Question 22:

The area of the sector of a circle of radius 10.5 cm is 69.3 cm². Find the central angle of the sector.

Answer:

Given:
Area of the sector = 63 cm²
Radius = 10.5 cm

Now,
Area of the sector $= \frac{{πr}^{2} θ}{360}$
$\Rightarrow 69.3 = \frac{22}{7} \times 10.5 \times 10.5 \times \frac{θ}{360} \Rightarrow θ = \frac{69.3 \times 7 \times 360}{22 \times 10.5 \times 10.5} \Rightarrow θ = 72^{\circ}$

∴ Central angle of the sector = $72^{\circ}$

Page No 820:

Question 23:

The perimeter of a certain sector of a circle of radius 6.5 cm in 31 cm. Find the area of the sector.

Answer:

Given:
Radius = 6.5 cm

Let O be the centre of the circle with radius 6.5 cm and OACBO be its sector with perimeter 31 cm.
Thus, we have:
OA + OB + arc AB = 31 cm

$\Rightarrow 6.5 + 6.5 + a r c A B = 31 \Rightarrow a r c A B = 31 - 13 \Rightarrow a r c A B = 18 cm$

Now,
Area of the sector OACBO = $\frac{1}{2} \times Radius \times Arc$
$= \frac{1}{2} \times 6.5 \times 18 = 58.5 {cm}^{2}$

Page No 820:

Question 24:

The radius of a circle is 17.5 cm. Find the area of the sector enclosed by two radii and an arc 44 cm in length.

Answer:

Given:
Radius = 17.5 cm
Length of the arc = 44 cm

Now,
Length of the arc $= \frac{2 πrθ}{360}$

$\Rightarrow 44 = 2 \times \frac{22}{7} \times 17.5 \times \frac{θ}{360} \Rightarrow θ = \frac{44 \times 7 \times 360}{44 \times 17.5} \Rightarrow θ = 144^{\circ}$
Also,
Area of the sector = $\frac{{πr}^{2} θ}{360}$

$= \frac{22}{7} \times 17.5 \times 17.5 \times 144 = 385 {cm}^{2}$

Page No 820:

Question 25:

Two circular pieces of equal radii and maximum area, touching each other are cut out from a rectangular cardboard of dimensions 14 cm ⨯ 7 cm. Find the area of the remaining cardboard. [CBSE 2013]

Answer:

We know that we can cut two circular pieces of equal radii and maximum area from the rectangular cardboard whose diameter is equal to the width of the rectangular cardboard.
∴ Radii of two circuar pieces = Half of the width of the rectangular cardboard = 3.5 cm
Now,
Area of remaining cardboard = Area of rectangular cardboard − 2 ⨯ Area of circular piece having radius 3.5 cm
$= 14 \times 7 - 2 (\frac{22}{7} \times 3.5 \times 3.5) = 98 - 77 = 21 {cm}^{2}$
Hence, the area of the remaining cardboard is 21 cm²

Page No 820:

Question 26:

In the given figure, ABCD is a square of side 4 cm. A quadrant of a circle of radius 1 cm is drawn at each vertex of the square and a circle of diameter 2 cm is also drawn. Find the area of the shaded region.
[Use $π$ = 3.14.]

Answer:

Area of the square ABCD = $(Side)^{2}$
            $= 4^{2} = 16 {cm}^{2}$

Area of the circle = ${πr}^{2}$
Radius = 1 cm
$Area = 3.14 \times (1)^{2} = 3.14 {cm}^{2}$

Area of the quadrant of one circle = $\frac{1}{4} {πr}^{2}$
         $= \frac{1}{4} \times 3.14 \times 1^{2} = 0.785 {cm}^{2}$

Area of the quadrants of four circles = $0.785 \times 4 = 3.14 {cm}^{2}$
Area of the shaded region = Area of the square $-$ Area of the circle $-$ Area of the quadrants of four circles
   = $16 - 3.14 - 3.14 = 9.72 {cm}^{2}$

Page No 820:

Question 27:

Answer:

Construction: Join OB

In right triangle AOB
OB² = OA² + AB²
= 20² + 20²
= 400 + 400
= 800
∴ OB² = 800
Area of the shaded region = Area of quadrant OPBQ − Area of Square OABC
$= \frac{1}{4} π {(OB)}^{2} - {(OA)}^{2} = \frac{1}{4} \times 3.14 \times 800 - 400 = 628 - 400 = 228 {cm}^{2}$
Hence, the area of the shaded region is 228 cm².

