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Page 1
IX CBSE MATHS
Solved problems
PIE TUTORIALS
9/2 CHOPASANI HOUSING BOARD, JODHPUR
FOR FREE NOTES & TEST PAPERS Visit www.pietutorials.com
MENSURATION
AREAS RELATED TO CIRCLES
1. In the adjoining figure ABC right angled triangle right angled at A. Semi circles
are drawn on the sides of the triangle ABC . Prove that area of the Shaded region
is equal to area of ABC
Ans: Refer CBSE paper 2008
2. The sum of the diameters of two circles is 2.8 m and their difference of
circumferences is 0.88m. Find the radii of the two circles (Ans: 77, 63)
Ans: d
1
+ d
2
= 2.8 m= 280cm
r
1
+r
2
= 140
2 (r
1
– r
2
) = 0.88m = 88cm
r
1
– r
2
=
2
88
=
44
7 88x
= 2 x 7 = 14
r
1
+r
2
= 140
r
1
-r
2
= 14
2r
1
= 154
r
1
=77
r
2
= 140 – 77 = 63
r
1
= 77 cm, r
2
= 63cm
3 Find the circumference of a circle whose area is 16 times the area of the circle
with diameter 7cm (Ans: 88cm)
Ans: R
2
= 16 r
2
R
2
= 16 r
2
R
2
= 16 x
2
7
x
2
7
= 49 x 4 R = 7 x 2 = 14cm
Circumference = 2 x
7
22
x 14 = 2 x 22 x 2 = 88 cm
A
C
B
Page 2
IX CBSE MATHS
Solved problems
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MENSURATION
AREAS RELATED TO CIRCLES
1. In the adjoining figure ABC right angled triangle right angled at A. Semi circles
are drawn on the sides of the triangle ABC . Prove that area of the Shaded region
is equal to area of ABC
Ans: Refer CBSE paper 2008
2. The sum of the diameters of two circles is 2.8 m and their difference of
circumferences is 0.88m. Find the radii of the two circles (Ans: 77, 63)
Ans: d
1
+ d
2
= 2.8 m= 280cm
r
1
+r
2
= 140
2 (r
1
– r
2
) = 0.88m = 88cm
r
1
– r
2
=
2
88
=
44
7 88x
= 2 x 7 = 14
r
1
+r
2
= 140
r
1
-r
2
= 14
2r
1
= 154
r
1
=77
r
2
= 140 – 77 = 63
r
1
= 77 cm, r
2
= 63cm
3 Find the circumference of a circle whose area is 16 times the area of the circle
with diameter 7cm (Ans: 88cm)
Ans: R
2
= 16 r
2
R
2
= 16 r
2
R
2
= 16 x
2
7
x
2
7
= 49 x 4 R = 7 x 2 = 14cm
Circumference = 2 x
7
22
x 14 = 2 x 22 x 2 = 88 cm
A
C
B
IX CBSE MATHS
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4. Find the area enclosed between two concentric circles of radii 3.5cm, 7cm. A
third concentric circle is drawn outside the 7cm circle so that the area enclosed
between it and the 7cm circle is same as that between two inner circles. Find the
radius of the third circle (Ans: 115.5 cm
2
r = 343 / 2 )
Ans: Area between first two circles = x 7
2
- x 3.5
2
= 49 - 12.25 -------------(1)
Area between next two circles = r
2
- x 7
2
= r
2
– 49 -----------------(2)
(1) & (2) are equal
49 - 12.25 = r
2
- 49
r
2
= 49 + 49 - 12.25
r
2
= 98 – 12.25 = 85.75
r
2
=
100
8575
=
4
343
r =
2
343
cm.
