Area Related to Circle Test

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Area Related to Circle Test - 7

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Question 1
1 / -0

A square circumscribes a circle. Find the area enclosed between the square and the circle, if radius of the circle is 10 cm.

A
86.71 cm²

B
85.71 cm²

C
314.29 cm²

D
214.29 cm²

Solution

Let ABCD be a square circumscribing a circle with centre O.
OE is the radius of the circle, which is 10 cm.
Side of the square = 2 OE = (2 10) cm = 20 cm
Thus, area of square = side side = 20 20 = 400 cm²
Area of circle with radius 10 cm = r²
= 10 10 = 314.29 cm²
Area of shaded region = 400 - 314.29 = 85.71 cm²
Question 2
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In the following figure, the diameter AB = 13 cm and BC = 5 cm. Then, the area of the shaded region is

A
42.25π sq. cm

B
(42.25π - 30) sq. cm

C
(42.25π - 15) sq. cm

D
(42.25π - 20) sq. cm

Solution

Area of the shaded region = Area of the circle - area of the triangle
Area of the circle = πr² = 42.25π sq. cm
Area of the triangle:
AB = 13 cm, which is the diameter of the circle.
Hence, angle ACB is a right angle.
Therefore, ACB is a right-angled triangle and AC = 12 cm.
So, area of the triangle = 30 sq cm.
Required area = (42.25π - 30) sq. cm
Question 3
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In a quadrant of a circle, a square is drawn as shown in the figure. If the area of the shaded region is 56 sq. cm, then the radius of the quadrant is

A
7 cm

B
14 cm

C
3.5 cm

D
2.1 cm

Solution

Area of the shaded region = Area of the quadrant - area of the square
Area of the quadrant =
Radius of the quadrant = Diagonal of the square
Side of the square =
Area of the square =
Area of the shaded region =
= 56

Solving for r, we get r = 14 cm
Question 4
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The area of a circle is 2500 m². If the minor sector covers 20% area of the circle, what is the area of the major sector?

A
500 m²

B
1000 m²

C
1500 m²

D
2000 m²

Solution

20% of the area is covered by the minor sector.
∴ Area covered by the major sector = (100 - 20)%
= 80%
∴ 80% of 2500 m²
=
= 2000 m²
Question 5
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The sum of areas of two circles touching internally is 500.5 sq. cm. If the distance between their centres is 3.5 cm, then the ratio of their radii is

A
1 : 2

B
3 : 2

C
4 : 3

D
3 : 4

Solution

Let the radii be X and Y. Then, their areas are πX² and πY². Distance between their centres is the difference between their radii.
Given:

Neglecting the negative value, we get Y = 7 and X = 10.5.
Ratio of their radii,
X : Y = 10.5 : 7 = 3 : 2
Question 6
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A circle having an area of 1386 cm² is made up of a wire. If the same wire is bent to make an equilateral triangle, what will be the area of the triangle? (Use pi = and = 1.73)

A
696.26 cm²

B
722.84 cm²

C
837.32 cm²

D
896.42 cm²

Solution

Area of a circle = πr²
1386 = × r²
= r²
441 = r²r =21 cm
Circumference of a circle = 2πr
Circumference of the circle = = = 132 cm
Circumference of the circle = Perimeter of the triangle = 132 cm
As the triangle is equilateral, sides of the triangle = 132 ÷ 3 = 44 cm
Area of the triangle = = = 837.32 cm²
Question 7
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In a square sheet of side 28 cm, four circular parts are cut as shown in the figure. Then, the area of the remaining sheet is

A
178 sq. cm

B
168 sq. cm

C
188 sq. cm

D
158 sq. cm

Solution

Radius of the circular part = 7 cm
Hence, area of the circular parts =
Area of the square = 784 sq. cm
Area of the remaining sheet = 784 - 616 = 168 sq. cm.
Question 8
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The cost of ploughing a circular field at 50 p per m² is Rs. 4928. What will be the cost of fencing the field at the rate of Rs. 12 per m?

A
Rs. 3936

B
Rs. 4045

C
Rs. 4254

D
Rs. 4224

Solution

Cost of ploughing = Rs. 4928
Area = ( 50 p = Rs. 0.5)
= 9856 m²
r² = 9856 r = 56 m
Circumference = 2r
= 2 56
= 352 m
Cost of fencing the field = 352 12
= Rs. 4224
Question 9
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As in the figure, a table fan of three wings in the shape of semicircles moving in a circle of radius 4.5 cm is shown. Find the area of remaining part of this circle which is not covered by the wings.

