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,

⇒ AC^{2} = AB^{2} + BC^{2}

= 24^{2} + 7^{2}

= 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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