5. Two circles touch externally. The sum of their areas is 58 cm
2
and the distance
between their centres is 10 cm. Find the radii of the two circles. (Ans:7cm, 3cm)
Ans: Sum of areas = r
2
+ (10 – r )
2
= 58
r
2
+ (100 – 20 r + r
2
) = 58
r
2
+ 100 – 20r + r
2
= 58
2r
2
– 20r +100 – 58 = 0
2r
2
– 20r +42 = 0
r
2
– 10r +21 = 0
(r-7), (r-3) = 0
r=7cm,3cm
6. From a sheet of cardboard in the shape of a square of side 14 cm, a piece in the
shape of letter B is cut off. The curved side of the letter consists of two equal
semicircles & the breadth of the rectangular piece is 1 cm. Find the area of the
remaining part of cardboard. (Ans: 143.5 cm
2
)
Ans: Area of remaining portion = Area of square – Area of 2 semi circles – Area of
rectangle
= 14 x 14 - x 3.5
2
– 14 x 1
=196 -
7
22
x 3.5 x 3.5 – 14
=196 – 38.5 – 14 = 143.5 cm
2
1cm 14 cm
Page 3
IX CBSE MATHS
Solved problems
PIE TUTORIALS
9/2 CHOPASANI HOUSING BOARD, JODHPUR
FOR FREE NOTES & TEST PAPERS Visit www.pietutorials.com
MENSURATION
AREAS RELATED TO CIRCLES
1. In the adjoining figure ABC right angled triangle right angled at A. Semi circles
are drawn on the sides of the triangle ABC . Prove that area of the Shaded region
is equal to area of ABC
Ans: Refer CBSE paper 2008
2. The sum of the diameters of two circles is 2.8 m and their difference of
circumferences is 0.88m. Find the radii of the two circles (Ans: 77, 63)
Ans: d
1
+ d
2
= 2.8 m= 280cm
r
1
+r
2
= 140
2 (r
1
– r
2
) = 0.88m = 88cm
r
1
– r
2
=
2
88
=
44
7 88x
= 2 x 7 = 14
r
1
+r
2
= 140
r
1
-r
2
= 14
2r
1
= 154
r
1
=77
r
2
= 140 – 77 = 63
r
1
= 77 cm, r
2
= 63cm
3 Find the circumference of a circle whose area is 16 times the area of the circle
with diameter 7cm (Ans: 88cm)
Ans: R
2
= 16 r
2
R
2
= 16 r
2
R
2
= 16 x
2
7
x
2
7
= 49 x 4 R = 7 x 2 = 14cm
Circumference = 2 x
7
22
x 14 = 2 x 22 x 2 = 88 cm
A
C
B
IX CBSE MATHS
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4. Find the area enclosed between two concentric circles of radii 3.5cm, 7cm. A
third concentric circle is drawn outside the 7cm circle so that the area enclosed
between it and the 7cm circle is same as that between two inner circles. Find the
radius of the third circle (Ans: 115.5 cm
2
r = 343 / 2 )
Ans: Area between first two circles = x 7
2
- x 3.5
2
= 49 - 12.25 -------------(1)
Area between next two circles = r
2
- x 7
2
= r
2
– 49 -----------------(2)
(1) & (2) are equal
49 - 12.25 = r
2
- 49
r
2
= 49 + 49 - 12.25
r
2
= 98 – 12.25 = 85.75
r
2
=
100
8575
=
4
343
r =
2
343
cm.
5. Two circles touch externally. The sum of their areas is 58 cm
2
and the distance
between their centres is 10 cm. Find the radii of the two circles. (Ans:7cm, 3cm)
Ans: Sum of areas = r
2
+ (10 – r )
2
= 58
r
2
+ (100 – 20 r + r
2
) = 58
r
2
+ 100 – 20r + r
2
= 58
2r
2
– 20r +100 – 58 = 0
2r
2
– 20r +42 = 0
r
2
– 10r +21 = 0
(r-7), (r-3) = 0
r=7cm,3cm
6. From a sheet of cardboard in the shape of a square of side 14 cm, a piece in the
shape of letter B is cut off. The curved side of the letter consists of two equal
semicircles & the breadth of the rectangular piece is 1 cm. Find the area of the
remaining part of cardboard. (Ans: 143.5 cm
2
)
Ans: Area of remaining portion = Area of square – Area of 2 semi circles – Area of
rectangle
= 14 x 14 - x 3.5
2
– 14 x 1
=196 -
7
22
x 3.5 x 3.5 – 14
=196 – 38.5 – 14 = 143.5 cm
2
1cm 14 cm
IX CBSE MATHS
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7. A piece of cardboard in the shape of a trapezium ABCD & AB || DE, BCD =
90
0
, quarter circle BFEC is removed. Given AB = BC = 3.5 cm, DE = 2 cm.