Page No 823:

Question 40:

In the given figure, APB and AQO are semicircles and AO = OB. If the perimeter of the figure is 40 cm, find the area of the shaded region. [CBSE 2015]

Answer:

Permieter of shaded region = Length of the arc AQO + Length of the arc APB + Length of OB
$\Rightarrow 40 = \frac{1}{2} \times 2 π (\frac{AO}{2}) + \frac{1}{2} \times 2 π (OB) + OB \Rightarrow 40 = \frac{11}{7} AO + \frac{22}{7} OB + OB \Rightarrow 40 = \frac{11}{7} OB + \frac{22}{7} OB + OB [∵ AO = OB] \Rightarrow 40 = \frac{40}{7} OB \Rightarrow OB = 7 cm$
Area of the shaded portion = Area of semicircle AQO + Area of semicircle APB
$= \frac{1}{2} π {(\frac{7}{2})}^{2} + \frac{1}{2} π {(7)}^{2} = \frac{1}{2} \times \frac{22}{7} \times {(\frac{7}{2})}^{2} + \frac{1}{2} \times \frac{22}{7} \times {(7)}^{2} = 96.25 {cm}^{2}$
Hence, the area of the shaded portion is 96.25 cm².

Page No 823:

Question 41:

Answer:

Question 13:

The length of an arc of a circle, subtending an angle of 54° at the centre, is 16.5 cm. Calculate the radius, circumference and area of the circle.

Answer:

Page No 832:

Question 23:

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

Answer:

Radius of the quadrant of the circle = 21 m
The shaded portion shows the part of the field the horse can graze.

Area of the grazed field = Area of the quadrant OPQ
                                 $= \frac{1}{4} of the circle having radius OP$
   $= \frac{1}{4} {πr}^{2} = \frac{1}{4} \times \frac{22}{7} \times 21 \times 21 = 346.5 m^{2}$

Total area of the field = $70 \times 52 = 3640 m^{2}$

Area left ungrazed = Area of the field $-$ Area of the grazed field
                               = $3640 - 346.5 = 3293.5 m^{2}$

Page No 832:

Question 24:

A horse is tethered to one corner of a field which is in the shape of an equilateral triangle of side 12 m. If the length of the rope is 7 m, find the area of the field which the horse cannot graze. Write the answer correct to 2 places of decimal.

Answer:

Side of the equilateral triangle = 12 m
Area of the equilateral triangle = $\frac{\sqrt{3}}{4} \times (Side)^{2}$
$= \frac{\sqrt{3}}{4} \times 12 \times 12 = 62.28 m^{2}$
Length of the rope = 7 m
Area of the field the horse can graze is the area of the sector of radius 7 m .Also, the angle subtended at the centre is 60 $°$

= $\frac{θ}{360} \times π r^{2}$

$= \frac{60}{360} \times \frac{22}{7} \times (7)^{2} = 25.67 m^{2}$

Area of the field the horse cannot graze = Area of the equilateral triangle $-$ Area of the field the horse can graze
$= 62.28 - 25.67 = 36.61 m^{2}$

Page No 832:

Question 25:

Four cows are tethered at the four corners of a square field of side 50 m such that each can graze the maximum unshared area. What area will be left ungrazed?

