Answer :


Let radius be r and the centre of the circle touching all the sides of a triangle be I i.e. incentre.


In the figure,


ID = IE = IF = r


Since ΔABC is a right angled triangle, B = 90°.


Also ID BC and AB BC.


ID || AB and ID || FB


Similarly, IF || BD


IFBD is a parallelogram.


FB = ID = r and BD = IF = r … (1)


Parallelogram IFBD is a rhombus.


Since B = 90°, parallelogram IFBD is a square.


By Pythagoras Theorem,


AC2 = AB2 + BC2


= 242 + 72


= 576 + 49


= 625


AC = 25


AB + BC + AC = 24 + 7 + 25


AF + FB + BD + DC + AC = 56


AE + r + r + CE + AC = 56


2r + (AE + CE) + AC = 56


2r + 2AC = 56


2r + 2(25) = 56


r + 25 = 28


r = 3


The radius of circle is 3.


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