Can we solve this?
In a rectangle ABCD of length 14 and breadth 7, an arc with centre A and radius 14 is drawn. Then the area of shaded region is
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1 solution
n
o
w
,
D
E
=
1
4
2
−
7
2
=
7
3
a
n
d
a
l
s
o
sin
θ
=
D
E
/
A
E
=
7
3
/
1
4
=
3
/
2
a
n
d
θ
=
6
0
s
o
⟹
9
0
−
θ
=
9
0
−
6
0
=
3
0
a
r
e
a
o
f
s
e
c
t
o
r
=
π
∗
r
∗
r
∗
θ
/
3
6
0
=
π
∗
1
4
∗
1
4
∗
3
0
/
3
6
0
=
5
1
.
3
1
a
r
e
a
o
f
A
D
E
=
b
∗
h
/
2
=
7
∗
7
√
(
3
)
/
2
=
4
2
.
3
4
a
r
e
a
o
f
r
e
c
t
a
n
g
l
e
=
l
∗
b
=
1
4
∗
7
=
9
8
a
r
e
a
o
f
s
h
a
d
e
d
r
e
g
i
o
n
=
a
r
r
e
c
t
a
n
g
l
e
−
(
a
r
s
e
c
t
o
r
+
a
r
A
D
E
)
=
9
8
−
(
5
1
.
3
1
+
4
2
.
4
3
)
=
9
8
−
9
3
.
7
4
=
4
.
2
6