CBSE 10th Maths -  Areas Related to Circles- Exercise 12.2

Solution:

It is given that the angle of the sector is 60°

We know that the area of sector = $(\dfrac{θ}{360°})× \pi r^2$

area of the sector with angle 60° = $(\dfrac{60°}{360°})× \pi r^2$ $cm^2$

= $(\dfrac{36}{6})\pi$ $cm^2$

= $6× \dfrac{22}{7}$ $cm^2$ = $\dfrac{132}{7}$ $cm^2$

Solution:

Circumference of the circle, C = 22 cm (given)

It should be noted that a quadrant of a circle is a sector which is making an angle of 90°

Let the radius of the circle = r

As C = 2$\pi$r = 22,

R = $\dfrac{22}{2} \pi$ cm = $\dfrac{7}{2}$ cm

area of the quadrant = $(\dfrac{θ}{360°}) × \pi r^2$

Here, θ = 90°

So, A = $(\dfrac{90°}{360°}) × \pi r^2$ $cm^2$

= $(\dfrac{49}{16}) \pi$ $cm^2$

= $\dfrac{77}{8}$ $cm^2$ = 9.6 $cm^2$

Solution:

Length of minute hand = radius of the clock (circle)

Radius (r) of the circle = 14 cm (given)

Angle swept by minute hand in 60 minutes = 360°

So, the angle swept by the minute hand in 5 minutes = 360° × $\dfrac{5}{60}$ = 30°

We know,

Area of a sector = $(\dfrac{θ}{360°}) × \pi r^2$

Now, the area of the sector making an angle of 30° = $(\dfrac{30°}{360°}) × \pi r^2$ $cm^2$

= $(\dfrac{1}{12}) × \pi14^2$

= $(\dfrac{49}{3})×(\dfrac{22}{7})$ $cm^2$

= $\dfrac{154}{3}$ $cm^2$

Solution:

Here, AB is the chord which is subtending an angle 90° at the centre O

It is given that the radius (r) of the circle = 10 cm

(i) minor segment

Area of minor sector = $(\dfrac{90}{360°})× \pi r^2$

= $(\dfrac{1}{4})×(\dfrac{22}{7})×10^2$

Or, the Area of the minor sector = 78.5 $cm^2$

Also, the area of ΔAOB = $\dfrac{1}{2}$×OB×OA

Here, OB and OA are the radii of the circle, i.e., = 10 cm

So, the area of ΔAOB = $\dfrac{1}{2}$×10×10

= 50 $cm^2$

Now, area of minor segment = area of the minor sector – the area of ΔAOB

= 78.5 – 50

= 28.5 $cm^2$

(ii) major sector

Area of major sector = Area of the circle – Area of he minor sector

= $(3.14×10^2)$-78.5

= 235.5 $cm^2$

Solution:

Given,

Radius = 21 cm

θ = 60°

(i) the length of the arc

Length of an arc = $\dfrac{θ}{360°}×Circumference(2 \pi r)$

Length of an arc AB = $(\dfrac{60°}{360°})×2×(\dfrac{22}{7})×21$

= $(\dfrac{1}{6})×2×(\dfrac{22}{7})×21$

Or Arc AB Length = 22cm

(ii) area of the sector formed by the arc

It is given that the angle subtended by the arc = 60°

So, the area of the sector making an angle of 60° = $(\dfrac{60°}{360°})×π r^2 cm^2$

= $\dfrac{441}{6}×\dfrac{22}{7} cm^2$

Or, the area of the sector formed by the arc APB is 231 $cm^2$

(iii) area of the segment formed by the corresponding chord

Area of segment APB = Area of sector OAPB – Area of ΔOAB

Since the two arms of the triangle are the radii of the circle and thus are equal, and one angle is 60°, ΔOAB is an equilateral triangle. So, its area will be $(\dfrac{\sqrt 3}{4})×a^2$ sq. Units.

