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IIT JEE Advanced 2006 Paper 1
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Section:
Mathematics
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© examsnet.com
Question : 82
Total: 120
In radius of a circle which is inscribed in an isosceles triangle, one of whose angle is 2π/3, is
√
3
, then the area of the triangle is
[JEE Adv 2006 P1]
4
√
3
12 -
7
√
3
12 +
7
√
3
none of these
Validate
Solution:
From the figure shown here, we have ∠A = 2π /3 and ∠B = ∠C. Now,
2
π
3
+ 2 ∠B = π
∠B =
π
6
r =
√
3
Now, tan 60° =
I
E
A
E
√
3
=
√
3
A
E
⇒ AE = 1 = AF
tan 150° =
I
E
B
E
1
−
1
√
3
1
+
1
√
3
=
√
3
B
E
BE =
√
3
(
√
2
+
1
)
(
√
2
−
1
)
=
√
3
(
√
3
+
1
)
2
2
=
√
3
(
3
+
1
+
2
√
3
)
2
BE =
√
3
(
2
+
√
3
)
= 3 +
2
√
3
AB = 1 + 3 +
2
√
3
= 4 +
2
√
3
Now, the area of ΔABC is
1
2
(
4
+
2
√
3
)
(
4
+
2
√
3
)
sin 120° =
1
2
(
4
+
2
√
3
)
2
.
√
3
2
=
√
3
4
(
4
+
2
√
3
)
2
=
√
3
(
4
+
3
+
4
√
3
)
=
7
√
3
+
12
© examsnet.com
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