Rs Aggarwal 2018 for Class 10 Math Chapter 1 - Real Numbers

Download PdfDownload Pdf

Share with your friends

Rs Aggarwal 2018 Solutions for Class 10 Math Chapter 1 Real Numbers are provided here with simple step-by-step explanations. These solutions for Real Numbers are extremely popular among Class 10 students for Math Real Numbers Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rs Aggarwal 2018 Book of Class 10 Math Chapter 1 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rs Aggarwal 2018 Solutions. All Rs Aggarwal 2018 Solutions for class Class 10 Math are prepared by experts and are 100% accurate.

Page No 819:

Question 1:

The difference between the circumference and radius of a circle is 37cm. Using $π = \frac{22}{7}$ , find the circumference of the circle. [CBSE 2013]

Answer:

Let the radius of the circle be r and circumference C.

Now,
$C - r = 37 \Rightarrow 2 π r - r = 37 \Rightarrow r (2 \times \frac{22}{7} - 1) = 37 \Rightarrow r = \frac{37 \times 7}{37} = 7 cm$
Now, $C = 2 π r = 2 \times \frac{22}{7} \times 7 = 44 cm$
Hence, the circumference of the circle is 44 cm.

Page No 819:

Question 2:

The circumference of a circle is 22 cm. Find the area of its quadrant. [CBSE 2012]

Answer:

Let the radius of the circle be r.
Now,
$Circumference = 22 \Rightarrow 2 π r = 22 \Rightarrow r = \frac{22 \times 7}{44} = \frac{7}{2} cm$
Now, Area of quadrant = $\frac{1}{4} π r^{2} = \frac{1}{4} \times \frac{22}{7} \times {(\frac{7}{2})}^{2} = \frac{77}{8} {cm}^{2}$
Hence, the area of the quadrant of the circle is $\frac{77}{8}$ cm².

Page No 819:

Question 3:

What is the diameter of a circle whose area is equal to the sum of the areas of two circles of diameters 10 cm and 24 cm? [CBSE 2012]

Answer:

Let the diameter of the required circle be d.
Now, Area of required circle = Area of circle having diameter 10 cm + Area of circle having diameter 24 cm

$\Rightarrow π {(\frac{d}{2})}^{2} = π {(\frac{10}{2})}^{2} + π {(\frac{24}{2})}^{2} \Rightarrow {(\frac{d}{2})}^{2} = 25 + 144 \Rightarrow {(\frac{d}{2})}^{2} = 13^{2} \Rightarrow \frac{d}{2} = 13 \Rightarrow d = 26 cm$
Hence, the diameter of the of the circle is 26 cm.

Page No 819:

Question 4:

If the area of the circle is numerically equal to twice its circumference then what is the diameter of the circle? [CBSE 2011]

Answer:

Let the diameter of the required circle be d.
Now, Area of circle = 2 ⨯ Circumference of the circle
$\Rightarrow π {(\frac{d}{2})}^{2} = 2 (2 π \times \frac{d}{2}) \Rightarrow {(\frac{d}{2})}^{2} = 2 d \Rightarrow d^{2} = 8 d \Rightarrow d^{2} - 8 d = 0 \Rightarrow d (d - 8) = 0 \Rightarrow d = 8 cm [∵ d \neq 0]$
Hence, the diameter of the of the circle is 8 cm.

Page No 819:

Question 5:

What is the perimeter of a square which circumscribes a circle of radius a cm? [CBSE 2011]

Answer:

We know that if a square circumscribes a circle, then the side of the square is equal to the diameter of the circle.
∴ Side of Square = 2a
Now, Perimeter of the square = 4 ⨯ Side of square = 4 ⨯ 2a = 8a cm
Hence, the perimeter of the square is 8a cm.

Page No 819:

Question 6:

Find the length of the arc of a circle of diameter 42 cm which subtends an angle of 60° at the centre. [CBSE 2012]

Answer:

We have $r = \frac{diameter}{2} = \frac{42}{2} = 21 cm and θ = 60 °$
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 length of the arc of the circle is 22 cm.