Calculate the area of remaining piece of cardboard. (Ans:6.125 cm
2
)
Ans: Area of remaining portion = Area of trap – Area of quadrant
=
2
1
x 3.5 (5.5 + 3.5) -
4
1
x
7
22
x3.5x3.5
= 15.75 -
2
25 . 19
= 15.75 – 9.625
= 6.125 cm
2
8. In the figure, ABCD is a square inside a circle with centre O. The Centre of the
square coincides with O & the diagonal AC is horizontal of AP, DQ are vertical &
AP = 45 cm, DQ = 25 cm. Find a) the radius of the circle b) side of square
c) area of shaded region (use =3.14 , 2 = 1.41)
Ans: a) 53cm
b) 39.48cm
c) 7252.26 cm
2
Self Practice
9. The area enclosed between two concentric circles is 770cm
2
. If the radius of the
outer circle is 21cm, find the radius of the inner circle. (Ans :14cm)
Ans: R
2
- r
2
= 770
(21
2
- r
2
) = 770
21
2
– r
2
=
22
770
x 7 =
2
70
x 7
r
2
= 441 -
2
490
=441 – 245=196
r= 14
r=14cm
10. A circular disc of 6 cm radius is divided into three sectors with central angles
120
0
, 150
0
,90
0
. What part of the circle is the sector with central angles 120
0
. Also
give the ratio of the areas of three sectors. (Ans:
3
1
(Area of the circle) 4 : 5 : 3)
Ans: Ratio of areas =
360
120
x 6
2
:
360
150
x 6
2
:
360
90
x 6
2
= 12 : 15 : 9
= 4 : 5 : 3
A
B
D
C
P
Q
Page 4
IX CBSE MATHS
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MENSURATION
AREAS RELATED TO CIRCLES
1. In the adjoining figure ABC right angled triangle right angled at A. Semi circles
are drawn on the sides of the triangle ABC . Prove that area of the Shaded region
is equal to area of ABC
Ans: Refer CBSE paper 2008
2. The sum of the diameters of two circles is 2.8 m and their difference of
circumferences is 0.88m. Find the radii of the two circles (Ans: 77, 63)
Ans: d
1
+ d
2
= 2.8 m= 280cm
r
1
+r
2
= 140
2 (r
1
– r
2
) = 0.88m = 88cm
r
1
– r
2
=
2
88
=
44
7 88x
= 2 x 7 = 14
r
1
+r
2
= 140
r
1
-r
2
= 14
2r
1
= 154
r
1
=77
r
2
= 140 – 77 = 63
r
1
= 77 cm, r
2
= 63cm
3 Find the circumference of a circle whose area is 16 times the area of the circle
with diameter 7cm (Ans: 88cm)
Ans: R
2
= 16 r
2
R
2
= 16 r
2
R
2
= 16 x
2
7
x
2
7
= 49 x 4 R = 7 x 2 = 14cm
Circumference = 2 x
7
22
x 14 = 2 x 22 x 2 = 88 cm
A
C
B
IX CBSE MATHS
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4. Find the area enclosed between two concentric circles of radii 3.5cm, 7cm. A
third concentric circle is drawn outside the 7cm circle so that the area enclosed
between it and the 7cm circle is same as that between two inner circles. Find the
radius of the third circle (Ans: 115.5 cm
2
r = 343 / 2 )
Ans: Area between first two circles = x 7
2
- x 3.5
2
= 49 - 12.25 -------------(1)
Area between next two circles = r
2
- x 7
2
= r
2
– 49 -----------------(2)
(1) & (2) are equal
49 - 12.25 = r
2
- 49
r
2
= 49 + 49 - 12.25
r
2
= 98 – 12.25 = 85.75
r
2
=
100
8575
=
4
343
r =
2
343
cm.
5. Two circles touch externally. The sum of their areas is 58 cm
2
and the distance
between their centres is 10 cm. Find the radii of the two circles. (Ans:7cm, 3cm)
Ans: Sum of areas = r
2
+ (10 – r )
2
= 58
r
2
+ (100 – 20 r + r
2
) = 58
r
2
+ 100 – 20r + r
2
= 58
2r
2
– 20r +100 – 58 = 0
2r
2
– 20r +42 = 0
r
2
– 10r +21 = 0
(r-7), (r-3) = 0
r=7cm,3cm
6. From a sheet of cardboard in the shape of a square of side 14 cm, a piece in the
shape of letter B is cut off. The curved side of the letter consists of two equal
semicircles & the breadth of the rectangular piece is 1 cm. Find the area of the
remaining part of cardboard. (Ans: 143.5 cm
2
)
Ans: Area of remaining portion = Area of square – Area of 2 semi circles – Area of
rectangle
= 14 x 14 - x 3.5
2
– 14 x 1
=196 -
7
22
x 3.5 x 3.5 – 14
=196 – 38.5 – 14 = 143.5 cm
2
1cm 14 cm
IX CBSE MATHS
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7. A piece of cardboard in the shape of a trapezium ABCD & AB || DE, BCD =
90
0
, quarter circle BFEC is removed. Given AB = BC = 3.5 cm, DE = 2 cm.
