1 of 3

CIRCLE

  • Sum based on Theorems –
  • Two tangents from an external point

to a circle are equal and

Radius is perpendicular to the tangent

2 of 3

A

B

O

D

C

Q

R

4 cm

4 cm

4 cm

6 cm

8 cm

x

x

A(ΔOBC)

×

1

2

28cm2

=

Sol:

2

=

BC

×

OD

×

1

2

14

×

4

=

=

2

×

1

2

(x + 6)

×

4

2(x + 6)cm2

=

=

2

×

1

2

(x + 8)

×

4

2(x + 8)cm2

=

6 cm

8 cm

A (ΔOAC)

×

1

2

=

AC

×

OQ

A (ΔOAB)

×

1

2

=

AB

×

OR

14 cm

(x + 6)

(x + 8)

D

6 cm

8 cm

Base

height

A(ΔOBC)

A (ΔOAC)

A (ΔOAB)

Q. A triangle ABC is drawn to circumscribe a circle of radius

4cm such that the segments BD and DC into which BC is

divided by the point of contact D are of lengths 8cm and

6cm respectively. Find the sides AB and AC.

Area of triangle = ½ × base × height

Consider ΔOAB

What can you say about BD and BR?

They are tangents from external point B

What can you say about AQ and AR?

Let, AQ = AR = x

They are tangents from external point A

What can you say about CD and CQ?

They are tangents from external point C

Let us find area of ΔOBC

We know, radius is perpendicular to tangent

Consider ΔOAC

Draw OC, OB, and OA

Whenever there is point of contact and centre, we can draw radius

Consider ΔOBC

CD

=

CQ

BD

=

BR

AQ

=

AR

[Tangents from an external point to a circle are equal in length]

Consider point C

We know, tangents from an external point to a circle are equal in length.

Consider point B

=

6 cm

=

8 cm

=

x cm

Consider point A

3 of 3

28

=

+

2x

+

12

+

2x

+

16

56

=

+

4x

A(ΔABC)

4

=

(14 + x)

...(i)

Semi-perimeter of ΔABC (s)

For ΔABC,

=

a + b + c

2

=

x + 6

2

=

(2x + 28)

2

(x + 14) cm

=

A(ΔABC)

A(ΔOBC)

=

+

A(ΔOAC)

+

A(ΔOAB)

28

=

+

2(x + 6)

+

2(x + 8)

A

B

O

D

C

Q

R

4 cm

4 cm

4 cm

6 cm

8 cm

x

x

6 cm

8 cm

14 cm

(x + 6)

(x + 8)

+

14

+

x + 8

Let a = AC = x + 6 ,

b = AB = x + 8,

and c = BC = 14.

=

2 (x + 14)

2

Q. A triangle ABC is drawn to circumscribe a circle of radius

4cm such that the segments BD and DC into which BC is

divided by the point of contact D are of lengths 8cm and

6cm respectively. Find the sides AB and AC.

Sol:

Now, let us consider ΔABC

ΔABC is made up of 3 triangles

ΔOBC,

ΔOAC,

ΔOAB

28cm2

=

A(ΔOBC)

2(x + 6)cm2

=

A(ΔOAC)

=

A(ΔOAB)

2(x + 8)cm2

Now let us apply heron’s

formula to find area of ΔABC