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Area Related to Circle Test - 6

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Area Related to Circle Test - 6

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Weekly Quiz Competition

Question 1
1 / -0

In the given figure, a quadrant of a circle with centre S, is occupied by a square PQRS. The area of the shaded portion will be what fraction of thearea of the circle, if the side of the square is 14 cm?

A

B

C

D

Solution

Side of square = Radius of circle = 14 cm

Area of square = 14 × 14 = 196 cm²

Area of circle = πr² = 196π cm²

area of circle = × 196π = 112π cm²
Area of shaded quadrant =
== 49π cm²Required fraction ==
Question 2
1 / -0

A circular field of area 5544 m², has a fence around it. If the same fence is to be used around a new rectangular field of length 115 m and area 9775 m², then how much more fence is required? [Take π =]

A
130 m

B
132 m

C
134 m

D
136 m

Solution

Area of the circle = 5544 m²
πr² = 5544
r² = (5544 × 7) ÷ 22
r² = 1764
r = 42 m
Length of the fence around the circular field = Circumference of the circle = 2πr
= [2 × 22 × 42] ÷ 7 = 264 m
Area of the rectangular field = 9775 m²
Length of the rectangular field = 115 m
Width of the rectangular field = 9775 ÷ 115 = 85 m
Perimeter of the rectangular field = 2 (L + W)
= 2 (115 + 85) = 400 m
Length of the fence required = 400 – 264
= 136 m
Question 3
1 / -0

In a quadrilateral park, two opposite sides having length 42 m and 48 m, are parallel and the distance between them is 18 m. If on all the corners of the park, arcs of the same radius, i.e. 3.5 m, are cut and areas enclosed within the arcs are kept without grass to make refreshment corners in the park, then what will be the area of the park which is covered by grass? [Take π =]

A
761.5 m²

B
770 m²

C
771.5 m²

D
772 m²

Solution

Required figure of the park:

As we can see that the park is in trapezium shape, so
Area of the park =× (PQ + RS) × PO
=× (42 + 48) × 18
= 810 m²
Area of the sector centred at P =
Area of the sector centred at Q =
Area of the sector centred at R =
Area of the sector centred at S =
Total area of the 4 sectors =
As PQRS is a quadrilateral, ∠P + ∠Q + ∠R + ∠S = 360°
Total area of the 4 sectors =
=× (3.5 × 3.5) = 38.5 m²
So, area of the park which is covered by grass = Area of park - Area of 4 sectors
= 810 – 38.5
= 771.5 m²
Question 4
1 / -0

Find the sum of the area of a circle and that of a square, if the circumference of the circle is 88 m and the side of the square is twice the radius of the circle.

A
600 m²

B
800 m²

C
1000 m²

D
1400 m²

Solution

Given: Circumference of the circle = 88 m

Now, side of the square (b) = 2 x r = 2 x 14 = 28 m
Therefore, sum of areas of circle and square
Question 5
1 / -0

A regular hexagonal ground is inscribed in a circular field which has a radius of 24 m. What is the area of the field left unused around the hexagonal ground in that circular field?

A
341.3 m²

B
312.19 m²

C
368.5 m²

D
365.9 m²

Solution

Radius of the circular ground = 24 m
Area of the circular ground = πr²
= 3.14 × 24 × 24 = 1808.64 m²
As hexagon is regular and inscribed in a circle, the side of the hexagon is the same as the radius of circle, i.e. 24 m.
So, △AOB is an equilateral triangle.
Area of △AOB =
=

=
= 1.732 × 144 = 249.408 m²
As there are 6 triangles in a hexagon,, so area of the hexagon = 6 × 249.408 = 1496.45 m²
Area of the field left unused around the hexagonal ground in that circular field = 1808.64 – 1496.45 = 312.19 m²
Question 6
1 / -0

In the figure given below, ABCD is a square with each side measuring 7 cm and a circle is inscribed in the square.

What is the total area of the shaded region?

