# Free Areas Related to Circles 02 Practice Test - 10th Grade

The circumference of a circular field is 528 cm. Then its radius is __________.

A.

42 cm

B.

84 cm

C.

72 cm

D.

56 cm

#### SOLUTION

Solution : B

The circumference of a circle with radius r is given by
2πr.

Hence,
2πr=528

r=5282π =84 cm

(π =227)

In the figure a small square is inserted inside the bigger square. The area of the shaded region of the square is__ sq.cm.

#### SOLUTION

Solution :

Here we need to find the area of the shaded part of the square. For finding the area of the shaded region we can find the area of the bigger square and the smaller square and then subtract the area of the smaller square from area of the larger square.

Area of the larger square =x2
= 82
= 64cm2

Area of the smaller square = b2
= 42
= 16 cm2

The area of the shaded region will be = Area of the bigger square – Area of the smaller square
= 64 – 16
= 48 cm2
Therefore the area of the shaded region will be 48cm2.

If the radius of a circle is 7π cm, then find the area of the circle (in square cm).

A.

154

B.

49π

C.

22

D.

49

#### SOLUTION

Solution : D

Area of a circle of radius r
=πr2

Area
=π7π7π=49 cm2

A chord of a circle of radius 12 cm subtends an angle of 120 at the centre. Find the area in cm2 of the corresponding segment of the circle. (Use π = 3.14 and 3 = 1.73)

___

#### SOLUTION

Solution :

The area of the sector = 120360×π×122 = 150.8 cm2

Perpendicular from chord is drawn to the centre bisects the chord. The angle subtended by each triangle at the centre is 60.

Height of perpendicular =  r x cos 60 = 12 x 0.5 = 6 cm.

Length of chord = 2 × r × sin 60 = 24 × 32= 20.6 cm

The area of triangle = 0.5 × 20.6 × 6 = 61.8 cm2

The area of segment = 150.8 - 61.8 = 89 cm2

If the circumference of a circle is 528 cm, then its area is ______ cm2.

(π=227)

A.

22176

B.

22576

C.

23176

D.

24576

#### SOLUTION

Solution : A

Circumference of the circle
=2πr =528 cm
(where r is the radius of the circle)

r=5282π =84 cm

Now, area of the circle = πr2

= 227842 = 22176 cm2

Find the area of the shaded region in the figure given below, if ABCD is a square of side 14 cm and APD and BPC are semicircles.
(Take π=227)

A.

45 cm2

B.

42 cm2

C.

60 cm2

D.

35 cm2

#### SOLUTION

Solution : B

Area of a circle
=πr2

From Figure, the diameter of circle is 14 cm. Two semi-circles make one full circle.

The area of one full circle is
=227×72=154 cm2

The total area of square
=142=196 cm2

The area of shaded portion = [Area of square- Area of full circle]
= 196 - 154 = 42cm2
Hence, area of shaded region
=42 cm2

The area of an equilateral ΔABC is 17320.5 cm2. A circle is drawn taking the vertex of the triangle as centre. The radius of the circle is half the length of the side of triangle. Find the area of the shaded region (in cm2) . (π = 3.14 , 3 =1.73205)

___

#### SOLUTION

Solution :

Area of shaded region = area of ΔABC - 3 (Area of sector BPR)

Let 'a' be the side of the equilateral ΔABC.

Using area of an equilateral triangle = 34a2,
34a2 = 17320.5
Solving, a2=17320.5×43
a2=17320.5×41.73205
a2=17320.5×417320.5×104
a=2×102
a = 200 cm.

Radius of the circles = 12×200 = 100 cm

Now, using area of a sector when the degree measure of the angle at the centre is θ = θ360πr2
Required area =17320.5 - 3[60360×3.14×1002 ]

Required area = 1620.5cm2

A drain cover is made from a square metal plate of side 40 cm and has  336  holes of radius 1 cm each drilled in it. Find the area in cm2 of the remaining square plate.
(Take π =227)

A.

253

B.

544

C.

636

D.

564

#### SOLUTION

Solution : B

Area of a square plate
=side2

Given length of the side of the square plate = 40 cm
Area of square plate
=402
=1600 cm2

Area of a circle
=π r2
There are 336 holes of radius 1 cm each.

Total area of circles
=336× 227 × 12
=1056 cm2

Remaining area =  [Area of square plate- Total area of circles]
=16001056
=544 cm2

Area of remaining square plate
=544 cm2

State true or false.
The area in cm2 of a sector of a circle with radius 6 cm and angle of the sector 60 is 20.

A.

True

B.

False

#### SOLUTION

Solution : B

Area of the sector of angle  θ = θ360 × πr2

A = 60360 × 3.14 × (62) = 18.85 cm2

The perimeter of a sector of circle with radius 5.7 m is 27.2 m. The length of the arc of sector (in m) is _______.

A.

13.4

B.

14.1

C.

16.9

D.

15.8

#### SOLUTION

Solution : D

Perimeter of sector = 2r + Length of the arc of sector

From the given data,

Length of arc of sector = 27.2 - 2 × 5.7

Length of arc of sector = 27.2 - 11.4 = 15.8 m.