Page No 819:

Question 7:

Find the diameter of the circle whose area is equal to the sum of the areas of two circles having radii 4 cm and 3 cm. [CBSE 2011]

Answer:

Let the diameter of the required circle be d.
Now, Area of the required circle = Area of circle having radius 4 cm + Area of circle having radius 3 cm

$\Rightarrow π {(\frac{d}{2})}^{2} = π {(4)}^{2} + π {(3)}^{2} \Rightarrow {(\frac{d}{2})}^{2} = 16 + 9 \Rightarrow {(\frac{d}{2})}^{2} = 25 = {(5)}^{2} \Rightarrow \frac{d}{2} = 5 \Rightarrow d = 10 cm$

Hence, the diameter of the circle is 10 cm.

Page No 819:

Question 8:

Find the area of a circle whose circumference is 8π. [CBSE 2014]

Answer:

Let the radius of the circle be r.
Now,
$Circumference = 8 π \Rightarrow 2 π r = 8 π \Rightarrow r = 4 cm$
Now, Area of circle= $π r^{2} = π \times 4^{2} = 16 π {cm}^{2}$
Hence, the area of the circle is 16π cm².

Page No 819:

Question 9:

Find the perimeter of a semicircular protractor whose diameter is 14 cm. [CBSE 2009]

Answer:

Perimeter of a semicircular protractor = Circumference of semicircular protractor + diameter of semicircular protractor
$= \frac{1}{2} (2 π r) + d = π r + d = π \frac{d}{2} + d = \frac{22}{7} \times 7 + 14 = 22 + 14 = 36 cm$
Hence, the perimeter of a semicircular protractor is 36 cm.

Page No 819:

Question 10:

Find the radius of a circle whose perimeter and area are numerically equal.

Answer:

Let the radius of the required circle be r.
Now, Area of circle = Perimeter of the circle
$\Rightarrow π r^{2} = 2 π \times r \Rightarrow r^{2} = 2 r \Rightarrow r^{2} - 2 r = 0 \Rightarrow r (r - 2) = 0 \Rightarrow r - 2 = 0 [∵ r \neq 0] \Rightarrow r = 2 units$
Hence, the radius of the circle is 2 units.

Page No 819:

Question 11:

The radii of two circles are 19 cm and 9 cm, Find the radius of the circle which has circumference equal to the sum of the circumferences of the two circles.

Answer:

Let the radius of the required circle be r.
Now, Circumference of the required circle = Circumference of circle having radius 19 cm + Circumference of circle having radius 9 cm

$\Rightarrow 2 π r = 2 π \times 19 + 2 π \times 9 \Rightarrow r = 19 + 9 \Rightarrow r = 28 cm$
Hence, the radius of the required circle is 28 cm.

Page No 819:

Question 12:

The radii of two circles are 8 cm and 6 cm. Find the radius of the circle having area equal to the sum of the areas of the two circles.

Answer:

Let the radius of the required circle be r.
Now, Area of the required circle = Area of circle having radius 8 cm + Area of circle having radius 6 cm

$\Rightarrow π r^{2} = π {(8)}^{2} + π {(6)}^{2} \Rightarrow r^{2} = 64 + 36 \Rightarrow r^{2} = 10^{2} \Rightarrow r = 10 cm$
Hence, the radius of the circle is 10 cm.

Page No 819:

Question 13:

Find the area of the sector of a circle having radius 6 cm and of angle 30°. [Take π = 3.14]

Answer:

We have $r = 6 cm and θ = 30 °$
Now, Area of sector = $\frac{θ}{360^{°}} \times π r^{2} = \frac{30 °}{360 °} \times 3.14 \times 36 = 9.42 {cm}^{2}$
Hence, the area of the sector of the circle is 9.42 cm².

Page No 820:

Question 14:

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

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 length of the arc of the circle is 22 cm.

Page No 820:

Question 15:

The circumferences of two circles are in the ratio 2: 3. What is the ratio between their areas?