Answer:

Each cow can graze a region that cannot be accessed by other cows.
∴ Radius of the region grazed by each cow = $\frac{50}{2} = 25 m$

Area that each cow grazes = $\frac{1}{4} \times π \times r^{2}$
$= \frac{1}{4} \times 3.14 \times 25 \times 25 = 490.625 {cm}^{2}$

Total area grazed = $4 \times 490.625 = 1963.49 m^{2}$
$Area of the square = (Side)^{2} = 50^{2} = 2500 {cm}^{2}$

Now,
Area left ungrazed = Area of the square $-$ Grazed area
= $2500 - 1963.49 = 536.51 m^{2}$

Page No 832:

Question 26:

In the given figure, OPQR is a rhombus, three of whose vertices lie on a circle with centre O. If the area of the rhombus is $32 \sqrt{3}$ , find the radius of the circle.

Answer:

Let the radius of the inscribed circle be r cm.
Given:
Area of the circle = 154 ${cm}^{2}$
We know:
Area of the circle = $π r^{2}$
$\Rightarrow 154 = \frac{22}{7} r^{2} \Rightarrow \frac{154 \times 7}{22} = r^{2} \Rightarrow r^{2} = 49 \Rightarrow r = 7$
In a triangle, the centre of the inscribed circle is the point of intersection of the medians and altitudes of the triangle. The centroid divides the median of a triangle in the ratio 2:1.
Here,
AO:OD = 2:1

Now,
Let the altitude be h cm.
We have:
$∠ A D B = 90^{\circ} O D = \frac{1}{3} A D O D = \frac{h}{3}$
$\Rightarrow h = 3 r \Rightarrow h = 21$

Let each side of the triangle be a cm.
$In the right - angled ∆ ADB, we have : A B^{2} = A D^{2} + D B^{2} a^{2} = h^{2} + {(\frac{a}{2})}^{2} 4 a^{2} = 4 h^{2} + a^{2} 3 a^{2} = 4 h^{2} a^{2} = \frac{4 h^{2}}{3} a = \frac{2 h}{\sqrt{3}} a = \frac{42}{\sqrt{3}}$
∴ Perimeter of the triangle = 3a
$= 3 \times \frac{42}{\sqrt{3}} = \sqrt{3} \times 42 = 72.66 cm$

Page No 833:

Question 36:

Four equal circles are described about the four corners of a square so that each touches two of the others, as shown in the figure. Find the area of the shaded region, if each side of the square measures 14 cm.

Answer:

Side of the square = 14 cm
Radius of the circle $= \frac{14}{2}$ = 7 cm
Area of the quadrant of one circle = $\frac{1}{4} {πr}^{2}$
            $= \frac{1}{4} \times \frac{22}{7} \times 7 \times 7 = 38.5 {cm}^{2}$
Area of the quadrants of four circles = $38.5 \times 4 = 154 {cm}^{2}$

Now,
Area of the square = ${(Side)}^{2}$
   $= 14^{2} = 196 {cm}^{2}$

Area of the shaded region = Area of the square $-$ Area of the quadrants of four circles
           = 196 $-$ 154
             = 42 cm²

Page No 833:

Question 37:

Four equal circles, each of radius 5 cm, touch each other, as shown in the figure. Find the area included between them.

Answer:

Radius = 5 cm
AB = BC = CD = AD = 10 cm
All sides are equal, so it is a square.
Area of a square = ${Side}^{2}$
Area of the square = $10^{2} = 100 {cm}^{2}$
Area of the quadrant of one circle = $\frac{1}{4} {πr}^{2}$

$= \frac{1}{4} \times \frac{22}{7} \times 5 \times 5 = 19.64 {cm}^{2}$
Area of the quadrants of four circles = $19.64 \times 4 = 78.57 {cm}^{2}$
Area of the shaded portion = Area of the square $-$ Area of the quadrants of four circles
$= 100 - 78.57 = 21.43 {cm}^{2}$

Page No 833:

Question 38:

Four equal circles, each of radius a units, touch each other. Show that the area between them is $(\frac{6}{7} a^{2})$ sq units.

