35. In the given figure, ABCD is a trapezium with D Q 60 degrees P43 cm-A B C AB || CD. The area of the shaded region is (d) 7 pi cm^ 2 (c) 9 pi cm^ 2 (b) 6 pi cm^ 2 (a) 3 pi cm^ 2
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Given: CE=CB=7cm
∴ CD=CE+ED=(7+4)cm=11cm
In Δ CLB, we have
sin60
∘
=
BC
BL
⇒
2
3
=
7
BL
⇒BL=
2
7
3
cm
∴ Area of trapezium =
2
1
(AB+CD)×BL=
2
1
(7+11)×
2
7
3
cm
2
=
2
63
3
cm
2
Area of sector BFEC =
360
∘
60
∘
×
7
22
×7
2
cm
2
=
3
77
cm
2
(Area of sector=
360
θ
×πr
2
)
∴ Required area = (
2
63
3
−
3
77
)cm
2
=(54.558−25.666)cm
2
=28.89cm
2
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