B
figure. (Take a =
Hint.
34. Find the perimeter and area of the shaded region in the given
AC2 = AB2 + BC2 = (144 + 256) = 400 = AC = 20 cm.
r = 10 cm.
(str + 12 + 16) cm = (3.14 × 10 + 28) cm = 59.4 cm
Area = ( x -3x12x16 )cm? = (x3.
x12x16 )cm? = 4x3.14x100 – 96 )cm?
[
16 cm
So,
12 cm
A
Perimeter =
с
= 61 cm
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Answer:
answer--is 61cm ................. ............. .
Step-by-step explanation:
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Solution
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Given that,
Radius of the small circle, OB=7 cm
Radius of second circle, OA=14 cm
and ∠AOC=40
o
We know that, the area of a sector that subtends an angle θ at the centre of the circle is
360
0
θ
×πr
2
where, θ is in degrees.
∴ Area of minor sector OBD=
360
40
×
7
22
×7×7[∵ π=
7
22
]
=
9
1
×22×7
=17.11 cm
2
Also, area of minor sector OAC=
360
40
×
7
22
×14×14
=
9
1
×22×2×14
=68.4 cm
2
Now, area of the shaded region = Area of sector OAC− Area of sector OBD
=68.4−17.1
=51.3 cm
2
Hence, the area of the shaded region is 51.3 cm
2
.
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