Answer:

When four circles touch each other, their centres form the vertices of a square. The sides of the square are 2a units.

Area of the square = ${(2 a)}^{2} = 4 a^{2} sq . units$

Area occupied by the four sectors
$= 4 \times \frac{90}{360} \times π \times a^{2} = π a^{2} sq . units$

Area between the circles = Area of the square $-$ Area of the four sectors
$= (4 - \frac{22}{7}) a^{2} = \frac{6}{7} a^{2} sq . units$

Page No 833:

Question 39:

Three equal circles, each of radius 6 cm, touch one another as shown in the figure. Find the area of enclosed between them.

Answer:

Join ABC. All sides are equal, so it is an equilateral triangle.
Now,
Area of the equilateral triangle = $\frac{\sqrt{3}}{4} \times {Side}^{2}$

   $= \frac{1.73}{4} \times 12 \times 12 = 62.28 {cm}^{2}$

$Area of the arc of the circle = \frac{60}{360} {πr}^{2} = \frac{1}{6} {πr}^{2} = \frac{1}{6} \times \frac{22}{7} \times 6 \times 6 = 18.86 {cm}^{2} Area of the three sectors = 3 \times 18.86 = 56.57 {cm}^{2}$


Area of the shaded portion = Area of the triangle $-$ Area of the three quadrants

$= 62.28 - 56.57 = 5.71 {cm}^{2}$

Page No 833:

Question 40:

If three circles of radius a each, are drawn such that each touches the other two, prove that the area included between them is equal to $\frac{4}{25} a^{2} .$

Answer:

When three circles touch each other, their centres form an equilateral triangle, with each side being 2a.

Area of the triangle = $\frac{\sqrt{3}}{4} \times 2 a \times 2 a = \sqrt{3} a^{2}$

Total area of the three sectors of circles = $3 \times \frac{60}{360} \times \frac{22}{7} \times a^{2} = \frac{1}{2} \times \frac{22}{7} \times a^{2} = \frac{11}{7} a^{2}$

Area of the region between the circles = $Area of the triangle - Area of the three sectors$
$= (\sqrt{3} - \frac{11}{7}) a^{2} = (1.73 - 1.57) a^{2} = 0.16 a^{2} = \frac{4}{25} a^{2}$

Page No 835:

Question 49:

The perimeter of the quadrant of a circle is 25 cm. Find its area. [CBSE 2012]

Answer:

Let the radius of the circle be r
Now, Perimeter of quadrant = $\frac{1}{4} (2 π r) + 2 r$
$\Rightarrow 25 = \frac{1}{2} \times \frac{22}{7} \times r + 2 r \Rightarrow 25 = \frac{25 r}{7} \Rightarrow r = 7 cm$
Area of quadrant = $\frac{1}{4} π r^{2} = \frac{1}{4} \times \frac{22}{7} \times 7 \times 7 = 38.5 {cm}^{2}$
Hence, the area of quadrant is 38.5 cm²

Page No 835:

Question 50:

A chord of a circle of radius 10 cm subtends a right angle at the centre. Find the area of the minor segment. [Use π = 3.14.] [CBSE 2012]

Answer:

Area of minor segment = Area of sector AOBC − Area of right triangle AOB
$= \frac{θ}{360 °} π {(OA)}^{2} - \frac{1}{2} \times OA \times OB = \frac{90 °}{360 °} \times 3.14 {(10)}^{2} - \frac{1}{2} \times 10 \times 10 = 78.5 - 50 = 28.5 {cm}^{2}$
Hence, the area of minor segment is 28.5 cm²

Page No 835:

Question 51:

The radius of a circular garden is 100 m. There is a road 10 m wide, running all around it. Find the area of the road and the cost of levelling
it at Rs 20 per m². [Use π = 3.14] [CBSE 2011]

Answer:

Area of the road = Area of outer circle − Area of inner circle
$= π R^{2} - π r^{2} = π [{(110)}^{2} - {(100)}^{2}] = 3.14 \times 2100 = 6594 m^{2}$
Cost of levelling the road = Area of the road ⨯ Rate
= 6594 ⨯ 20
= Rs 131880