Answer:

Let the the radii of the two circles be r and R, the circumferences of the circles be c and C and the areas of the two circles be a and A.
Now,
$\frac{c}{C} = \frac{2}{3} \Rightarrow \frac{2 π r}{2 π R} = \frac{2}{3} \Rightarrow \frac{r}{R} = \frac{2}{3}$
Now, the ratio between their areas is given by
$\frac{a}{A} = \frac{π r^{2}}{π R^{2}} = {(\frac{r}{R})}^{2} = {(\frac{2}{3})}^{2} = \frac{4}{9}$
Hence, the ratio between their areas is 4 : 9.

Page No 820:

Question 16:

The areas of two circles are in the ratio 4: 9. What is the ratio between their circumferences?

Answer:

Let the radii of the two circles be r and R, the circumferences of the circles be c and C and the areas of the two circles be a and A.
Now,
$\frac{a}{A} = \frac{4}{9} \Rightarrow \frac{π r^{2}}{π R^{2}} = {(\frac{2}{3})}^{2} \Rightarrow \frac{r}{R} = \frac{2}{3}$
Now, the ratio between their circumferences is given by
$\frac{c}{C} = \frac{2 π r}{2 π R} = \frac{r}{R} = \frac{2}{3}$
Hence, the ratio between their circumferences is 2 : 3.

Page No 820:

Question 17:

A square is inscribed in a circle. Find the ratio of the areas of the circle and the square.

Answer:

Let the side of the square be a and radius of the circle be r
We know that if a circle circumscribes a square, then the diameter of the circle is equal to the diagonal of the square.
$∴ \sqrt{2} a = 2 r \Rightarrow a = \sqrt{2} r$
Now,
$\frac{Area of circle}{Area of square} = \frac{π r^{2}}{a^{2}} = \frac{{πr}^{2}}{{(\sqrt{2} r)}^{2}} = \frac{π r^{2}}{2 r^{2}} = \frac{π}{2}$
Hence, the ratio of the areas of the circle and the square is π : 2

Page No 820:

Question 18:

The circumference of a circle is 8 cm. Find the area of the sector whose central angle is 72°.

Answer:

Let the radius of the circle be r.
Now,
$Circumference = 8 \Rightarrow 2 π r = 8 \Rightarrow r = \frac{14}{11} cm$
We have $r = \frac{14}{11} cm and θ = 72 °$
Area of sector = $\frac{θ}{360^{°}} \times π r^{2} = \frac{72 °}{360 °} \times \frac{22}{7} \times {(\frac{14}{11})}^{2} = 1.02 {cm}^{2}$
Hence, the area of the sector of the circle is 1.02 cm².

Disclaimer : If we take the circumference of the circle is 8 cm then the area of the sector will be 1.02 cm². But if we take the circumference of the circle is 88 cm then the area of the sector will be 123.2 cm²

Page No 820:

Question 19:

A pendulum swings through an angle of 30° and describes an arc 8.8 cm in length. Find the length of the pendulum.

Answer:

Given:
Length of the arc = 8.8 cm
And,
$θ = 30^{\circ}$

Now,
Length of the arc = $\frac{2 πrθ}{360}$
$\Rightarrow 8.8 = \frac{2 \times \frac{22}{7} \times r \times 30}{360} \Rightarrow r = \frac{8.8 \times 360 \times 7}{44 \times 30} \Rightarrow r = 16.8 cm$

∴ Length of the pendulum = 16.8 cm

Page No 820:

Question 20:

The minute hand of a clock is 15 cm long. Calculate the area swept by it in 20 minutes.

Answer:

Angle inscribed by the minute hand in 60 minutes = $360^{\circ}$
Angle inscribed by the minute hand in 20 minutes = $\frac{360}{60} \times 20 = 120^{\circ}$

We have:
$θ = 120^{\circ} and r = 15 cm$

∴ Required area swept by the minute hand in 20 minutes = Area of the sector with r = 15 cm and $θ = 120^{\circ}$
$= \frac{{πr}^{2} θ}{360}$

$= 3.14 \times 15 \times 15 \times \frac{120}{360} = 235.5 {cm}^{2}$

Page No 820:

Question 21:

A sector of 56°, cut out from a circle, contains 17.6 cm². Find the radius of the circle.

