# Free Areas Related to Circles 03 Practice Test - 10th Grade

### Question 1

The area in cm2 of the shaded region if the radius of the smaller circles are 1 cm each is 6.28 cm2.

True

False

#### SOLUTION

Solution :B

From figure, area equal to one full circle exists in the semi-circle.

Area of larger semi-circle = 3.14×222 = 6.28 cm2

Area of smaller semi-circles = 2 x 3.14×122 = 3.14 cm2

Area of shaded portion = 3.14 cm2

### Question 2

In the figure, two small circles touch each other externally at the centre of a bigger circle and these small circles are touched internally by a bigger circle with radius 4 cm. Find the area of the shaded region in cm2.

27

30

22

35

#### SOLUTION

Solution :A

From the figure, two circles of radius 4 cm exist in the rectangle.

Their total area = 2πr2 = 2 × 3.14 x 16 = 100.48 cm2

From figure

Length of rectangle is 4r = 16 cm

Breadth of rectangle is 2r = 8 cm

Area of rectangle = 16 x 8 = 128 cm2Required area = [Area of rectangle - Total area of circles]

= 128 - 100.48 = 27.52 cm2.

∴ The area of the shaded region

= 27.52 cm2.

### Question 3

In figure, ABC is a quadrant of a circle of radius 14 cm and a semicircle is drawn with BC as diameter. Find the area of the shaded region (in cm2).

#### SOLUTION

Solution :Area of quadrant = 90360×227×14×14 = 154 cm2

Area of △ABC = 0.5 x 142 = 98 cm2

^{ }The diameter = 14 × √2 = 19.8 cm.

Area of semi-circle = 0.5 3.14×9.92 = 153.95 cm2

The area of segment = 154 - 98 = 56 cm2

The required area = 153.95 - 56 = 97.95cm2.

### Question 4

State true or false.

A wire in the shape of a square of perimeter 88 cm is bent so as to form a circular ring. The radius of the ring formed will be 7 cm.

True

False

#### SOLUTION

Solution :B

Perimeter of square and circle is same.

88 = 2 π × r

= 2 227×rr = 14 cm.

### Question 5

Figure depicts a racing track whose left and right ends are semi-circular. The distance between the two inner parallel line segments is 60 m and they are each 106 m long. If the track is 10 m wide, find the area of the track (in m2).

#### SOLUTION

Solution :Area of track = Area of Field - Area inside the track

The area of the track =106 x 80 —106 x 60 + 2×12×227×(402−302)

=106 x 20 + 227 × (70) × (10)

=2120 +700 x 227 = 2120 + 2200

= 4320 m2

### Question 6

State true or false.

If the diameter of a semi-circular protractor is 14 cm then the perimeter will be 36 cm(Approx.)

True

False

#### SOLUTION

Solution :A

Diameter = 14 cm

Perimeter of semi circle = π × r = 227 x 7 = 21.99 cm

Total perimeter of protractor is = 21.99 + 14 = 35.99 cm ~ 36 cm.

### Question 7

In the figure shown, the segment CEDC is called the

#### SOLUTION

Solution :In the figure shown, the segment CFDC which is bigger is called the major segment and the

segment CEDC which is smaller is called the minor segment.

### Question 8

Find the area of the shaded region (in cm2) as shown in figure of the two concentric circles with centre O and radius 7 cm and 14 cm respectively. Given ∠AOC = 40∘.

(use π = 227)

42.1 cm2

51.32 cm2

67.8 cm2

96.5 cm2

#### SOLUTION

Solution :B

Given: radius for sector OAC = 14 cm and angle subtended = 40∘ and

radius for sector OBD = 7 cm and angle subtended = 40∘

Area of Sector = x∘360∘×πr2

Required area = [Area of sector OAC – Area of sector OBD]

=40360×227×142–40360×227×72

= 68.42 - 17.1

= 51.32 cm2

∴ Area of shaded region = 51.32 cm2

### Question 9

A car is provided with 2 non overlapping wipers. The structure of each wiper is as shown in the figure. The brown blade actually wipes the glass sweeping through an angle of 120∘. If the length of the wiper is 35 cm and length of the blade is 28 cm, find the total area(in cm2) cleaned for each sweep.

#### SOLUTION

Solution :

As the wiper sweeps, area covered by each wiper is indicated in brown color below.

Area cleaned by each wiper = Area of sector AOB – Area of sector COD

Area of the sector of angle Ɵ = (θ360∘)×π×r2

Area of sector AOB - Area of sector COD = 120360×227×352−(120360)×227×(72) = 1232 cm2

So, the total area cleaned by both the wipers = 2 × 1232 cm

^{2 }= 2464 cm2

### Question 10

A paper is in the form of a rectangle ABCD where AB = 22 cm and BC = 14 cm. A semicircle portion with BC as diameter is cut off. Find the area of the remaining paper in cm2.

221

210

231

240

#### SOLUTION

Solution :C

Area of rectangle = 22×14=308cm2

Area of semicircle

= 12×227×(7)2=77cm2

Required area = [Area of rectangle - Area of semicircle]

= 308 – 77 = 231cm2