If one of the sides of a right angled triangle, whose all sides have integral values, is #12#, then what is the largest possible radius of its incircle?

A. #3#
B. #4#
C. #6#
D. #5#

1 Answer
Jun 15, 2017

Answer:

Answer is D.

Explanation:

Incircle in a right angled triangle is very special case. Infact it can be proved easily using the following figure (by joining incenter to the vertices of the triangle).

https://donsteward.blogspot.com/2010/12/incircle-of-right-angled-triangle.html

that radius of incircle is #(ab)/(a+b+c)# or #(a+b-c)/2#

As one of the side is #12#, let the other side be #x# and then hypotenuse is #sqrt(144+x^2# and radius of incircle is

#r(x)=(12+x-sqrt(144+x^2))/2# or #6+1/2(x-sqrt(144+x^2))#

Further if all sides are integers if #h# is hypotenuse and we have one side as #x#, then #144=h^2-x^2=(h+x)(h-x)# and sum of possible factors of #144# has to be even as #h# and #x# are integers.

Possible solutions are #2xx72#, #4xx36#, #6xx24#, #8xx18# and #12xx12#. The latter is not admissible as it results in one side to be #0#. The first four results give values of #(h,x)# as #(37,35)#, #(20,16)# #(15,9)# and #(13,5)#.

and for these we get #r={5,4,3,2}#

Hence, largest incircle possible is of radius #5# and dimensions of triagle are #(12,35,37)# and answer is D.