Answer:

Area of the sector =17.6 cm²
Area of the sector $= \frac{{πr}^{2} θ}{360}$

$\Rightarrow 17.6 = \frac{22}{7} \times r^{2} \times \frac{56}{360} \Rightarrow r^{2} = \frac{17.6 \times 7 \times 360}{22 \times 56} \Rightarrow r^{2} = 36 \Rightarrow r = 6 cm$

∴ Radius of the circle = 6 cm

Page No 820:

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:

From a rectangular sheet of paper ABCD with AB = 40 cm and AD = 28 cm, a semicircular portion with BC as diameter is cut off. Find
the area of the remaining paper. [CBSE 2012]

Answer:

We know that the opposite sides of rectangle are equal
∴ AD = BC = 28 cm
Now, Radius of semicircular portion = $\frac{1}{2} BC = 14 cm$
∴ Area of remaining paper = Area of rectangular sheet −Area of semicircular portion
$= 40 \times 28 - \frac{1}{2} (\frac{22}{7} \times 14 \times 14) = 1120 - 308 = 812 {cm}^{2}$
Hence, the area of the remaining paper is 812 cm²

Page No 821:

Question 28:

In the given figure, OABC is a square of side 7 cm. If COPB is a quadrant of a circle with centre C find the area of the shaded region.

Answer:

Area of shaded region = Area of square OABC − Area of quadrant COPB having radius OC
$= {(Side)}^{2} - \frac{1}{4} (π \times r^{2}) = {(7)}^{2} - \frac{1}{4} [\frac{22}{7} \times 7^{2}] = 49 - 38.5 = 10.5 {cm}^{2}$
Hence, the area of the shaded region is 10.5 cm²

Page No 821:

Question 29:

In the given figure, three sectors of a circle of radius 7 cm, making angles of 60°, 80° and 40° at the centre are shaded. Find the area of the shaded region. [CBSE 2012]

Answer:

Area of the shaded region = Area of sector having central angle 60^∘ + Area of sector having central angle 80^∘ + Area of sector having central angle 40^∘
$= [\frac{60 °}{360^{°}} \times π {(7)}^{2}] + [\frac{80 °}{360^{°}} \times π {(7)}^{2}] + [\frac{40 °}{360^{°}} \times π {(7)}^{2}] = π {(7)}^{2} (\frac{60 °}{360^{°}} + \frac{80 °}{360^{°}} + \frac{40 °}{360^{°}}) = π {(7)}^{2} (\frac{180 °}{360^{°}}) = \frac{22}{7} {(7)}^{2} (\frac{1}{2}) = 77 {cm}^{2}$
Hence, the area of the shaded region is 77 cm².

Page No 821:

Question 30:

In the given figure, PQ and AB are respectively the arcs of two concentric circles of radii 7 cm and 3.5 cm with centre O. If ∠POQ = 30°, find the area of the shaded region. [CBSE 2012]

Answer:

Area of the shaded portion = Area of sector OPQ − Area of sector OAB
$= \frac{30 °}{360^{°}} \times π {(7)}^{2} - \frac{30 °}{360^{°}} \times π {(3.5)}^{2} = \frac{22}{7} \times \frac{1}{12} [{(7)}^{2} - {(3.5)}^{2}] = \frac{22}{7} \times \frac{1}{12} [{(7)}^{2} - {(\frac{7}{2})}^{2}] = \frac{22}{7} \times \frac{1}{12} \times \frac{147}{4} = \frac{77}{8} {cm}^{2}$
Hence, the area of the shaded portion is $\frac{77}{8}$ cm².

Page No 821:

Question 31:

In the given figure, find the area of the shaded region, if ABCD is a square of side 14 cm and APD and BPC are
semicircles. [CBSE 2012]

Answer:

Area of the shaded region = Area of Square ABCD − (Area of semicircle APD + Area of semicircle BPC)
$= {(14)}^{2} - [\frac{1}{2} π {(\frac{14}{2})}^{2} + \frac{1}{2} π {(\frac{14}{2})}^{2}] = {(14)}^{2} - \frac{22}{7} {(7)}^{2} = 42 {cm}^{2}$
Hence, the area of the shaded region is 42 cm².