A
5.25 cm²

B
6.25 cm²

C
7.54 cm²

D
9.6 cm²

Solution

Diameter of the circle = 7 cm
Radius = cm
Area of the square = 7 cm 7 cm = 49 cm²
Area of the circle = cm²= 38.5 cm²
Difference between the areas = 49 cm² – 38.5 cm² = 10.5 cm²
Area of the shaded region = cm² = 5.25 cm²Hence, (1) is the correct option.
Question 7
1 / -0

The sum of the area of a circular park and the total area including the jogging path around that is 2550.465 m². Also, the area of the park including the jogging path and the area of the park are in the ratio 5 : 4. What is the width of the jogging path? [Take π = 3.14]

A
2.24 m

B
3.24 m

C
1.9 m

D
3.6 m

Solution

Let the area of the park = 4x
Let the area of the park including the jogging path = 5x
Sum = 5x + 4x = 2550.465
9x = 2550.465
x = 283.385 m²
Let the radius of the park excluding the jogging path be r and the radius of the park including the jogging park be R.
Area of the park excluding the jogging path = 4 × 283.385 = 1133.54 m²
πr²= 1133.54
r² = 1133.54 ÷ 3.14
r =
r = 19 m
Area of the park including the jogging path = 5 × 283.385 = 1416.925 m²
πR² = 1416.925
R² = 1416.925 ÷ 3.14 = 451.25
R =
R = 21.24 m (Approx.)
Difference = 21.24 – 19 = 2.24 m
So, the width of the jogging path = 2.24 m
Question 8
1 / -0

What will be the area of the minor sector, if a chord of a circle of radius 63 cm subtends an angle of 120° at the centre of the circle? [Take π =]

A
4128 cm²

B
4142 cm²

C
4132 cm²

D
4158 cm²

Solution

Radius of circle r = 63 cm
Angle subtended at centre θ = 120°
Area of the minor sector =
=
= [22 × 63 × 63] ÷ [7 × 3]
= 22 × 9 × 21
Area = 4158 cm²
Question 9
1 / -0

In the circle given below, find the area of the shaded region, if the radius of the circle is 14 cm.

A
850 cm²

B
518 cm²

C
420 cm²

D
450 cm²

Solution

Area of circle = = 22 2 14 = 616 cm²
And, area of triangle = = 98 cm²
Area of the shaded region = 616 - 98 = 518 cm²
Question 10
1 / -0

Calculate the area of the remaining portion, after cutting two quadrants (quarter circles), each of radius 13 cm, from the two corners of a rectangle measuring 35 cm by 25 cm.

A
365.57 cm²

B
437.5 cm²

C
537.5 cm²

D
609.43 cm²

Solution

Let the rectangle be ABCD in which AB = 35 cm and BC = 25 cm.

Area of rectangle ABCD = AB BC
= 35 cm 25 cm = 875 cm²
Area of the two quadrants = 2
= cm²
= cm²
= 265.57 cm²
Required area = (875 – 265.57) cm²
= 609.43 cm²
Question 11
1 / -0

Find the length of chord AB of a circle of radius 5 cm if point A meets the largest chord of the circle on its circumference at points B and C respectively and AC = 6 cm.

A
6 cm

B
3 cm

C
5 cm

D
8 cm

Solution

Given:
Radius of circle = 5 cm

AC = 6 cm
Diameter is the biggest chord of the circle.
Therefore, (Angle in a semi-circle is a right angle)
In BAC, using Pythagoras theorem,

Hence, this is the required solution.
Question 12
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 13
1 / -0

In the given parallelogram PQRS, if the area of the shaded part of the circle with centre S and radius 14 cm is 462 cm², then
(Take = 22/7)

A
∠QRS = 90°

B
∠PSR = 120°

C
∠QRS = 80°

D
∠PSR = 150°

Solution

Area of sector = ,where is exterior angle PSR.

462 =

= 270°

If = 270°, ∠PSR = 360 - 270° (Whole turn = 360°)
∠PSR = 90°
∠PSR + ∠QRS = 180° (Adjacent angles of a parallelogram add up to 180°)
90° + ∠QRS = 180°
∠QRS = 90°
Question 14
1 / -0

In the given figure, a circle is drawn with B as the centre and AC as the diameter; two semicircles are drawn with AB and BC as diameters. If the perimeter of the shaded portion is 280 cm, what will be the radius of the semicircle whose diameter is AB?

A
22 cm

B
14.5 cm

C
12.25 cm

D
8cm

Solution

Let the radius of the circle be R.
Radius of the semicircle with diameter AB =
Circumference of the circle with diameter AC = 2R
Circumference of the semicircle with diameter AB = (R/2) + R

According to the question,
Perimeter of the one shaded part =
Considering both the shaded parts,

Perimeter of shaded portion =

280 =

280 =

= 280

R(3 + 2) = 280

R() = 280

R = 280

R =

R = 24.5 cm

So, radius of the semicircle with diameter AB = 12.25 cm
Question 15
1 / -0

A circular park has a concrete road 7 m wide build around its perimeter. If its inner perimeter is 22 m, what will be the ratio of the inner diameter to the outer diameter?