A
35.84 cm²

B
39.74 cm²

C
43.68 cm²

D
47.98 cm²

Solution

Radius of bigger circle = 4.5 cm
Area = R²= (4.5)²
Area of one semi-circle = πr² = ×
Area of three similar semicircles = × (4.5)²
Area of shaded region = (4.5)² - (4.5)²
= (4.5)² = (4.5)² × = × 4.5 × 4.5 ×
Area of shaded region = 39.74 cm².
Question 10
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Two identical circles intersect such that their centres and the points at which they intersect form a square of side 1 cm. The area (in sq. cm) of the portion that is common to the two circles is

A

B

C

D
- 1

Solution

Area of common portion of AOBO` = 2 Area of segment AOB

= = = ( - 1) sq. cm
Question 11
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The ratio of the circumference to the area of the shaded region of the following figure, if the radius of the sectors is 7 cm, is

A

B

C

D

Solution

Let the radius of the semicircle be r.
Then, area of the given figure =
Circumference of the given figure =
Ratio of the circumference to the area =
When r = 7, the ratio becomes .
Question 12
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The circumference of a circle is equal to the perimeter of a square. If the perimeter of the square is 44 cm, then what is the difference between the area of the circle and that of the square?

A
33 cm²

B
28 cm²

C
24 cm²

D
20 cm²

Solution

Circumference of circle = Perimeter of square
2r = 44
2 = 44 r = 7 cm
Area of circle =
=
= 154 cm²
Perimeter of square = 44
Side of square = 11 cm
Area of square = (11)²
= 121 cm²
Difference = 154 – 121
= 33 cm²
Question 13
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A truck covers 66 m in 15 seconds. If the area of the tyre of the truck is 616 cm², then how many revolutions will the tyre complete in 72 seconds?

A
336

B
360

C
372

D
396

Solution

Area of the tyre of the truck = 616 cm²
Area of a circle = π r²616 = × r²
616 × 7 = 22r²
4312 = 22r²
r² = 4312 ÷ 22
r² = 196
r =
r = 14 cm

Circumference of the tyre = 2πr
= 2 × × 14
= 88 cm
= 0.88 m
Number of revolutions made by the tyre in 15 seconds = 66 ÷ 0.88 = 75
Number of revolutions made in 1 second = 75 ÷ 15 = 5
Number of revolutions made in 72 seconds = 5 × 72 = 360
Question 14
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Find the area of the shaded portion in the given figure.

A
(334 – 4) cm²

B
(324 – 16) cm²

C
(420 – 16) cm²

D
(400 – 6) cm²

Solution

In the diagram,
Area of rectangle ABCD = L W
= 30 14
= 420 cm²
Now, EF = GH = 30 – (5 + 5) – (4 + 4) = 12 cm
EH = FG = 14 – (3 + 3) = 8 cm
Area of EFGH = 12 8 = 96 cm²
Area of two semi-circles = 2
=
= 16 cm²
Now, area of the shaded portion
= Area of rectangle ABCD – (area of rectangle EFGH + area of two semi-circles)
= 420 m² – (96 + 16) cm²
= (324 – 16) cm²
Question 15
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A is driving his truck in the rain. He uses its wipers to clean the glass. The wipers are sweeping through an angle of 105° each time. If the total area cleaned by the wipers in one sweep is 594 cm², then what is the length of the blade of the wiper?

A
18 cm

B
21 cm

C
14 cm

D
20 cm

Solution

Area swept by two blades = 2 × × πr²
594 = 2 × × πr²
594 ××× = r²
324 = r²
18 = r
r = 18 cm
Question 16
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The given figure consists of 4 small semi-circles and 2 big semi-circles. If small semi-circles are equal in radii and bigger semi-circles are also equal in radii, find the perimeter and area of the shaded portion of the figure, given that the radius of each bigger semi-circle is 20 cm.

A
420.17 cm and 941.86 cm²

B
228.57 cm and 942.86 cm²

C
140 cm and 900.06 cm²

D
122 cm and 941.68 cm²

Solution

Radius of each bigger semi-circle = R = 20 cm
Radius of each smaller semi-circle = r = = 10 cm (As small semi-circles are equal in radii)
As the shaded portion includes 1 bigger semi-circle and 2 smaller semi-circles;
Perimeter of the shaded portion
=
=
=
= 228.57 cm
Area of the shaded portion =
=
=
=
=
= 942.86 cm²
Question 17
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A square park is surrounded by 4 semi-circular pitches as shown in the figure. If the circumference of each semi-circular pitch is 33 m, find the cost of spraying pesticide on the square park and 4 semi-circular pitches at Rs. 4.5/m².