Page No 821:

Question 32:

In the given figure, the shape of the top of a table is that of a sector of a circle with centre O and ∠AOB = 90°. If AO = OB = 42 cm,· then find the perimeter of the top of the table. [CBSE 2012]

Answer:

We have $r = 42 cm and θ = 360 ° - 90 ° = 270 °$
Perimeter of the top of the table = Length of the major arc AB + Length of OA + Length of OB
$= \frac{θ}{360^{°}} \times 2 π r + 42 + 42 = \frac{270 °}{360 °} \times 2 \times \frac{22}{7} \times 42 + 84 = 198 + 84 = 282 cm$

Hence, the perimeter of the top of the table is 282 cm.

Page No 821:

Question 33:

In the given figure, ABCD is a square of side 7 cm, DPBA and DQBC are quadrants of circles each of the radius 7 cm. Find the area of shaded region.

Answer:

Area of the shaded portion = (Area of quadrant DPBA + Area of quadrant DQBC) − Area of Square ABCD
$= [\frac{1}{4} π {(7)}^{2} + \frac{1}{4} π {(7)}^{2}] - {(7)}^{2} = \frac{1}{2} \times \frac{22}{7} {(7)}^{2} - 49 = 28 {cm}^{2}$
Hence, the area of the shaded portion is 28 cm².

Page No 822:

Question 34:

In the given figure, OABC is a quadrant of a circle of radius 3.5 cm with centre O. If OD = 2 cm, find the area of the shaded portion.

Answer:

Area of the right-angled $∆$ COD = $\frac{1}{2} \times b \times h$
   = $\frac{1}{2} \times 3.5 \times 2 = 3.5 {cm}^{2}$

Area of the sector AOC = $\frac{θ}{360} \times π \times r^{2}$
                                       $= \frac{90}{360} \times \frac{22}{7} \times {(3.5)}^{2} = 9.625 {cm}^{2}$

Area of the shaded region = Area of the $∆$ COD $-$ Area of the sector AOC

                                           $= 9.625 - 3.50 = 6.125 {cm}^{2}$

Page No 822:

Question 35:

Find the perimeter of the shaded region in the figure, if ABCD is a square of side 14 cm and APB and CPD are semicircles. [CBSE 2011]

Answer:

Permieter of shaded region = Length of the arc APB + Length of the arc CPD + Length of AD + Length of BC
$= \frac{1}{2} \times 2 π r + \frac{1}{2} \times 2 π r + 14 + 14 = 2 π r + 28 = 2 \times \frac{22}{7} \times \frac{14}{2} + 28 = 72 cm$

Hence, the perimeter of the shaded region is 72 cm.

Page No 822:

Question 36:

In a circle of radius 7 cm, a square ABCD is inscribed. Find the area of the circle which is outside the square. [CBSE 2011]

Answer:

Let the diagonal of the square be d.
We know that if a circle circumscribes a square, then the diameter of the circle is equal to the diagonal of the square.
∴ d = 2 ⨯ 7 = 14 cm
Now,
Area of required region = Area of circle − Area of square
$= π r^{2} - \frac{1}{2} d^{2} = \frac{22}{7} \times {(7)}^{2} - \frac{1}{2} \times {(14)}^{2} = 56 {cm}^{2}$

Hence, the required area is 56 cm² .

