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7/29/2019 10 Mathematics Area Related to Circle Test 01 Answer
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CBSE TEST PAPER-01
CLASS - Mathematics (Area related to circle)
[ANSWERS]
Ans1. (B)
Ans2. (D)
Ans3. (B)
Ans4. (D)
Ans5 14d cm=
7r cm=
Circumference 2 r=
222 7 44
7cm= =
Ans6 Diameter of pond = 17.5cm
R = 8.75m
Width = 3.5m
Outer radius = R = 8.75 + 3.5 = 12.25m
Now
Area of path 2 2R r =
[ )( )R r R r= +
[ ) ( )
2
2212.25 8.75 12.25 8.75
7
22 66 721 3.50
7 2
33 7 231m
= +
= =
= =
Ans7 We know that
Area of sector2
360r
=
60 , 6r= =
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Required area ( )260 22
6360 7
=
222 6 132 618
7 7 7cm
= = =
Ans8 Let r be the radius of the circle by the given condition,
we have
Area of sector5
8OAPB = of the area of the circle
2 5360
360 18
xr =
100x =
Ans9 since for Rs. 24 The length of fencing = 1metre
for Rs 5280, the length of fencing1
5280 22024
m= =
Perimeter i.e. circumference of the field = 220m
Let r be radius of the field.
2 220r =
2 22 220 7220
7 40
5 7 35
r r
m
= =
= =
Area of the field = ( )22 2 2
35 1225r r m = = =
Rate = Rs 0.50 Per m2
Total cast of plareghing the field
( )1225 22 1
1225 0.507 2
Rs Rs
=
[ )175 11 1925Rs Rs= =
Ans10 Angle covered by minute hand in 60 minutes 360=
Angle covered in 1 minutes360
660
= =
Angle covered in 5 minutes = 6 5
30=
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We know that area swept by the minute hand during this
period
=Area of sector with sector angle 30
( )2
2
30 2214 14
360 360 7
22 22 728
7 3
151
3
r
cm
=
= =
=
Ans11 Area between two consecutive ribs of the umbrella
= Area of the sector with angle360
8made at centre
( )
( )
2
2 2
2
360/845
360
1 22 1145 2025
8 7 28
795.58 .
cm
cm nearly
=
= =
=
Ans12 Area of the shaded region
= Area of the semi-circle with O as centre and OQ as radius- ar
of .......( )PQR i
Since QR is a diameter passing thro the centre O of the
circle
90RPQ = [Angle of semi-circle]
2 2 2 2 2 27 24 49 576 625 25QR PR PQ = + = + = + = =
25QR =
diameter of the circle = 25cm
25
2r cm=
area of the semi circle21
2r=
2
1 22 25 25
2 7 2 2
11 625
28cm
=
=
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Also area of1
2PQR PR PQ =
2
17 24 7 12
2
84cm
= =
=
Hence area of the shaded region
2
11 625 11 625 84 2884
28 28
6875 2352 4523
28 28
126.644cm
= =
= =
=
Ans13 Area of shaded region = area of 3ABC (Area of sector
BPR)
Let a be the side of the equilateral
2 2
2
3 1.7320517320.5 17320.5
4 4
4 17320.54 10000
1.73205
2 100 200
a a
a
a
= = =
= =
= =
Required area = ( )260
17320.5 3 100
360
2
117320.5 3.14 10000
2
17320.5 1.57 10000
17320.5 15700 1620.5cm
=
=
= =
