Prove that if the hypotenuse of a right triangle is h and the radius of its incircle is r, then its area is r(h + r).

Let ABC be a right-angled triangle, right angle at B such that

AB = base = ‘a’ units

BC = perpendicular = ‘b’ units

And AC = hypotenuse = ‘h’ units

Also, Incircle of ΔABC is drawn of radius r

To Prove: area(ΔABC) = r(h + r)

Construction: Draw OD⊥AB, OF⊥BC and OE⊥AC

Proof:

Consider quadrilateral OFBD,

OF = OD = r [radii of incircle]

and ∠OFB = ∠ODB = ∠DBF = 90°

⇒ ∠DOF = 90° [Angle sum property of quadrilateral]

Hence, OFBD is a square [adjacent sides are equal and all angles are 90°]

⇒ OF = OD = BF = BD = ‘r’

Also,

AB = ‘a’

⇒ AD + BD = a

⇒ AD + r = a

⇒ AD = a – r

Now, AD = AE [tangents drawn from an external point to a circle are equal]

⇒ AE = a – r …[1]

Similarly,

⇒ CE = b - r …[2]

Adding [1] and [2], we get

AE + CE = a + b - 2r

⇒ AC + 2r = a + b

⇒ a + b = h + 2r ….[3]

Now,

area(ΔABC) = area(ΔOAB) + area(ΔOAC) + area(ΔOBC)

⇒ area(ΔABC) = r(h + r)

Hence Proved.