Page No 822:

Question 37:

In the given figure, APB and CQD are semicircles of diameter 7 cm each, while ARC an BSD are semicircles of diameter 14 cm each. Find the
(i) perimeter, (ii) area of the shaded region. [CBSE 2011]

Answer:

(i) Perimeter of shaded region = Length of the arc APB + Length of the arc ARC + Length of the arc BSD + Length of the arc CQD
$= \frac{1}{2} \times 2 π r_{1} + \frac{1}{2} \times 2 π r_{2} + \frac{1}{2} \times 2 π r_{3} + \frac{1}{2} \times 2 π r_{4} = \frac{1}{2} \times 2 π (\frac{7}{2}) + \frac{1}{2} \times 2 π (7) + \frac{1}{2} \times 2 π (7) + \frac{1}{2} \times 2 π (\frac{7}{2}) = 2 π (\frac{7}{2}) + 2 π (7) = 2 π (\frac{7}{2} + 7) = 2 \times \frac{22}{7} \times \frac{21}{2} = 66 cm$

(ii) Area of shaded region = Area of the arc ARC + Area of the arc BSD − (Area of the arc APB + Area of the arc CQD)
$= \frac{1}{2} \times π {(r_{1})}^{2} + \frac{1}{2} \times π {(r_{2})}^{2} - [\frac{1}{2} \times π {(r_{3})}^{2} + \frac{1}{2} \times π {(r_{4})}^{2}] = \frac{1}{2} \times π {(7)}^{2} + \frac{1}{2} \times π {(7)}^{2} - [\frac{1}{2} \times π {(\frac{7}{2})}^{2} + \frac{1}{2} \times π {(\frac{7}{2})}^{2}] = π {(7)}^{2} - π {(\frac{7}{2})}^{2} = π (49 - \frac{49}{4}) = \frac{22}{7} \times \frac{147}{4} = 115.5 {cm}^{2}$

Page No 822:

Question 38:

In the given figure, PSR, RTQ and PAQ are three semicircles of diameter 10 cm, 3 cm and 7 cm respectively. Find the perimeter of shaded region. [Use π= 3.14] [CBSE 2014]

Answer:

Perimeter of shaded region = Length of the arc PAQ + Length of the arc PSR + Length of the arc RTQ
$= \frac{1}{2} \times 2 π r_{1} + \frac{1}{2} \times 2 π r_{2} + \frac{1}{2} \times 2 π r_{3} = \frac{1}{2} \times 2 π (\frac{7}{2}) + \frac{1}{2} \times 2 π (\frac{10}{2}) + \frac{1}{2} \times 2 π (\frac{3}{2}) = \frac{7}{2} π + 5 π + \frac{3}{2} π = \frac{7}{2} π + \frac{3}{2} π + 5 π = 5 π + 5 π = 10 π = 31.4 cm$
Hence, the perimeter of shaded region is 31.4 cm.

Page No 823:

Question 39:

In the given figure, a square OABC is inscribed in a quadrant OPBQ of a circle. If OA = 20 cm, find the area of the shaded region. [Use π = 3.14.] [CBSE 2014]

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:

Find the area of a quadrant of a circle whose circumference is 44 cm. [CBSE 2011]

Answer:

Let the radius of the circle be r.
Now,
$Circumference = 44 \Rightarrow 2 π r = 44 \Rightarrow r = 7 cm$
Now,
Area of quadrant = $\frac{1}{4} π r^{2} = \frac{1}{4} \times \frac{22}{7} \times {(7)}^{2} = 38.5 {cm}^{2}$
Hence, the area of the quadrant of the circle is 38.5 cm².

Page No 823:

Question 42:

In the given figure, find the area of the shaded region, where ABCD is a square of side 14 cm and all circles are of the same diameter.

Answer:

Area of the square = ${Side}^{2} = 14^{2} = 196 sq . cm$

Area of the circles = $4 \times π \times 3.5 \times 3.5 = 154 sq . cm$

Area of the shaded region = Area of the square $-$ Area of four circles
$= 196 - 154 = 42 {cm}^{2}$

Page No 823:

Question 43:

Find the area of the shaded region in the given figure, if ABCD is a rectangle with sides 8 cm and 6 cm and O is the centre of the circle. [CBSE 2014]

Answer:

In right triangle ABC
AC² = AB² + BC²
= 8² +6²
= 64 + 36
= 100
∴ AC² = 100
⇒ AC = 10 cm
Now, Radius of circle(OA) $= \frac{1}{2} AC = 5 cm$
Area of the shaded region = Area of circle − Area of rectangle OABC
$= π {(OA)}^{2} - AB \times BC = \frac{22}{7} \times {(5)}^{2} - 8 \times 6 = 78.57 - 48 = 30.57 {cm}^{2}$
Hence, the area of the shaded region is 30.57 cm².

