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KVPY SA Exam 05-Nov-2017 Question Paper
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© examsnet.com
Question : 62
Total: 80
In an isosceles trapezium, the length of one of the parallel sides, and the lengths of the non-parallel sides are all equal to 30. In order to maximize the area of the trapezium, the smallest angle should be
π
6
π
4
π
3
π
2
Validate
Solution:
In ∆ABF, y = 30 cosθ
Area is
A
=
1
2
[
60
+
60
cos
θ
]
(
30
sin
θ
)
=
30
(
1
+
cos
θ
)
(
30
sin
θ
)
=
900
(
sin
θ
+
sin
θ
cos
θ
)
For maximum or minimum
d
A
d
θ
=
900
[
cos
θ
+
(
−
sin
2
θ
+
cos
θ
)
]
=
0
cos
θ
=
1
+
cos
2
θ
+
cos
2
θ
=
0
2
cos
2
θ
+
cos
θ
−
1
=
0
(
2
cos
θ
−
1
)
(
cos
θ
+
1
)
=
0
cos
θ
=
1
2
or
cos
θ
=
−
1
(not possible)
θ = 60°
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