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AP EAMCET Engineering 2013 Solved Paper
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© examsnet.com
Question : 108
Total: 160
In any
Δ
A
B
C
,
r
1
r
2
+
r
2
r
3
+
r
3
r
1
is equal to
Δ
2
r
2
Δ
r
2
Δ
r
Δ
2
Validate
Solution:
In
Δ
A
B
C
r
1
r
2
+
r
2
r
3
+
r
3
r
1
=
Δ
s
−
a
.
Δ
s
−
b
+
Δ
s
−
b
.
Δ
s
−
c
+
Δ
s
−
c
.
Δ
s
−
a
=
Δ
2
(
s
−
a
)
(
s
−
b
)
(
s
−
c
)
[
s
−
c
+
s
−
a
+
s
−
b
]
=
Δ
2
3
s
−
(
a
+
b
+
c
)
(
s
−
a
)
(
s
−
b
)
(
s
−
c
)
(
∵
2
s
=
a
+
b
+
c
)
=
Δ
2
(
3
s
−
2
s
)
(
s
−
a
)
(
s
−
b
)
(
s
−
c
)
=
Δ
2
.
s
(
Δ
2
s
)
[
∵
Δ
2
s
=
(
s
−
a
)
(
s
−
b
)
(
s
−
c
)
]
=
s
2
.
Δ
2
Δ
2
=
Δ
2
(
Δ
s
)
2
=
Δ
2
r
2
[
∵
r
=
Δ
s
]
© examsnet.com
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