WizakoBaskar wrote:

For a right triangle the quickest formula to find the radius of the inscribed circle, r = s - h, where s is the semi perimeter and h is the hypotenuse (I will provide the derivation for this as a reply to this post)

Please refer to the diagram 1 in the attachment

ABC is a right triangle right angled at B. AC is the hypotenuse.

Objective to find radius of inscribed circle.

PQR are points where the inscribed circle touches the sides of the triangle.

AR and AQ are equal tangents drawn to the circle from A.

Similarly, BP and BR are equal tangents drawn to the circle from B and CP and CQ are equal tangents drawn to the circle from C.

Hypotenuse AC, h = AQ + QC

We can replace AQ with AR and QC with CP. So, h = AR + CP ... (1)

Refer to diagram 2.

OR is perpendicular to AB and OP is perpendicular to BC (tangent and radius will be at right angles at the point of tangency)

The triangle is right angled at B. So, angle B is 90 degrees. So, BROP is a square whose side equals radius of the inscribed circle.

So, BR = BP = r (the radius of the inscribed circle).

If side AB measures c units, AR = AB - BR = c - r ... (2)

If side BC measures a units, CP = BC - BP = a - r ... (3)

Substitute (2) and (3) in (1)

h = (c - r) + (a - r)

Or 2r = a + c - h

Add and subtract h on the right side

2r = a + c + h - h - h

So, r = (a + c + h)/2 - (2h)/2

(a + c + h)/2 is the semi perimeter of the triangle.

Or r = s - h

Attachments

right-triangle-inradius-proof.png [ 29.54 KiB | Viewed 654 times ]

_________________

An IIM C Alumnus - Class of '94

GMAT Tutor at Wizako GMAT Classes & Online Courses