A
1 : 3

B
2 : 5

C
3 : 7

D
4 : 5

Solution

Let the radius of the inner and outer circles be r₁ and r₂, respectively.
Therefore, r₂ = r₁ + 7m
Given, 2πr₁ = 22
r₁ =

= =

= 3.5 m
r₂ = 7 + 3.5 cm = 10.5 cm
Inner diameter = r₁ × 2 = 7 cm
Outer diameter = r₂ × 2 = 21 cm
Ratio = 7 : 21 = 1 : 3
Question 16
1 / -0

James is driving his car in the rain. He uses its wipers to clean its glass. If the length of its each blade of the wiper is 21.5 cm and they are sweeping through an angle of 105°, then what will be the total area cleaned at each sweep of the blades?

A
847.5 cm²

B
761.8 cm²

C
652.2 cm²

D
423.7 cm²

Solution

Total area cleaned at each sweep of the blades = 2 × Area of that sector = 2 × × πr²
= 2 × × × (21.5)²
= 847.5 cm²
Question 17
1 / -0

Milli has a square cardboard whose area is 196 cm². She wants to cut a semicircle from this square cardboard. If she manages to cut a semicircle along one of the sides of the square cardboard, then which of the following represents the radius of the semicircle?

A
4 cm

B
7 cm

C
11 cm

D
14 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 .
The area of the given square = a²= 196 cm²a =
So, the length of one side of the square = 14 cm
And the radius of the given semicircle = = = 7 cm
Question 18
1 / -0

A and B have two threads and they used these threads to prepare circles. A prepares a circle whose radius is 6.7 cm and B prepares a circle whose radius is 18.4 cm. They decide to prepare a bigger circle using these two threads. What is the area of the new circle?

A
1980 cm²

B
2134 cm²

C
2676 cm²

D
3127 cm²

Solution

Circumference of the first circle = 2πr₁ = 2 × π × 6.7
Circumference of the second circle = 2πr₂ = 2 × π × 18.4
According to the question:
Circumference of the bigger circle = Circumference of the first circle + Circumference of the second circle
2πR = 2 × π × 6.7 + 2 × π × 18.4
R = 6.7 + 18.4
R = 25.1 cm
Area of the bigger circle = πR²
= × (25.1)²
= 1980 cm²
Question 19
1 / -0

Maddison is asked to find the length of the arc of that sector of a circle whose area is 9π cm². It is also given that the diameter of this circle is 18 cm. According to this information, which of the following represents the value of the length of the arc?

A
2π cm

B
π cm

C
4π cm

D
9π cm

Solution

If r is the radius of a circle and is the angle subtended by an arc at the centre of the circle, then the length of the arc is × 2πr and the area of the sector of angle is .
Thus, the area of the given sector = = 9π cm²
= = 40°
Thus, the length of the given arc = × 2 × π × 9
= 2π cm
Question 20
1 / -0

The area of the shaded part in the given figure is 40 cm². Which of the following represents the area of the un-shaded part, which is ΔAOB?

A
651 cm²

B
542 cm²

C
231 cm²

D
191 cm²

Solution

Area of the sector with angle =
Area of ΔAOB = Area of the sector with angle - Area of the minor segment

Thus, area of the sector with angle 60° = × π × 21 × 21 cm²
= 231 cm²
Thus, the area of ΔAOB = 231 - 40 = 191 cm² [ Area of the minor segment = 40 cm²]
Question 21
1 / -0

A train is travelling at a speed of 108 km per hour. If the area of a wheel of the train is 18,634 cm², then how many complete revolutions does each wheel make in 13 minutes?

A
4835

B
3899

C
4105

D
4262

Solution

Area of a wheel = 18,634 cm²πr²= 18,634
r² = 18,634 ×
r² = 5929
r = 77 cm

Circumference of the wheel = 2πr
2 × × 77 = 484 cm

Speed of the train = 108 km per hour

= cm min^-1 = 1,80,000 cm min^-1
Distance travelled by the train in 13 minutes = 1,80,000 × 13 = 23,40,000 cm
Let the number of revolutions of the wheel of the train be n.
n × Distance travelled in 1 revolution = Distance travelled in 13 minutes
n × 484 = 23,40,000
n = 23,40,000 ÷ 484
n = 4834.7 = 4835 (Approximate)
Question 22
1 / -0

Match the shaded regions given in the figures below with their areas.

A
A - (i), B - (ii), C - (iii)

B
A - (iii), B - (ii), C - (i)

C
A - (ii), B - (i), C - (iii)

D
A - (ii), B - (iii), C - (i)

Solution

(A) If a is one side of a square, then the area of the square is a². If D and r are the diameter and the radius of a semicircle, respectively, then the area of the semicircle is r², where r = .

