## Book: RS Aggarwal - Mathematics

### Chapter: 12. Circles

#### Subject: Maths - Class 10th

##### Q. No. 41 of Multiple Choice Questions (MCQ)

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41
##### 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

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

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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