The area of segment APB = $231-(\dfrac{\sqrt 3}{4})×(OA)^2$

= $231-(\dfrac{\sqrt 3}{4})×21^2$

Or, the area of segment APB = $[231-\dfrac{441× \sqrt 3}{4}] cm^2$

Solution:

Given,

Radius = 15 cm

θ = 60°

So,

Area of sector OAPB = $(\dfrac{60°}{360°}) × πr^2$ $cm^2$

= $\dfrac{225}{6} π$ $cm^2$

Now, ΔAOB is equilateral as two sides are the radii of the circle and hence equal and one angle is 60°

So, Area of ΔAOB = $(\dfrac{\sqrt3}{4}) ×a^2$

Or, $(\dfrac{\sqrt3}{4}) ×15^2$

Area of ΔAOB =$ 97.31$ $cm^2$

Now, the area of minor segment APB = Area of OAPB – Area of ΔAOB

Or, the area of minor segment APB = $((\dfrac{225}{6})π – 97.31)$ $cm^2$ = 20.43 $cm^2$

And,

Area of major segment = Area of the circle – Area of the segment APB

Or, area of major segment = $(π×15^2)$ – 20.4 = 686.06 $cm^2$

Solution:

Radius, r = 12 cm

Now, draw a perpendicular OD on chord AB, and it will bisect chord AB.

So, AD = DB

Now, the area of the minor sector = $(\dfrac{θ}{360°})×πr^2$

= $(\dfrac{120}{360})×(\dfrac{22}{7})×12^2$

= 150.72 $cm^2$

Consider the ΔAOB,

$\angle OAB$= 180°-(90°+60°) = 30°

Now, cos 30° = $\dfrac{AD}{OA}$

$\dfrac{\sqrt3}{2} = \dfrac{AD}{12}$

Or, AD = $6\sqrt3$ cm

We know OD bisects AB. So,

AB = 2×AD = $12\sqrt3$ cm

Now, sin 30° = $\dfrac{OD}{OA}$

Or, $\dfrac{1}{2} = \dfrac{OD}{12}$

OD = 6 cm

So, the area of ΔAOB = $\dfrac{1}{2}$ × base × height

Here, base = AB = $12\sqrt3$ and

Height = OD = 6

So, area of ΔAOB = $\dfrac{1}{2}×12\sqrt3×6 = 36\sqrt3$ cm = 62.28 $cm^2$

area of the corresponding Minor segment = Area of the Minor sector – Area of ΔAOB

= 150.72 $cm^2$– 62.28 $cm^2$ = 88.44 $cm^2$

Solution:

As the horse is tied at one end of a square field, it will graze only a quarter (i.e. sector with θ = 90°) of the field with a radius 5 m.

Here, the length of the rope will be the radius of the circle, i.e. r = 5 m

It is also known that the side of the square field = 15 m

(i) the area of that part of the field in which the horse can graze.

Area of circle = $πr^2 = \dfrac{22}{7} × 5^2$ = 78.5 $m^2$

Now, the area of the part of the field where the horse can graze = $\dfrac{1}{4}$ (the area of the circle)

= $\dfrac{78.5}{4}$ = 19.625 $m^2$

(ii) the increase in the grazing area if the rope were10 m long instead of 5 m. (Use p = 3.14)

If the rope is increased to 10 m,

Area of circle will be = $πr^2 =\dfrac{22}{7}×10^2$ = 314 $m^2$

Now, the area of the part of the field where the horse can graze = $\dfrac{1}{4}$ (the area of the circle)

= $\dfrac{314}{4}$ = 78.5 $m^2$

increase in the grazing area = 78.5 $m^2 $ – 19.625 $m^2 $ = 58.875 $m^2$

Solution:

Diameter (D) = 35 mm

Total number of diameters to be considered= 5

Now, the total length of 5 diameters that would be required = 35×5 = 175

Circumference of the circle = 2πr

Or, C = $πD = \dfrac{22}{7}$×35 = 110

Area of the circle = $πr^2$

Or, A = $(\dfrac{22}{7})×(\dfrac{35}{2})^2 = \dfrac{1925}{2} mm^2$

(i) the total length of the silver wire required

Total length of silver wire required = Circumference of the circle + Length of 5 diameter

= 110+175 = 285 mm

(ii) the area of each sector of the brooch

Total Number of sectors in the brooch = 10

So, the area of each sector = $\dfrac{total \: area \: of \: the \: circle}{number \: of \: sectors}$

Area of each sector = $(\dfrac{1925}{2}) × \dfrac{1}{10} = \dfrac{385}{4}$ $mm^2$

Solution:

The radius (r) of the umbrella when flat = 45 cm

So, the area of the circle (A) = $πr^2 = (\dfrac{22}{7})×(45)^2 =6364.29 cm^2$

Total number of ribs (n) = 8

The area between the two consecutive ribs of the umbrella = $\dfrac{A}{n} = \dfrac{6364.29}{8} cm^2$

Or, The area between the two consecutive ribs of the umbrella = 795.5 $cm^2$