Page No 835:

Question 52:

The area of an equilateral triangle is $49 \sqrt{3}$ cm² . Taking each angular point as centre, circles are drawn with radius equal to half the length of the side of the triangle. Find the area of the triangle not included in the circles. [Take π = 1.73] [CBSE 2009]

Answer:

$Area of equilateral triangle = 49 \sqrt{3} \Rightarrow \frac{\sqrt{3}}{4} \times {(Side)}^{2} = 49 \sqrt{3} \Rightarrow {(Side)}^{2} = 7^{2} \times 2^{2} \Rightarrow Side = 14 cm$

Radius of the circle = Half of the side of the triangle = 7 cm
Area of triangle not included in the circle = Area of triangle − Area of 3 sectors having central angle 60^∘
$= 49 \sqrt{3} - 3 \times \frac{60 °}{360 °} \times \frac{22}{7} \times 7 \times 7 = 84.77 - 77 = 7.77 {cm}^{2}$
Hence, the required area is 7.77 cm²

Page No 835:

Question 53:

A child draws the figure of an aeroplane as shown. Here, the wings ABCD and FGHI are parallelograms, the tail DEF is an isosceles triangle, the cockpit CKI is a semicircle and CDFI is a square. In the given figure, BP ⊥ CD, HQ ⊥ FI and EL ⊥ DF. If CD = 8 cm, BP = HQ = 4 cm and DE = EF = 5 cm, find the area of the whole figure.

Answer:

CD = 8 cm
BP = HQ = 4 cm
DE = EF = 5 cm
Area of the parallelogram ABCD = $B \times H$
           $= 8 \times 4 = 32 sq . cm$

Area of parallelogram FGHI = $B \times H$
   $= 8 \times 4 = 32 sq . cm$
Area of the square = ${Side}^{2}$
   = $8^{2} = 64 sq . cm$

In $∆$ ELF, we have:
${EL}^{2} = 5^{2} - 4^{2} {EL}^{2} = 9 EL = 3 cm$

Area of $△$ DEF = $\frac{1}{2} \times B \times H$
$= \frac{1}{2} \times 8 \times 3 = 12 sq . cm$

Area of the semicircle = $\frac{1}{2} {πr}^{2}$

          $= \frac{1}{2} \times \frac{22}{7} \times 16 = 25.12 sq . cm$
∴ Total Area = Area of the parallelogram ABCD + Area of the parallelogram FGHI + Area of the triangle DEF + Area of the semicircle CKI + Area of the square
Total Area = 165.12 cm²

Page No 835:

Question 54:

A circular disc of radius 6 cm is divided into three sectors with central angles 90°,120° and 150°. What part of the whole circle is the sector with central angle 150°? Also, calculate the ratio of the areas of the three sectors.

Answer:

Area of sector having central angle 150° = $= \frac{150 °}{360 °} π {(6)}^{2} = \frac{5}{12} \times Area of circular disc$

Now, Area of sector having central angle 90° : Area of sector having central angle 120° : Area of sector having central angle 150°
$= \frac{90 °}{360 °} π {(6)}^{2} : \frac{120 °}{360 °} π {(6)}^{2} : \frac{150 °}{360 °} π {(6)}^{2} = \frac{1}{4} : \frac{1}{3} : \frac{5}{12} = 3 : 4 : 5$

Page No 835:

Question 55:

A round table cover has six equal designs as shown in the given figure. If the radius of the cover is 35 cm, then find the total area of the design.

Answer:

Join each vertex of the hexagon to the centre of the circle.

The hexagon is made up of six triangles.