Calculate the area of remaining piece of cardboard. (Ans:6.125 cm
2
)
Ans: Area of remaining portion = Area of trap – Area of quadrant
=
2
1
x 3.5 (5.5 + 3.5) -
4
1
x
7
22
x3.5x3.5
= 15.75 -
2
25 . 19
= 15.75 – 9.625
= 6.125 cm
2
8. In the figure, ABCD is a square inside a circle with centre O. The Centre of the
square coincides with O & the diagonal AC is horizontal of AP, DQ are vertical &
AP = 45 cm, DQ = 25 cm. Find a) the radius of the circle b) side of square
c) area of shaded region (use =3.14 , 2 = 1.41)
Ans: a) 53cm
b) 39.48cm
c) 7252.26 cm
2
Self Practice
9. The area enclosed between two concentric circles is 770cm
2
. If the radius of the
outer circle is 21cm, find the radius of the inner circle. (Ans :14cm)
Ans: R
2
- r
2
= 770
(21
2
- r
2
) = 770
21
2
– r
2
=
22
770
x 7 =
2
70
x 7
r
2
= 441 -
2
490
=441 – 245=196
r= 14
r=14cm
10. A circular disc of 6 cm radius is divided into three sectors with central angles
120
0
, 150
0
,90
0
. What part of the circle is the sector with central angles 120
0
. Also
give the ratio of the areas of three sectors. (Ans:
3
1
(Area of the circle) 4 : 5 : 3)
Ans: Ratio of areas =
360
120
x 6
2
:
360
150
x 6
2
:
360
90
x 6
2
= 12 : 15 : 9
= 4 : 5 : 3
A
B
D
C
P
Q
IX CBSE MATHS
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Area of sector of central angle 120
o
=
o
o
360
120
x r
2
(i.e.)
3
1
of area of the circle.
11. If the minute hand of a big clock is 1.05 m long, find the rate at which its tip is
moving in cm per minute. (Ans:11cm/min)
Ans: Self Practice
12. ABC is a right angled triangle in which A = 90
0
. Find the area of the shaded
region if AB = 6 cm, BC=10cm & I is the centre of the Incircle of ABC.
(Ans:
7
80
sq.cm)
Ans: A =90
0
BC = 10cm; AB = 6cm;
AC = 6cm
Area of the =
2
1
x 6 x 8= 24 cm
2
Let the Radius of the Incircle be r
2
1
x 10 x r +
2
1
x 8 x r +
2
1
x 6 x r = 24
2
1
r [10 + 8 + 6] = 24
r= 2 cm
Area of circle = r
2
=
7
22
x 2 x 2 =
7
88
cm
2
Area of shaded region = 24 -
7
88
=
7
88 168
=
7
80
cm
2
13. Find the perimeter of the figure, where AED is a semi-circle and ABCD is a
rectangle. (Ans : 76cm)
Ans: Perimeter of the fig = 20 + 14 + 20 + length of the arc (AED)
Length of Arc = ( x r) =
7
22
x7 = 22cm
Perimeter of the figure = 76 cm
A
I
B C
20cm 20cm
14cm
A
C B
E
D
Page 5
IX CBSE MATHS
Solved problems
PIE TUTORIALS
9/2 CHOPASANI HOUSING BOARD, JODHPUR
FOR FREE NOTES & TEST PAPERS Visit www.pietutorials.com
MENSURATION
AREAS RELATED TO CIRCLES
1. In the adjoining figure ABC right angled triangle right angled at A. Semi circles
are drawn on the sides of the triangle ABC . Prove that area of the Shaded region
is equal to area of ABC
Ans: Refer CBSE paper 2008
2. The sum of the diameters of two circles is 2.8 m and their difference of
circumferences is 0.88m. Find the radii of the two circles (Ans: 77, 63)
Ans: d
1
+ d
2
= 2.8 m= 280cm
r
1
+r
2
= 140
2 (r
1
– r
2
) = 0.88m = 88cm
r
1
– r
2
=
2
88
=
44
7 88x
= 2 x 7 = 14
r
1
+r
2
= 140
r
1
-r
2
= 14
2r
1
= 154
r
1
=77
r
2
= 140 – 77 = 63
r
1
= 77 cm, r
2
= 63cm
3 Find the circumference of a circle whose area is 16 times the area of the circle
with diameter 7cm (Ans: 88cm)
Ans: R
2
= 16 r
2
R
2
= 16 r
2
R
2
= 16 x
2
7
x
2
7
= 49 x 4 R = 7 x 2 = 14cm
