The boundary of a circle is called its .................

1

(a)

circumference

Area of sector with angle q° and radius r is ...............

1

(a)

\(\frac { \pi { r }^{ 2 }{ q }^{ ° } }{ { 360 }^{ ° } } \)

If the ratio of the circumferences of two circles is 3 : 1, then find the ratio of their areas.

1

(a)

\(\frac { Circumference\ of\ one\ circle({ C }_{ 1 }) }{ Circumference\ of\ other\ circle({ C }_{ 2 }) } =\frac { 3 }{ 1 } \\ \Rightarrow \frac { { 2\pi r }_{ 1 } }{ { 2\pi r }_{ 2 } } =\frac { 3 }{ 1 } \Rightarrow \frac { { r }_{ 1 } }{ { r }_{ 2 } } =\frac { 3 }{ 1 } \)
Now, ratio of their areas \(=\frac { { \pi r }_{ 1 }^{ 2 } }{ { \pi r }_{ 2 }^{ 2 } } =\frac { { r }_{ 1 }^{ 2 } }{ { r }_{ 2 }^{ 2 } } =\left( \frac { { r }_{ 1 } }{ { r }_{ 2 } } \right) ^{ 2 }=\left( \frac { 3 }{ 1 } \right) ^{ 2 }=\frac { 9 }{ 1 } \)

For a race of 1540 m, find the number of rounds one have to take on a circular track of radius 3.5 m.

1

(a)

n \(\times\) Circumference of circular track = Length of race
\(\Rightarrow n\times (2\pi r)=1540\)
where, n = number of rounds taken on a circular track
\(\Rightarrow n\times 2\times \frac { 22 }{ 7 } \times 3.5=1540\quad \quad [\because r=3.5cm]\\ \Rightarrow n=\frac { 1540 }{ 2\times \frac { 22 }{ 7 } \times 3.5 } =\frac { 1540 }{ 22\times 2\times 3.5 } \times 7=70\)
Hence, the number of rounds is 70.

The sum of circumference and the radius of a circle is 51 cm. Find the radius of circle.

1

(a)

Because Sum of circumference of a circle and radius = 51 cm
\(\Rightarrow \ 2\pi r+r=51\\ \Rightarrow \ r(2\pi +1)=51\\ \Rightarrow \ r\left( 2\times \frac { 22 }{ 7 } +1 \right) =51\\ \Rightarrow\ r\times \frac { 51 }{ 7 } =51\Rightarrow r=7cm\)
Hence, the radius of the circle is 7 cm.

The area of a circle is 2464 cm ² . Find the diameter of circle.

1

(a)

56 cm

Two coins of diameter 2 cm and 4 cm respectively are kept one over the other as shown in the figure, find the area of the shaded ring shaped region in square cm.

1

(a)

\(\because\) Area of circle = \(\pi { r }^{ 2 }\)
\(\therefore\) Area of the shaded region = \(\pi \)(2)² -  \(\pi \)(1)²
4\(\pi \) - \(\pi \) = 3\(\pi \) sq units.

What is the name of a line which intersects a circle at two distinct points?

1

(a)

A line intersecting the circle at two distinct points is called a secant.

A thin wire is in the shape of a circle of radius 77 cm. It is bent into a square. Find the side of the square \(\left( Taking,\pi =\frac { 22 }{ 7 } \right) \)

1

(a)

Perimeter of the circle = Perimeter of square 
Let side of square be x cm.
2\(\pi\) r = 4x
\(\Rightarrow \quad 2\times \frac { 22 }{ 7 } \times 77=4x\)
\(\therefore \quad x=\frac { 2\times 22\times 11 }{ 4 } =121\)
Side of the square = 121 cm.

If the circumference of a circle increases from 4\(\pi\) to 8\(\pi\), then what about its area?

1

(a)

Circumference of the circle = 4\(\pi\) cm \(\Rightarrow\) r = 2 cm.
Increase circumference = 8\(\pi\) cm. \(\Rightarrow\) r = 4 cm.
Area of the 1^st circle = \(\pi\) \(\times\) (2)²= 4\(\pi\) cm²
Area of the new circle = \(\pi\) (4)²= 16\(\pi\) = 4 \(\times\) 4\(\pi\)
\(\therefore\)  Area of the new circle = 4 times the area of first circle.

Area of a circle is the portion enclosed under perimeter.

1

(a) False

(b) True

The perimeter of a semicircle is \(\pi r+2r\).

1

(a) False

(b) True

The diameter of a circle whose area is equal to the sum of the areas of the two circles of radill 24 cm and 7 cm is 31 cm.

1

(a) True

(b) False

OAB is a quadrant of a circle. Then, perimeter of OAB is 25 cm

1

(a) False

(b) True

Area of a major segment is always more than area of minor segment.

1

(a) False

(b) True

Tick the correct answer in the following: Area of a sector of angle \( \theta\) (in degree) of a circle with radius R is
(a) \({\theta \over 180^o}\times 2\pi R\) 
(b) \({\theta \over 180^o}\times \pi R^2\) 
(c) \({\theta \over 360^o}\times 2\pi R\) 
(d) \({\theta \over 720^o}\times 2\pi R^2\)

1

(a)

We know that area of sector of angle θ = \(\frac{\theta}{360^{\circ}} \times \pi R^{2}\)
Area of sector of angle P = \(\frac{p}{360^{\circ}}\left(\pi R^{2}\right)\)
\(=\left(\frac{p}{720^{\circ}}\right)\left(2 \pi R^{2}\right)\)
Hence, p/720 x 2Π R² is the correct answer.