$Total area of the design = Area of the circle - Area of six triangles = [\frac{22}{7} \times 35 \times 35] - [6 \times (\frac{\sqrt{3}}{4} \times 35 \times 35)] = (35 \times 35) [\frac{22}{7} - (\frac{6 \times 1.732}{4})] = 1225 (\frac{22}{7} - 2.598) = 1225 \times 0.542 = 663.95 {cm}^{2}$

A chord of a circle of radius 10 cm subtends a right angle at the centre. The area of the minor segments (given, π = 3.14) is
(a) 32.5 cm²
(b) 34.5 cm²
(c) 28.5 cm²
(d) 30.5 cm²

Answer:

Area of minor segment = Area of sector AOBC − Area of right triangle AOB
$= \frac{90 °}{360 °} π {(OA)}^{2} - \frac{1}{2} \times OA \times OB = \frac{1}{4} \times 3.14 \times {(10)}^{2} - \frac{1}{2} \times 10 \times 10 = 78.5 - 50 = 28.5 {cm}^{2}$
Hence, the correct answer is option (c).

Page No 848:

Question 25:

In a circle of radius 21 cm, an arc subtends an angle of 60° at the centre. The length of the arc is
(a) 21 cm
(b) 22 cm
(c) 18.16 cm
(d) 23.5 cm

Answer:

We have $r = 21 cm and θ = 60 °$
Length of arc = $\frac{θ}{360^{°}} \times 2 π r = \frac{60 °}{360 °} \times 2 \times \frac{22}{7} \times 21 = 22 cm$
Hence, the correct answer is option (b)

Page No 848:

Question 26:

In a circle of radius 14 cm, an arc subtends an angle of 120° at the centre. If $\sqrt{3}$ = 1.73 then the area of the segment of the circle is
(a) 120.56 cm²
(b) 124.63 cm²
(c) 118.24 cm²
(d) 130.57 cm²

Answer:

Radius of the circle, r = 14 cm
Draw a perpendicular OD to chord AB. It will bisect AB.
∠A = 180° − (90° + 60°) = 30°

$\cos 30 ° = \frac{AD}{OA} \Rightarrow \frac{\sqrt{3}}{2} = \frac{AD}{14} \Rightarrow AD = 7 \sqrt{3} \Rightarrow AB = 2 \times AD = 14 \sqrt{3} cm \sin 30 ° = \frac{OD}{OA} \Rightarrow \frac{1}{2} = \frac{OD}{14} \Rightarrow OD = 7 cm$

Area of minor segment = Area of sector OAPB − Area of triangle AOB
$= \frac{θ}{360 °} π {(OA)}^{2} - \frac{1}{2} \times OD \times AB = \frac{120 °}{360 °} \times \frac{22}{7} {(14)}^{2} - \frac{1}{2} \times 7 \times 14 \sqrt{3} = 205.33 - 84.77 = 120.56 {cm}^{2}$
Hence, the correct answer is option (a).

Page No 852:

Question 1:

In the given figure, a square OABC has been inscribed in the quadrant OPBQ. If OA = 20 cm, then the area of the shaded region is

(a) 214 cm²
(b) 228 cm²
(c) 242 cm²
(d) 248 cm²

Answer:

(b) 228 cm²
Join OB.
Now, OB is the radius of the circle.
$We have : O B^{2} = O A^{2} + A B^{2} [By Pythagoras' theorem] \Rightarrow O B^{2} = \{{(20)}^{2} + {(20)}^{2}\} {cm}^{2}$

$\Rightarrow O B^{2} = (400 + 400) {cm}^{2} \Rightarrow O B^{2} = 800 {cm}^{2} \Rightarrow O B = 20 \sqrt{2} cm$

Hence, the radius of the circle is $20 \sqrt{2} cm$ .
Now,
Area of the shaded region = Area of the quadrant $-$ Area of the square OABC
$= |(\frac{1}{4} \times 3.14 \times 20 \sqrt{2} \times 20 \sqrt{2}) - (20 \times 20)| {cm}^{2} = |(\frac{1}{4} \times \frac{314}{100} \times 800) - 400| {cm}^{2} = (628 - 400) {cm}^{2} = 228 {cm}^{2}$

Page No 852:

