CIRCLE
of two tangents drawn from
an external point to a circle are equal.
Sol:
[Tangents drawn
from an external point to
a circle are equal length]
AD
=
AF
=
x
BD
=
BE
=
y
CE
=
CF
=
z
Now,
AB
=
12 cm
x
+
y
=
12
y
+
z
=
8
z
+
x
=
10
x + y
+
y + z
+
z + x
=
12
+
8
+
10
2
(x + y + z)
=
30
x
+
y
+
z
=
15
A
B
D
F
E
C
x
x
y
z
z
y
12cm
10cm
8cm
Let, AD = AF = x
Let, BD = BE = y
Let, CE = CF = z
Q. A circle is inscribed in a ΔABC having sides 8 cm, 10 cm and
12 cm. Find AD, BE and CF.
2x
+
2y
+
2z
=
30
Consider point A
What can you say about AD and AF?
They are tangents from external point A
Consider point B
BD and BE are tangents from external point B
Consider point C
CE and CF are tangents from external point C
…(i)
…(ii)
…(iii)
Adding (i), (ii) and (iii)
We know that, tangents from an external point to a circle are equal in length
Let us add (i), (ii) and (iii)
AD
+
DB
=
12
Similarly,
BE + CE = 8cm
∴ y + z = 8cm
CF + AF = 10cm
∴ z + x = 10cm
∴
∴
∴
∴
∴
∴
∴
Q. A circle is inscribed in a ΔABC having sides 8 cm, 10 cm and 12 cm. Find AD, BE and CF.
Sol:
x
+
y
+
z
=
15
12
+
z
=
15
z
=
3
+
+
=
x
y
z
15
x
+
8
=
15
x
=
7
+
+
z
x
y
=
15
y
+
10
=
15
y
=
5
Hence, AD = 7 cm, BE = 5 cm and CF = 3 cm.
A
B
D
F
E
C
12cm
8cm
x
x
y
z
z
y
x
+
y
=
12,
y
+
z
=
8,
and z
+
x
=
10
CF
=
3cm
AD
=
7cm
BE
=
5cm
3
3
7
7
5
5
∴
∴
∴
10cm
∴
∴
∴
∴
∴
∴