Thus, the diameter of the semicircle = NO = 7 cm
Radius of the semicircle = cm
Area of the semicircle = cm²
= cm² = cm²
Now, the area of the square = 7² cm² = 7 7 = 49 cm²
Area of the shaded portion = (49 – ) cm²= 49 cm²
= 49 cm²
(B) If 'a' is the side of the square, then area of the square = a²
If 'r' is the radius of the circle, then area of a quadrant of the circle =

Thus, the area of the square OABC = (5 cm)² = 5 cm 5 cm = 25 cm²

And the radius of the circle = OA = 5 cm
Thus, the area of the quadrant of the circle = cm²
= × × 5 × 5
= × × 25 cm²

= cm²
Hence, the area of the shaded region = area of square OABC - x area of circle =
(C) In an equilateral triangle, all the sides are equal. If D and r are the diameter and the radius of the semicircle, respectively, then r = and the area of the semicircle is .

So, in the given figure, AB = BC = CA = 6 cm
Thus, the diameter of the semicircle = BC = 6 cm
Radius of the semicircle = 6 ÷ 2 cm = 3 cm
Therefore, the area of the shaded portion = Area of the semicircle
= cm²
Question 23
1 / -0

A circle has a diameter of 64 cm. An equilateral triangle with the largest possible side length is drawn inside the circle and the corners of the triangle are touching the circle. If Priya decides to cut the triangle out, then what will be the area of the remaining part?

A
cm²

B
cm²

C
cm²

D
cm²

Solution

Radius = Diameter ÷ 2 = 64 ÷ 2 = 32 cm
Let AD be the median of ΔABC,
and O be centre of the circle.

AO = × AD = 32

AD = 48 cm

In ΔABD,
AB² = AD² + BD²AB² = (48)² +
= (48)²

AB =

= cm

Area of equilateral triangle ΔABC =

= × 32 × 32 × 3 = 96 × 8 ×
= 768 cm²
Area of circle = πr²= × (32)²= × 1024

= cm²If Priya decides to cut the triangle out, then the area of the remaining part = area of circle – area of ΔABC

= ( - 768 ) cm²
Question 24
1 / -0

P is trying to make a drawing. He has a rectangle ABCD with dimensions 6 cm × 4 cm. He joined a semicircle whose diameter is the same as the length of this rectangle. What is the area of this drawing?

A
38.14 cm²

B
45.16 cm²

C
51.24 cm²

D
61.62 cm²

Solution

If l and b are the length and breadth of a rectangle, respectively, then the area of the rectangle is l × b.
If D and r are the diameter and the radius of a semicircle, respectively, then
Area of the semicircle = , where r =

Diameter of the semicircle = AD = 6 cm
Radius of the semicircle = 6 ÷ 2 = 3 cm
Area of the semicircle = cm²
Area of the rectangle = 4 × 6 = 24 cm²
Area of the given figure = Area of the semicircle + Area of the rectangle
= cm²
= 38.14 cm²
Question 25
1 / -0

Which of the following are the correct answers according to the given statements?

(I) P has a circle whose circumference is 44.87 cm. He wants to find out the area of a quadrant of this circle. What will be the area of the quadrant of this circle?
(II) A has a square inside a circle whose diameter is 14.8 cm. If the square has each side of 12.8 cm, then what will be the area of the shaded part in the given diagram?

(III) If the perimeter of a circle is c, the which of the expressions represents the radius of the circle?

A
(I) 20.08 cm² (II) 83.26 cm² (III)

B
(I) 35.06 cm² (II) 63.84 cm² (III)

C
(I) 40.05 cm² (II) 8.26 cm² (III)

D
(I) 80.08 cm² (II) 7.26 cm²(III)

Solution

(I) Circumference of the circle = 44.87
2πr = 44.87
r =

Quadrant of the circle will subtend an angle of 90° at the centre of the circle.
Area of quadrant = × πr²
= × πr²
= 0.25 × π
= 40.05 cm²

(II) Area of the square = a × a = 12.8 × 12.8 = 163.84 cm²
Radius of the circle = d ÷ 2 = 14.8 ÷ 2 = 7.4 cm
Area of the circle =
= × 7.4 × 7.4 = 172.10 cm²
Area of the shaded part = Area of the circle - Area of the square
= 172.10 – 163.84 = 8.26 cm²

(III) If r is the radius of the circle, then the perimeter of circle is 2πr.
Thus, the perimeter of the given circle = 2πr = c
2πr = c
r =
Radius of the given circle =

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