Question 2:

The diameter of a wheel is 84 cm. How many revolutions will it make to cover 792 m?
(a) 200
(b) 250
(c) 300
(d) 350

Answer:

(c) 300

Let d cm be the diameter of the wheel.
We know:
Circumference of the wheel $= π \times d$
$= (\frac{22}{7} \times 84) cm = 264 cm$

Now,
Number of revolutions to cover 792 m $= (\frac{792 \times 1000}{264})$
=300

Page No 852:

Question 3:

The area of a sector of a circle with radius r, making an angle of x° at the centre is
(a) $\frac{x}{180} \times 2 π r$
(b) $\frac{x}{180} \times π r^{2}$
(c) $\frac{x}{360} \times 2 π r$
(d) $\frac{x}{360} \times π r^{2}$

Answer:

(d) $\frac{x}{360} \times π r^{2}$

The area of a sector of a circle with radius r making an angle of $x °$ at the centre is $\frac{x}{360} \times π r^{2}$ .

Page No 852:

Question 4:

In the given figure, ABCD is a rectangle inscribed in a circle having length 8 cm and breadth 6 cm. If π = 3.14, then the area of the shaded region is

(a) 264 cm²
(b) 266 cm²
(c) 272 cm²
(d) 254 cm²

Answer:

All options are incorrect; the correct answer is 30.5 cm.

Join AC.
Now, AC is the diameter of the circle.
$We have : A C^{2} = A B^{2} + B C^{2} [By Pythagoras' theorem] \Rightarrow A C^{2} = \{{(8)}^{2} + {(6)}^{2}\} {cm}^{2} \Rightarrow A C^{2} = (64 + 36) {cm}^{2} \Rightarrow A C^{2} = 100 {cm}^{2} \Rightarrow A C = 10 cm$
∴ Radius of the circle $= \frac{10}{2} cm$
=5 cm
Now,
Area of the shaded region = Area of the circle with radius 5 cm $-$ Area of the rectangle ABCD
$= |(3.14 \times 5 \times 5) - (8 \times 6)| {cm}^{2} = |(\frac{314}{100} \times 25) - 48| {cm}^{2} = (\frac{157}{2} - 48) {cm}^{2} = \frac{61}{2} {cm}^{2} = 30.5 {cm}^{2}$

Page No 852:

Question 5:

The circumference of a circle is 22 cm. Find its area.

Answer:

Let r cm be the radius of the circle.
Now,
Circumference of the circle:
$2 π r = 22 \Rightarrow 2 \times \frac{22}{7} \times r = 22 \Rightarrow r = (22 \times \frac{7}{44}) cm \Rightarrow r = \frac{7}{2}$

Also,
Area of the circle $= π r^{2}$
$= (\frac{22}{7} \times \frac{7}{2} \times \frac{7}{2}) {cm}^{2} = \frac{77}{2} {cm}^{2} = 38.5 {cm}^{2}$

Page No 852:

Question 6:

In a circle of radius 21 cm, an arc subtends an angle of 60° at the centre. Find the length of the arc.

Answer:

Let ACB be the given arc subtending at an angle of $60 °$ at the centre.
Now, we have:
$r = 21 cm and θ = 60 °$
∴ Length of the arc ACB $= \frac{2 π r}{360}$
$= (2 \times \frac{22}{7} \times 21 \times \frac{60}{360}) = 22 cm$

Page No 852:

Question 10:

In the given figure, the sectors of two concentric circles of radii 7 cm and 3.5 cm are shown. Find the area of the shaded region.