A bicycle wheel makes 5000 revolutions in moving 11 km.  Find the diameter of the wheel. (use \(\pi ={22\over 7}\))

1

(a)

Distance covered in 5000 revolutions = 11 km.
Distance covered in 1 revolution
Distance covered in 1 revolution \(={11000\over 5000}m={11\over 5}m\)
Distance covered in 1 revolution = circumference of the wheel
\(\Rightarrow\)\(2\ \pi r ={11\over 5} \Rightarrow 2\times {22\over 7}\times r = {11\over 5}\)
\(\Rightarrow\)  \(r \Rightarrow {11\over 5}\times 7\times {1\over 2\times 22}={7\over 20}m\)
\(\therefore\) Diameter \(=2\times r=2\times {7\over 20}={7\over 10}\times 100 cm=70cm\)

In the given figure, the area of the shaded region between two concentric circle is \(286 cm^2.\) If the difference of the radii of the two circles is 7 cm, find the sum of their radii. [Use \(\pi ={22\over 7}\)]

1

(a)

Let radius of outer circle = \(R_1\) and radius of Inner circle = \(R_2\)
A.T.Q.  \(\pi R_1^2-\pi R^2_2=286\)
\(\Rightarrow\) \(\pi (R^2_1-R^2_2)=286 \Rightarrow {22\over 7}\times (R_1-R_2)(R_1+R_2)=286\)
\(\Rightarrow\) \({22\over 7}\times 7\times (R_1+R_2)=286 \Rightarrow R_1+R_2=13 cm\)

A horse is placed for grazing inside are rectangular field 70 m by 52 m and is tethered to one corner by a rope 21 m long. On how much area can it graze?

1

(a)

Area of portion that horse can graze area of the shaded portion.
Shaded portion is a sector of radius 21 m = length of the rope 
Angle of this sector = angle of the corners of the rectangle = 90°
Area the shaded portion that horse can graze 
= \(\frac { \theta }{ 360° } \times { \pi r }^{ 2 }=\frac { 90° }{ 360° } \times \frac { 22 }{ 7 } \times { (21) }^{ 2 }{ m }^{ 2 }\)
= \(\frac { 1 }{ 4 } \times \frac { 22 }{ 7 } \times 21\times 21 \ { m }^{ 2 } \ = \ \frac { 11 }{ 2 } \times 3\times 21 \ { m }^{ 2 }=346.5 \ \ { m }^{ 2 }\)

The diameter of a wheel of a bus is 90 cm which makes 315 revolutions per minute. Determine its speed in km/h. \([Use\ \pi = {22\over 7}]\)

1

(a)

d = 90 cm, circumference = \(\pi\)d = 90\(\pi\) cm = Distance covered in 1 revolution 
Number of revolutions per minute = 315 
ஃ Number of revolution per hour = 315 x 60 cm 
Distance covered in 1 hour = 90\(\pi\) x 315 x 60 cm 
= 90 x \({22\over7}\times{315\times60\over100000}km={9\times22\times45\times6\over1000}km\)
∴ Speed of the bus = 53.46 Km/h

The circumference of the wheel of an engine of a train is \(4{2\over7}m\).  If it makes seven revolutions in 4 seconds then find the speed of the train. \([use \ \pi={22\over 7}]\)

1

(a)

Distance covered by train in 1 revolution 
= Circumference of the wheel 
= 4\(2\over7\) m = \(30\over7\)3
ஃ Distance covered in 4 seconds 
= 7 x \(30\over7\)=30 m
ஃ Distance covered in 1 second 
= \(30\over4\) = 7.5 m
ஃ Speed = 7.5 m/s.

The circumference of a circular field and perimeter of a square field are equal.If the area of the square field is 484m², then find the length of the diameter of the circular field.

1

(a)

Area of square field = 484 m² 
∴ Side x Side = 22 x 22 
⇒ Side = 22 m 
∴ Perimeter of square field = 4 X 22 = 88 m 
Since circumference of circular field = Perimeter of square field
\(\Rightarrow \pi d=88\Rightarrow \frac { 22 }{ 7 } d=88\)
\(\Rightarrow d=88\times \frac { 7 }{ 22 } =28m\)
Hence, the length of the diameter of the circular field is 28 m.

If area of circle is 301.84 cm², then find the radius of circle.

1

(a)

9.8 cm

A chord of a circle of radius 28 cm subtends an angle 45° at centre of the circle, find the area of the minor segment.

1

(a)

30.77 cm²

In the fig., find the perimeter of shaded region where ADC, AEB and BFC are semicircles on diameters AC, AB and BC respectively.

3

(a)

In the fig., 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.

3

(a)

In given figure, an equilateral triangle has been inscribed in a circle of radius 6 cm.  Find the area of the shaded region.  \([Use\ \pi=3.14]\) 

3

(a)

In fig., APB and AQO are semicircle, and AO = OB. If the perimeter of the figure is 40 cm, find the area of the shaded region. [ Use \(\pi\) = 22 / 7]

3

(a)

How long will Onkar take to run \(12{1\over2}\) rounds at the rate of 3.3km/h, around a circular track of area 5544m²?

3

(a)

An elastic belt is placed round the rim of a pulley of radius 5 cm. One point on the belt is pulled directly away from the centre O of the pulley until it is at P, 10 cm away from O. Find the length of the belt that is in contact with the rim of the pulley. Also find the shaded area . ( Use \(\pi =3.14,\sqrt { 3 } =1.73)\) 

5

(a)