Question

19. ABC is a right angled triangle, right
angled at B with AB= 12 cm and AC=
20 cm. A circle with centre O, has been
inscribed inside the triangle. The radius of
the inscribed circle is​

Answer 1

Answer:

In △ABC,

⇒ ∠B=90

o

[ Given ]

⇒ AB=12cm and AC=13cm [ Given ]

Here, O is center of a circle and x is a radius.

⇒ (AC)

2

=(AB)

2

+(BC)

2

[ By Pythagoras theorem ]

⇒ (13)

2

=(12)

2

+(BC)

2

⇒ 169=144+(BC)

2

⇒ (BC)

2

=25

∴ BC=5cm

Now, AB,BC and CA are tangents to the circle at P,N and M respectively.

∴ OP=ON=OM=x [ Radius of a circle ]

⇒ Area of △ABC=

2

1

×BC×AB

=

2

1

×5×12

=30cm

2

Area of △ABC= Area of △OAB+ Area of △OBC+ Area of △OCA

⇒ 30=

2

1

x×AB+

2

1

x×BC+

2

1

x×CA

⇒ 30=

2

1

x(AB+BC+CA)

⇒ x=

AB+BC+CA

2×30

⇒ x=

12+5+13

60

⇒ x=

30

60

∴ x=2cm

solution