Answer:

(i) Area of the quadrant OACB $= (\frac{1}{4} \times \frac{22}{7} \times 3.5 \times 3.5) {cm}^{2}$
$= (\frac{1}{4} \times \frac{22}{7} \times \frac{35}{10} \times \frac{35}{10}) {cm}^{2} = \frac{77}{8} {cm}^{2} = 9.625 {cm}^{2}$

(ii) Area of the shaded region = Area of the quadrant OACB $-$ Area of $∆ A O D$
$= |(\frac{77}{8}) - (\frac{1}{2} \times 3.5 \times 2)| {cm}^{2} = (\frac{77}{8} - \frac{35}{10}) {cm}^{2} = \frac{49}{8} {cm}^{2} = 6.125 {cm}^{2}$

Page No 853:

Question 14:

In the given figure, ABCD is a square each of whose sides measures 28 cm. Find the area of the shaded region.

Answer:

Let r be the radius of the circle.
Thus, we have:
$r = \frac{28}{2} cm$
=14 cm
Now,
Area of the shaded region = (Area of the square ABCD) $-$ 4(Area of the sector where $r = 14 cm and θ = 90 °$ )
$= |(28 \times 28) - 4 (\frac{22}{7} \times 14 \times 14 \times \frac{90}{360})| {cm}^{2} = |784 - 4 (154)| {cm}^{2} = (784 - 616) {cm}^{2} = 168 {cm}^{2}$

Page No 853:

Question 15:

In the given figure, an equilateral triangle has been inscribed in a circle of radius 4 cm. Find the area of the shaded region.

Answer:

Draw $O D ⊥ B C$ .
Because $∆ A B C$ is equilateral, $∠ A = ∠ B = ∠ C = 60 °$ .
Thus, we have:

$∠ O B D = 30 ° \Rightarrow \frac{O D}{O B} = \sin 30 ° \Rightarrow \frac{O D}{O B} = \frac{1}{2} \Rightarrow O D = (\frac{1}{2} \times 4) cm [∵ O B = radius] \Rightarrow O D = 2 cm$

${BD}^{2} = ({OB}^{2} - {OD}^{2}) [B y Pythagoras' theorem] \Rightarrow {BD}^{2} = (4^{2} - 2^{2}) {cm}^{2} \Rightarrow {BD}^{2} = (16 - 4) {cm}^{2} \Rightarrow {BD}^{2} = 12 {cm}^{2} \Rightarrow BD = 2 \sqrt{3} cm$

Also,
$BC = 2 \times BD = (2 \times 2 \sqrt{3}) cm = 4 \sqrt{3} cm$
∴ Area of the shaded region = (Area of the circle) $-$ (Area of $∆ A B C$ )
$= |(3.14 \times 4 \times 4) - (\frac{\sqrt{3}}{4} \times 4 \sqrt{3} \times 4 \sqrt{3})| {cm}^{2} = |50.24 - (12 \times 1.73)| {cm}^{2} = (50.24 - 20.76) {cm}^{2} = 29.48 {cm}^{2}$

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Question 19:

Four cows are tethered at the four corners of a square field of side 50 m such that the each can graze the maximum unshared area. What area will be left ungrazed?

Answer:

Let r be the radius of the circle.
Thus, we have:
$r = \frac{50}{2} m$
= 25 m
Area left ungrazed = (Area of the square) $-$ 4(Area of the sector where $r = 25 m and θ = 90 °$ )

$= |(50 \times 50) - 4 (3.14 \times 25 \times 25 \times \frac{90}{360})| m^{2} = |2500 - (4 \times \frac{1962.5}{4})| m^{2} = (2500 - 1962.5) m^{2} = 537.5 m^{2}$

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Question 20:

A square tank has an area of 1600 cm². There are four semicircular plots around it. Find the cost of turfing the plots at Rs 12.50 per m².

Answer:

Let a m be the side of the square.
Area of the square $= a^{2}$
Thus, we have:
$a^{2} = 1600 \Rightarrow a = 40$
Area of the plots = 4(Area of the semicircle of radius 20 m)
$= |4 (\frac{1}{2} π r^{2})| m^{2} = |4 (\frac{1}{2} \times 3.14 \times 20 \times 20)| m^{2} = 2512 m^{2}$

∴ Cost of turfing the plots at $12.50 {per m}^{2} = Rs (2512 \times 12.50)$
= Rs 31400

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