5$

#### Explanation:

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 $D \left(R = 5\right)$

Let $a$ be the smallest side =$12$
$\implies R = \frac{12 b}{12 + b + c} = 5$

$\implies c = \frac{7 b - 60}{5}$ ...............(1)

As the triangle is right-angled, $\implies {c}^{2} = {a}^{2} + {b}^{2}$
$\implies {c}^{2} = 144 + {b}^{2}$ ....... (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)

$R = \frac{a \times b}{a + b + c} = \frac{12 \times 35}{12 + 35 + 37} = \left(\frac{420}{84}\right) = 5$,

Hence, option $D$ is the answer.

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Footnote : the solution below is a much better one. A big "thank you" to $\mathrm{dk}$_$c h$ 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 $c \mathmr{and} b$ 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 $c \mathmr{and} b$ 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 $c = 37 \mathmr{and} b = 35$.