**Question:**

ABC is a right angled triangle, right angled at B such that BC = 6 cm and AB = 8 cm. A circle with centre O is inscribed in Δ ABC. The radius of the circle is

(a) 1 cm

(b) 2 cm

(c) 3 cm

(d) 4 cm

**Solution:**

Let us first put the given data in the form of a diagram.

Let us first find out AC using Pythagoras theorem.

$A C^{2}=A B^{2}+B C^{2}$

$A C^{2}=8^{2}+6^{2}$

$A C^{2}=64+36$

$A C^{2}=100$

$A C=10$

Also, we know that tangents drawn from an external point will be equal in length. Therefore we have the following,

*BL = BM* …… (1)

*CM = CN* …… (2)

*AL = AN* …… (3)

We have found that,

*AC* = 10

That is,

*AN + NC* = 10

But from (2) and (3), we can say

*AL + MC* = 10 …… (4)

It is given that,

*AB* = 8

*BC* = 6

Therefore,

*AB + BC* =14

Looking at the figure, we can rewrite this as,

*AL + LB + BM + MC* = 14

(*AL + MC*) + (*BM + BL*) = 14

Using (1) and (3) we can write the above equation as,

10 + 2*BL* = 14

*BL* = 2Consider the quadrilateral, *BLOM*, we have,

*BL = BM *(From (1))

*OL = OM*(Radii of the same circle)

$\angle L B M=90^{\circ}$ (Given data)

$\angle B L O=90^{\circ}$ (Radii is always perpendicular to the tangent at the point of contact)

$\angle B M O=90^{\circ}$ (Radii is always perpendicular to the tangent at the point of contact)

Since the sum of all angles of a quadrilateral will be equal to , we have,

$\angle L B M+\angle L O M+\angle B L O+\angle B M O=360^{\circ}$

$90^{\circ}+90^{\circ}+90^{\circ}+\angle L O M=360^{\circ}$

$\angle L O M=90^{\prime}$

Since all the angles of the quadrilateral are equal to and since adjacent sides are equal, the quadrilateral *BLOM* is a square. We know that all the sides of a square are of equal length.

We have found *BL* = 2

Therefore,

*OM* = 2

*OM* is nothing but the radius of the circle.

Therefore, radius of the circle is 2 cm.

Hence the correct answer to the question is option (b).

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