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Question Number 46637 by rahul 19 last updated on 29/Oct/18

Commented by tanmay.chaudhury50@gmail.com last updated on 29/Oct/18

$o k . . .$
Commented by rahul 19 last updated on 29/Oct/18

$A n s . i s D . . . . s i n c e i t i s a s k i n g f o r$ $l a r g e s t p o s s i b l e r a d i u s . . . .$
Commented by rahul 19 last updated on 29/Oct/18

$f o r e g : i f s i d e s a r e 12, 16, 20 t h e n r = 4 . . .$
	Answered by tanmay.chaudhury50@gmail.com last updated on 29/Oct/18

	$s i d e s a r e 5, 12, 13$ $\frac{1}{2} \times 5 \times r + \frac{1}{2} \times 12 \times r + \frac{1}{2} \times 13 \times r = \frac{1}{2} \times 5 \times 12$ $30 r = 60 r = 2$
	Answered by MJS last updated on 31/Oct/18

	$these are all possible triangles :$ $9 12 15 r_{i} = 3$ $12 16 20 r_{i} = 4$ $5 12 13 r_{i} = 2$ $12 35 37 r_{i} = 5$ $\Rightarrow answer is D$
	Commented by rahul 19 last updated on 31/Oct/18

	$S i r, h o w y o u c h e c k e d t h e s e a r e o n l y$ $p o s s i b l e t r i a n g l e s . . . . ?$
	Commented by MJS last updated on 31/Oct/18
	$a^2 +b^2 =c^2 with a, b, c ∈N\{0} we can set c=12 and b=a+n with n∈N a^2 +(a+n)^2 =144 ⇒ a=((√(288−n^2 ))/2)−(n/2)∈N ⇒ no solution we can set b=12 and c=a+n with n∈N a^2 +144=(a+n)^2 ⇒ a=((72)/n)−(n/2)∈N ⇒ n∈{2, 4, 6, 8}$
	$a^{2} + b^{2} = c^{2} with a, b, c \in N ∖ {0}$ $we can set c = 12 and b = a + n with n \in N$ $a^{2} + {(a + n)}^{2} = 144$ $\Rightarrow a = \frac{\sqrt{288 - n^{2}}}{2} - \frac{n}{2} \in N \Rightarrow no solution$ $we can set b = 12 and c = a + n with n \in N$ $a^{2} + 144 = {(a + n)}^{2}$ $\Rightarrow a = \frac{72}{n} - \frac{n}{2} \in N \Rightarrow n \in {2, 4, 6, 8}$
	Commented by rahul 19 last updated on 31/Oct/18

	$p e r f e c t!$ $T h a n k s s i r!$
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