AREAS RELATED
TO CIRCLE
and Right-angled triangle
A
B
C
Q. In the adjoining given figure, ABC is a right angled triangle at A.
find the area of the shaded region if AB = 6cm, BC = 10cm and I is
the center of incircle of ΔABC.
Sol.
Applying Pythagoras theorem in ΔABC, we have
BC2
=
AB2
+ AC2
AC2
=
BC2
– AB2
AC2
=
100
– 36
AC2
=
64
AC
=
8 cm
Area of ΔABC =
1
2
×
AB
×
AC
∴
Area of ΔABC =
1
2
×
6
×
8
Area of ΔABC =
24cm2
10 cm
Area of shaded region =
ar(ΔABC)
– Area of circle
?
✔
Area of triangle =
1
2
×
Product of
perpendicular sides
Area of ΔABC =
1
2
×
×
AC
∴
?
8 cm
3
∴
∴
AC2
=
102
– 62
∴
I
I
What is formula to find area of triangle?
?
AB
6 cm
✔
∴
∴
∴
To find: AC
Q. In the adjoining given figure, ABC is a right angled triangle at A.
find the area of the shaded region if AB = 6cm, BC = 10cm and I is
the center of incircle of ΔABC.
Sol.
Let radius of the incircle be ‘r’ cm .
Area of ΔABC =
Area of ΔBIC
+ Area of ΔAIC
+ Area of ΔAIB
24
=
1
2
×
(BC ×
+
(AC ×
1
2
+
(AB ×
1
2
24
=
1
2
×
(BC + AC + AB)
24
=
1
2
×
r ×
(10 +
8 +
6)
24
=
12r
r
=
2
A
B
C
6 cm
10 cm
I
r
r
r
Area of shaded region =
ar(ΔABC)
– Area of circle
?
✔
Let us draw radius
Let us draw AI, BI, & CI
ΔABC is divided into
three triangle
ar(ΔABC) =
24cm2
1
2
Area of triangle =
×
b
×
h
r)
r)
r)
8 cm
Area of circle = πr2
=
22
7
×
2
×
2
=
88
7
2
2
2
1
2
24
=
Area of circle
×
r ×
24
12
cm2
What is formula to find area of circle?
πr2
?
ΔAIB
, ΔBIC
, ΔAIC
∴
∴
∴
∴
∴
∴
✔
r
Area of the shaded region =
Area of ΔABC –
Area of circle
Q. In the adjoining given figure, ABC is a right angled triangle at A.
find the area of the shaded region if AB = 6cm, BC = 10cm and I is
the center of incircle of ΔABC.
Sol.
=
24
–
80
7
cm2
A
B
C
6 cm
10 cm
I
2
2
2
Area of shaded region =
ar(ΔABC)
– Area of circle
ar(ΔABC) =
24cm2
Area of circle =
88
7
88
7
=
8 cm
∴
80
7
cm2
Area of the shaded region is
✔
✔
168
7
=
–
88