Page No 823:

Question 44:

A wire is bent to form a square enclosing an area of 484 cm².Using the same wire, a circle is formed. Find the area of the circle.

Answer:

Area of the circle = 484 cm²
Area of the square = ${Side}^{2}$
$\Rightarrow 484 = {Side}^{2} \Rightarrow 22^{2} = {Side}^{2} \Rightarrow Side = 22 cm$
Perimeter of the square = $4 \times Side$
Perimeter of the square = $4 \times 22$
= 88 cm
Length of the wire = 88 cm
Circumference of the circle = Length of the wire = 88 cm
Now, let the radius of the circle be r cm.
Thus, we have:
$2 πr = 88 \Rightarrow 2 \times \frac{22}{7} \times r = 88 \Rightarrow r = 14$

Area of the circle = ${πr}^{2}$
$= \frac{22}{7} \times 14 \times 14 = 616 {cm}^{2}$

Thus, the area enclosed by the circle is 616 cm².

Page No 823:

Question 45:

A square ABCD is inscribed in a circle of radius r. Find the area of the square.

Answer:

Let the diameter of the square be d and having circumscribed circle of radius r.
We know that if a circle circumscribes a square, then the diameter of the circle is equal to the diagonal of the square.
∴ d = 2r
Now,
$Area of square = \frac{1}{2} d^{2} = \frac{1}{2} {(2 r)}^{2} = 2 r^{2} sq units$
Hence, the area of the square ABCD is 2r² sq units.

Page No 824:

Question 46:

The cost of fencing a circular field at the rate of Rs 25 per metre is Rs 5500. The field is to be ploughed at the rate of 50 paise per m² . Find the cost of ploughing the field. [Take $π = \frac{22}{7}$ ].

Answer:

$Circumference = \frac{Total cost of fencing}{Rate of fencing} = \frac{5500}{25} = 220$
Let the radius of the circle be r.
Now,
$Circumference = 220 \Rightarrow 2 π r = 220 \Rightarrow r = 35 cm$
Now,
Area of field = $π r^{2} = \frac{22}{7} \times {(35)}^{2} = 3850 {cm}^{2}$
Cost of ploughing = Rate ⨯ Area of field = 0.5 ⨯ 3850 = Rs 1925
Hence, the cost of ploughing the field is Rs 1925.

Page No 824:

Question 47:

A park is in the form of a rectangle 120 m by 90 m. At the centre of the park there is a circular lawn as shown in the figure. The area of the park excluding the lawn is 2950 m². Find the radius of the circular lawn.

Answer:

Area of the rectangle = $l \times b$
           $= 120 \times 90 = 10800 sq . m$

Area of the park excluding the lawn = 2950 m²

Area of the circular lawn = Area of the park $-$ Area of the park excluding the lawn
                                          = 10800 $-$ 2950
                                          = 7850 m²
Area of the circular lawn = $π r^{2}$
$\Rightarrow 7850 = \frac{22}{7} \times r^{2} \Rightarrow \frac{7850 \times 7}{22} = r^{2} \Rightarrow r^{2} = 2497.72 \Rightarrow r = 49.97 Or, r \approx 50 m$

Thus, the radius of the circular lawn is 50 m.

Page No 824:

Question 48:

In the given figure, PQRS represents a flower bed. If OP = 21 m and OR = 14 m, find the area of the flower bed.

Answer:

Area of the flower bed is the difference between the areas of sectors OPQ and ORS.

$Area of the flower bed = \frac{θ}{360} \times π ({PO}^{2} - {OR}^{2}) = \frac{90}{360} \times \frac{22}{7} (21^{2} - 14^{2}) = \frac{1}{4} \times \frac{22}{7} \times 35 \times 7 = 192.5 m^{2}$

Page No 824:

Question 49:

In the given figure, O is the centre of the bigger circle, and AC is its diameter. Another circle with AB as diameter is drawn. If AC = 54 cm and BC = 10, find the area of the shaded region.

