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

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

True

B.

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

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. A.

27

B.

30

C.

22

D.

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 cm2

Required area = [Area of rectangle -  Total area of circles]
= 128 - 100.48 = 27.52 cm2.

The area of the shaded region
= 27.52 cm2.

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.

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.

A.

True

B.

False

#### SOLUTION

Solution : B

Perimeter of square and circle is same.

88 = 2 π × r
=  2 227×r

r = 14 cm.

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×(402302)

=106 x 20 + 227 ×  (70) × (10)

=2120 +700 x 227 = 2120 + 2200

= 4320 m2

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

A.

True

B.

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.

In the figure shown, the segment CEDC is called the ___ segment. #### 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. 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) A.

42.1 cm2

B.

51.32 cm2

C.

67.8 cm2

D.

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 = x360×πr2
Required area = [Area of sector OAC – Area of sector OBD]
=40360×227×14240360×227×72
= 68.42 - 17.1
= 51.32 cm2
Area of shaded region = 51.32 cm2

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= 2464 cm2

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.

A.

221

B.

210

C.

231

D.

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