Q. 415.0( 7 Votes )

In a right triangle ABC, right - angled at B, BC = 12 cm and AB = 5 cm. The radius of the circle inscribed in the triangle is
A. 1 cm

B. 2 cm

C. 3 cm

D. 4 cm

Answer :


Let AB, BC and AC touch the circle at points P, Q and R respectively.


As ABC is a right triangle,


By Pythagoras Theorem


[i.e. (hypotenuse)2 = (perpendicular)2 + (base)2]


(AC)2 = (BC)2 + (AB)2


(AC)2 = (12)2 + (5)2


(AC)2 = 144 + 25 = 169


AC = 13 cm


Let O be the center of circle, Join OP, OQ and PR


Let the radius of circle be r,


We have


r = OP = OQ = OR


[radii of same circle] [1]


Now,


ar(ABC) = ar(AOB) + ar(BOC) + ar(AOC)


As we know,


Area of triangle is 1/2 × Base × Height (Altitude)


Now,


OP AB


[Tangents drawn at a point on circle is perpendicular to the radius through point of contact]


OP is the altitude in AOB


OQ BC


[Tangents drawn at a point on circle is perpendicular to the radius through point of contact]


OQ is the altitude in BOC


OR AC


[Tangents drawn at a point on circle is perpendicular to the radius through point of contact]


OR is the altitude in AOC


So, we have


1/2 × BC × AB = (1/2 × AB × OP) + (1/2 × BC × OQ) + (1/2 × AC × OR)


12(5) = 5(r) + 12(r) + 13(r) [Using 1]


60 = 30r


r = 2 cm

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