Circumference = 2 x
7
22
x 14 = 2 x 22 x 2 = 88 cm
A
C
B
IX CBSE MATHS
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4. Find the area enclosed between two concentric circles of radii 3.5cm, 7cm. A
third concentric circle is drawn outside the 7cm circle so that the area enclosed
between it and the 7cm circle is same as that between two inner circles. Find the
radius of the third circle (Ans: 115.5 cm
2
r = 343 / 2 )
Ans: Area between first two circles = x 7
2
- x 3.5
2
= 49 - 12.25 -------------(1)
Area between next two circles = r
2
- x 7
2
= r
2
– 49 -----------------(2)
(1) & (2) are equal
49 - 12.25 = r
2
- 49
r
2
= 49 + 49 - 12.25
r
2
= 98 – 12.25 = 85.75
r
2
=
100
8575
=
4
343
r =
2
343
cm.
5. Two circles touch externally. The sum of their areas is 58 cm
2
and the distance
between their centres is 10 cm. Find the radii of the two circles. (Ans:7cm, 3cm)
Ans: Sum of areas = r
2
+ (10 – r )
2
= 58
r
2
+ (100 – 20 r + r
2
) = 58
r
2
+ 100 – 20r + r
2
= 58
2r
2
– 20r +100 – 58 = 0
2r
2
– 20r +42 = 0
r
2
– 10r +21 = 0
(r-7), (r-3) = 0
r=7cm,3cm
6. From a sheet of cardboard in the shape of a square of side 14 cm, a piece in the
shape of letter B is cut off. The curved side of the letter consists of two equal
semicircles & the breadth of the rectangular piece is 1 cm. Find the area of the
remaining part of cardboard. (Ans: 143.5 cm
2
)
Ans: Area of remaining portion = Area of square – Area of 2 semi circles – Area of
rectangle
= 14 x 14 - x 3.5
2
– 14 x 1
=196 -
7
22
x 3.5 x 3.5 – 14
=196 – 38.5 – 14 = 143.5 cm
2
1cm 14 cm
IX CBSE MATHS
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7. A piece of cardboard in the shape of a trapezium ABCD & AB || DE, BCD =
90
0
, quarter circle BFEC is removed. Given AB = BC = 3.5 cm, DE = 2 cm.
Calculate the area of remaining piece of cardboard. (Ans:6.125 cm
2
)
Ans: Area of remaining portion = Area of trap – Area of quadrant
=
2
1
x 3.5 (5.5 + 3.5) -
4
1
x
7
22
x3.5x3.5
= 15.75 -
2
25 . 19
= 15.75 – 9.625
= 6.125 cm
2
8. In the figure, ABCD is a square inside a circle with centre O. The Centre of the
square coincides with O & the diagonal AC is horizontal of AP, DQ are vertical &
AP = 45 cm, DQ = 25 cm. Find a) the radius of the circle b) side of square
c) area of shaded region (use =3.14 , 2 = 1.41)
Ans: a) 53cm
b) 39.48cm
c) 7252.26 cm
2
Self Practice
9. The area enclosed between two concentric circles is 770cm
2
. If the radius of the
outer circle is 21cm, find the radius of the inner circle. (Ans :14cm)
Ans: R
2
- r
2
= 770
(21
2
- r
2
) = 770
21
2
– r
2
=
22
770
x 7 =
2
70
x 7
r
2
= 441 -
2
490
=441 – 245=196
r= 14
r=14cm
10. A circular disc of 6 cm radius is divided into three sectors with central angles
120
0
, 150
0
,90
0
. What part of the circle is the sector with central angles 120
0
. Also
give the ratio of the areas of three sectors. (Ans:
3
1
(Area of the circle) 4 : 5 : 3)
Ans: Ratio of areas =
360
120
x 6
2
:
360
150
x 6
2
:
360
90
x 6
2
= 12 : 15 : 9
= 4 : 5 : 3
A
B
D
C
P
Q
IX CBSE MATHS
Solved problems
PIE TUTORIALS
9/2 CHOPASANI HOUSING BOARD, JODHPUR
FOR FREE NOTES & TEST PAPERS Visit www.pietutorials.com
Area of sector of central angle 120
o
=
o
o
360
120
x r
2
(i.e.)
