# Free Areas Related to Circles 01 Practice Test - 10th Grade

### Question 1

On a square handkerchief of side 50 cm, seven non-intersecting circles designs of each radius 7 cm are made. Find the area of the remaining portion of the handkerchief in cm2.

#### SOLUTION

Solution :Area of a square = side2

Side of the square = 50 cm

Area of square Handkerchief = 502 = 2500 cm2

Area of circles = π×r2

Radius of the circle = 7 cmTotal area of circles = 7 x 227 x 72 = 1078cm2

Remaining area = 2500 - 1078 = 1422 cm2

### Question 2

The given figure is a sector of a circle of radius 20 cm. Find the perimeter of the sector.

(Take π = 3.14)

55.25 cm

60.93 cm

65.48 cm

70.17

#### SOLUTION

Solution :B

The circumference i.e , perimeter of a sector of angle P∘ of a circle with radius R is given by

P∘360 ×2πr +2R=P∘360 × 2π R+2R

=60∘360 × 2π (20)+2(20)

= 20.93 + 40= 60.93 cm

### Question 3

The area of a sector (in cm2 )of a circle with radius 6 cm if angle of the sector is 70∘ will be

#### SOLUTION

Solution :Given that radius r = 6 cm

θ = 70∘

Area of sector of an angle θ with radius r

=θ360 × π × (r)2

= 70360 × 227 × 62

= 22 cm2

### Question 4

ABCD is a flower bed. If OA = 21m and OC = 14m. Find the area of the bed in m2.

180.30

921

190

192.50

#### SOLUTION

Solution :D

Area of a sector with angle θ is given by =θ360° × π × r2 ( where θ is the angle made by the sector )

Required area = Area of sector OABO – Area of sector OCDOThe angle formed by the sector OABO and sector OCDO = 90°

=90360×227×(21)2–90360×227×(14)2

=14×227×{(21)2–(14)2}

=14×227×{(35)(7)} [a2−b2=(a+b)(a−b)]

=192.50 m2

### Question 5

State true or false.

If the radius of a circle is 7√π cm, then the area of the circle in cm2 is 49π2.

True

False

#### SOLUTION

Solution :B

A = πr2

Given r = 7√π cmA=π×7√π×7√π

=49 cm2

### Question 6

A circle is inserted inside a square, such that the edge of the circle touches the sides of the square.If X = 14 cm, the area of the shaded portion will be

#### SOLUTION

Solution :

Here we are asked to find the area of the shaded area. The area of the shaded area should be the difference between the area of the square and the area of the circle.

Here we are given the side of the square, x = 14 cm

The area of the square is = side2 = x2 = 142= 196cm2

^{ }Therefore the radius of the circle will be, r =d2

r = 142 = 7 cm

Area of the circle,

A=π×r2 (use π = 22/7)

= (227)×72 = 154 cm2

Area of the shaded portion = Area of the square - Area of the circle

= 196 - 154 =42 cm2

Therefore the area of the shaded portion will be 42 cm2

### Question 7

State true or false.

Area of a circle is inversely proportional to the radius of circle.

True

False

#### SOLUTION

Solution :B

Area of a circle is directly proportional to the square of the radius of circle.

### Question 8

An umbrella has 8 ribs which are equally spaced. Assuming the umbrella to be a flat circle of radius 42 cm, find the area in cm2 between the two consecutive ribs of the umbrella.

#### SOLUTION

Solution :Radius = 42 cm

8 ribs implies angle subtend between consecutive ribs = 3608 = 45∘ .

Area between consecutive ribs = θ360×π×(radius)2

45360 × 227 × 422 = 693 cm2

### Question 9

Tick the correct answer in the following:

Area of a sector of angle p (in degrees) of a circle with radius R is

#### SOLUTION

Solution :D

Area of a sector of angle p=p360×πR2=p360×22×πR2=p720×2πR2

### Question 10

OACB is a quadrant of a circle with centre O and radius 7 cm. It OD = 4cm, then find area of shaded region.

#### SOLUTION

Solution :For the quadrant θ = 90°

Area of the quadrant = θ360°×π(r)2

Area of quadrant OACB = 90360 × 227 (7)2 = 772Area of shaded region = Area of quadrant OACB - area of ΔOAD

= 772 - 12 × 7 × 4

= 24.5 cm2