A
Rs. 5133

B
Rs. 5103

C
Rs. 5101

D
Rs. 5003

Solution

Circumference of the given semi-circle = 33 m
Circumference of a semi-circle = πr
So, 33 =× r
33 × 7 = 22r
231 = 22r
r = 231 ÷ 22
r = 10.5 m
Area of the semi-circle =
=
= 173.25 m²
Since there are four semi-circles, total area of the semi-circles = 173.25 × 4 = 693 m²By looking at the given figure we get that the diameter of the semi-circle makes the side of the square park.
Diameter of the semi-circle = r × 2
= 10.5 × 2 = 21 m
Area of the square park = (side)²= (21)²= 441²
Total area = Area of the semi-circular pitches + area of the square park = 693 + 441 = 1134 m²
Cost of spraying pesticide on the square park and 4 semi-circular pitches = Rs. (4.5 × 1134) = Rs. 5103
Question 18
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The area of a square is 196 cm². What will be the radius of the largest semicircle that can be drawn completely inside on one of its sides?

A
98 cm

B
14 cm

C
7 cm

D
28 cm

Solution

If the area of a square is 'a²', then the length of one of its sides is 'a'. The radius (r) of a semicircle drawn completely inside on one of its sides is equal to 'a/2'.

Thus, the area of the given square = a² = 196 cm²
a =
So, the length of one side of the square = 14 cm
And the radius of the given semicircle = a/2 = 14/2 = 7 cm
Question 19
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There is a circular playground which is divided into four sections. These four sections are used by different persons. Half of the playground is used by the cricket coach. Rest of the playground is divided into three sections, where one is in the shape of a triangle. The biggest side of the triangle has same length as the diameter of the circular playground and the other two sides of the triangle are 48 m and 14 m. If all the vertices of the triangle touch the circle, then what will be the area of the shaded sections? (Answer up to two decimal places.)

A
2992.57 m²

B
3433.75 m²

C
3678.98 m²

D
3592.57 m²

Solution

Here, AC = 48 m, BC = 14 m and AOB is the diameter of the circle.
So, P = 90° (Angle in the semi-circle)
Therefore, ΔABC is a right triangle.
AB² = BC² + AC²
AB² = 48² + 14²
AB² = 2304 + 196
AB² = 2500
AB = 50 m
Now, area of the semi-circle = × π × r²
= × 22 ÷ 7 × 50 × 50
= 3928.57 m²
Area of ΔABC =× 48 × 14 = 336 m²
So, area of the shaded sections = 3928.57 – 336 = 3592.57 m²
Question 20
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Anu prepares two playgrounds for pets, one is circular and another is rectangular. The areas of the circular and rectangular playgrounds are 22,176 cm² and 55,440 cm², respectively. She wants to create a fence around the playgrounds. If the length of the rectangular playground is 13.86 m, then how much more wire does she have to buy for the rectangular playground than for the circular playground?

A
20.15 m

B
18.44 m

C
23.24 m

D
28.94 m

Solution

Area of the circular playground = πr²
22,176 = πr²
22,176 ×= r²
7056 = r²
r = 84 cm
Amount of wire used for the circular playground = Circumference of the circle = 2πr
= 2 ×× 84
= 528 cm --- (1)
13.86 m = 13.86 × 100 = 1386 cm
Area of the rectangular playground = l × b
55,440 = 1386 × b
55,440 ÷ 1386 = b
40 cm = b
Perimeter of the rectangle = 2(l + b)
= 2(1386 + 40)
= 2(1426)
= 2852 cm --- (2)
Length of the extra wire for rectangular playground = 2852 – 528 [from equations (1) and (2)] = 2324 cm
2324 cm = 2324 ÷ 100 = 23.24 m
Question 21
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The perimeter of a circle is the same as the perimeter of a triangle with all the sides of same size and the perimeter of the triangle is the same as the perimeter of a square which is 528 cm.

(1) What will be the area of a circle which has twice the diameter of the given circle?
(2) If there is a circle inscribed in the square, then what will be the area of the inscribed circle?
(3) What will be the difference between the area of the triangle and the area of the square?

Note: Give answer up to two decimal places and .