3
1
of area of the circle.
11. If the minute hand of a big clock is 1.05 m long, find the rate at which its tip is
moving in cm per minute. (Ans:11cm/min)
Ans: Self Practice
12. ABC is a right angled triangle in which A = 90
0
. Find the area of the shaded
region if AB = 6 cm, BC=10cm & I is the centre of the Incircle of ABC.
(Ans:
7
80
sq.cm)
Ans: A =90
0
BC = 10cm; AB = 6cm;
AC = 6cm
Area of the =
2
1
x 6 x 8= 24 cm
2
Let the Radius of the Incircle be r
2
1
x 10 x r +
2
1
x 8 x r +
2
1
x 6 x r = 24
2
1
r [10 + 8 + 6] = 24
r= 2 cm
Area of circle = r
2
=
7
22
x 2 x 2 =
7
88
cm
2
Area of shaded region = 24 -
7
88
=
7
88 168
=
7
80
cm
2
13. Find the perimeter of the figure, where AED is a semi-circle and ABCD is a
rectangle. (Ans : 76cm)
Ans: Perimeter of the fig = 20 + 14 + 20 + length of the arc (AED)
Length of Arc = ( x r) =
7
22
x7 = 22cm
Perimeter of the figure = 76 cm
A
I
B C
20cm 20cm
14cm
A
C B
E
D
IX CBSE MATHS
Solved problems
PIE TUTORIALS
9/2 CHOPASANI HOUSING BOARD, JODHPUR
FOR FREE NOTES & TEST PAPERS Visit www.pietutorials.com
14. Find the area of shaded region of circle of radius =7cm, if AOB=70
o
, COD=50
o
and EOF=60
o
.
(Ans:77cm
2
)
Ans: Ar( Sector AOB + Sector COD + Sector OEF)
=
360
70
x 7
2
+
360
50
x 7
2
+
360
60
x 7
2
49 (
36
7
+
36
5
+
36
6
) = 49 x
36
18
=
2
49
x
7
22
= 77 cm
2
15. What is the ratio of the areas of sectors I and II ? (Ans:4:5)
Ans: Ratio will be
360
120
r
2
:
360
150
r
2
12
4
:
12
5
= 4:5
16. Find the area of shaded region, if the side of square is 28cm and radius of the
sector is ½ the length of side of square.
(Ans:1708cm)
Ans: Area of shaded region is
2 (
360
270
) x 14 x 14 + 28 x 28
2 x
4
3
x
7
22
x 14 x14 + 784
924 + 784 = 1708 cm
2
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FAQs on Solved Examples - Areas Related to Circle, Maths, Class 8
| 1. What is the formula to calculate the area of a circle? |  |
| 2. How do you find the circumference of a circle? | |
Ans. The formula to calculate the circumference of a circle is C = 2πr, where C is the circumference and r is the radius of the circle. π is a mathematical constant that represents the ratio of the circumference of a circle to its diameter, which is approximately equal to 3.14.
| 3. What is the relationship between the radius and diameter of a circle? | |
Ans. The diameter of a circle is twice the length of its radius. In other words, if you multiply the radius by 2, you get the diameter. Conversely, if you divide the diameter by 2, you get the radius.
| 4. How do you calculate the radius of a circle if you know its area? | |
Ans. The formula to calculate the radius of a circle if you know its area is r = √(A/π), where r is the radius and A is the area of the circle. π is a mathematical constant that represents the ratio of the circumference of a circle to its diameter, which is approximately equal to 3.14.
| 5. Can the area of a circle be negative? | |
Ans. No, the area of a circle cannot be negative. The area of a circle is always a positive value because it is a measure of the amount of space enclosed by the circle. If the radius of the circle is negative, then the area would still be positive because the formula for the area of a circle involves squaring the radius, which makes it positive.
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