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

Answer 1

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

Let us first find out AC using Pythagoras theorem.

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 + 2BL = 14

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

BL = BM (From (1))

OL = OM(Radii of the same circle)

(Given data)

(Radii is always perpendicular to the tangent at the point of contact)

(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,

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).