A

(1) (2) (3)

22,176 cm² 41,661.36 cm² 3255.65 cm²

B

(1) (2) (3)

44,352 cm² 48,761.37 cm² 3654.41 cm²

C

(1) (2) (3)

88,704 cm² 13,690.29 cm² 4026.88 cm²

D

(1) (2) (3)

90,774 cm² 13,690.29 cm² 4566.65 cm²

Solution

Perimeter of the circle = Perimeter of the triangle = Perimeter of the square = 528 cm
(1)
Perimeter of the circle = 528
2πr = 528
r =×× 528
r = 84 cm
Diameter of the circle = 2 × r
= 2 × 84
= 168 cm
So, diameter of the new circle = 2 × diameter of the first circle
= 2 × 168
= 336 cm
Radius of the new circle = 336 ÷ 2 = 168 cm
Area of the new circle = πr²
=× 168 × 168
= 88,704 cm²

(2) Perimeter of the square = 528 cm
4s = 528
s = 528 ÷ 4
s = 132 cm
Diameter of inscribed circle = Side of the square = 132 cm

Radius of the circle = 132 ÷ 2 = 66 cm
Area of the inscribed circle = πr²
=× 66 × 66
= 13,690.29 cm²

(3) Perimeter of the square = 528 cm
4s = 528
s = 528 ÷ 4
s = 132 cm
Area of the square = s × s
= 132 × 132
= 17,424 cm²
Perimeter of the triangle = 528 cm
3a = 528
a = 528 ÷ 3
a = 176 cm
Area of the equilateral triangle =
=
= × 176 × 176
= 7744

= 7744 × 1.73
= 13,397.12 cm²

Difference = 17,424 – 13,397.12
= 4026.88 cm²
Question 22
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Find the area of the shaded region if radius is 21 cm and area of triangle OAB is ^.

A

B

C

D

Solution

Area of the circle =
= 22 3 21 cm²
= 1386 cm²
Area of the shaded region = Area of the circle - Area of AOB =

Question 23

1 / -0

Match the shapes in Column I with the area of the shaded region given in Column II

	Column I	Column II
(M)	The lengths of sides AB and BC are 8 cm and 4 cm, respectively.	(i) 40 cm²
(N)	The area of the triangle AOB is 191 cm².	(ii) 112 cm²
(O)	ABCD is a square with each side 14 cm.	(iii) 8(4 – ) cm²

M - (ii); N - (iii); O - (i)

M - (i); N - (ii); O - (iii)

M - (ii); N - (i); O - (iii)

M - (iii); N - (i); O - (ii)

Solution

(M)
Area of the rectangle = 8 x 4 cm² = 32 cm²

And area of the semicircle = = = cm²Thus, area of the shaded portion = (32 – 8) cm²= 8(4 – ) cm²
(N)

Area of minor segment (shaded region) = Area of sector with angle - Area of triangle AOB

Area of the sector with angle =
Area of minor segment (shaded region) = - 191

Area of minor segment (shaded region) = 231 - 191
Area of minor segment (shaded region) = 40 cm²

(O)
We can split the figure as shown:

Area of this shaded region = Area of the sector - Area of triangle

Area of the shaded region = - = 56 cm²

Total area of the shaded region = 2 × 56 cm² = 112 cm²

Question 24
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ABCD is a rectangle with AD = 1 unit. DPF and CQF are two equal arcs drawn with A and B as centres respectively. E is the midpoint of CD. Another arc with E as centre touches the two arcs DPF and CQF at P and Q, respectively. What is the area of the shaded portion?

A
2 - (3 - 2) π sq. units

B
2 - ( - ) π sq. units

C
2 - ( - 1)²sq. units

D
2 - π (2 - ) sq. units

Solution

As AF and BF are both radii of equal arcs, AF = BF = AD = 1 unit.
∴ AB = 2 units
Area of the rectangle = 2 sq. units

Note that BE is the diagonal of square FBCE.
Therefore, BE = units
∴ EQ = BE - BQ = - 1 (∵ BQ = BC = 1)
∴ Area of semicircle with centre E = ( - 1)² sq. units
Also, total area of the two segments with centres A and B = 2 × ××1² = sq. units
∴ Area of the shaded portion = 2 - [1 + ( - 1)²] = 2 - [2 - 2 + 2] = 2 - (2 - ) sq. units
Question 25
1 / -0

In a semi-circular sheet of radius 2r, a circular area and 2 semi-circles of radius r are cut as shown in the figure. If AB = BC = BD = 2r, then what is the area of the shaded region?

A

B

C

D

Solution

Required area = Area of semicircle ADC - 2(Area of smaller semi-circles) - Area of the circle
Area of semi-circle ADC =
Area of smaller semi-circle =
Area of the circle:
Let the centre of the circle be E and those of the semicircles be F and G, as shown in the figure.

If the radius of the circle is x, then ED = x, FE = r + x, FG = 2r and BD = 2r
Hence, DE = 2r - EB

Hence, the area of the circle is .
Thus, area of the shaded region
=