Answer:

We have:
OA = OC = 27 cm
AB = AC $-$ BC
= 54 $-$ 10
= 44
AB is the diameter of the smaller circle.
Thus, we have:
Radius of the smaller circle = $\frac{AB}{2} = \frac{44}{2} = 22 cm$
Area of the smaller circle = ${πr}^{2}$
            $= \frac{22}{7} \times 22 \times 22 = 1521.14 {cm}^{2}$

Radius of the larger circle = $\frac{AC}{2} = \frac{54}{2} = 27 cm$
Area of the larger circle = ${πr}^{2}$
$= \frac{22}{7} \times 27 \times 27 = 2291.14 {cm}^{2}$
∴ Area of the shaded region = Area of the larger circle $-$ Area of the smaller circle
   = 2291.14 $-$ 1521.14
                 = 770 cm²

Page No 824:

Question 50:

From a thin metallic piece in the shape of a trapezium ABCD in which AB || CD and ∠BCD = 90°, a quarter circle BFEC is removed. Given, AB = BC = 3.5 cm and DE = 2 cm, calculate the area of remaining (shaded) part of metal sheet. [CBSE 2011]

Answer:

Since, BFEC is a quarter of a circle.
Hence, BC = EC = 3.5 cm
Now, DC = DE + EC = 2 + 3.5 = 5.5 cm
Area of shaded region = Area of the trapezium ABCD − Area of the quadrant BFEC
$= \frac{1}{2} \times (AB + DC) \times BC - \frac{1}{4} \times π {(EC)}^{2} = \frac{1}{2} \times (3.5 + 5.5) \times 3.5 - \frac{1}{4} \times \frac{22}{7} \times {(3.5)}^{2} = 6.125 {cm}^{2}$
Hence, the area of the shaded region is 6.125 cm² .

Page No 824:

Question 51:

Find the area of the major segment APB of a circle of radius 35 cm and ∠AOB = 90°, as shown in the given figure.

Answer:

Area of minor segment = Area of sector AOBC − Area of right triangle AOB
$= \frac{90 °}{360 °} \times π {(OA)}^{2} - \frac{1}{2} \times OA \times OB = \frac{1}{4} \times \frac{22}{7} \times {(35)}^{2} - \frac{1}{2} \times 35 \times 35 = \frac{1}{4} \times \frac{22}{7} \times {(35)}^{2} - \frac{1}{2} \times 35 \times 35 = 962.5 - 612.5 = 350 {cm}^{2}$
Area of major segment APB = Area of circle − Area of minor segment
$= π {(OA)}^{2} - 350 = \frac{22}{7} \times {(35)}^{2} - 350 = 3850 - 350 = 3500 {cm}^{2}$
Hence, the area of major segment is 3500 cm²

Page No 831:

Question 1:

The circumference of a circle is 39.6 cm. Find its area.

Answer:

Circumference = 39.6 cm
We know:
Circumference of a circle = $2 π r$
$\Rightarrow 39.6 = 2 \times \frac{22}{7} \times r \Rightarrow \frac{39.6 \times 7}{2 \times 22} = r \Rightarrow r = 6.3 cm$

Also,
Area of the circle = $π r^{2}$
$= \frac{22}{7} \times 6.3 \times 6.3 = 124.74 {cm}^{2}$

Page No 831:

Question 2:

The area of a circle is 98.56 cm². Find its circumference.

Answer:

Let the radius of the circle be r.
Now,
$Area = 98.56 \Rightarrow π r^{2} = 98.56 \Rightarrow \frac{22}{7} \times r^{2} = 98.56 \Rightarrow r = 5.6$
Now,
Circumference = $2 π r = 2 \times \frac{22}{7} \times 5.6 = 35.2 cm$
Hence, the circumference of the circle is 35.2 cm.

Page No 831:

Question 3:

The circumference of a circle exceeds its diameter by 45 cm. Find the circumference of the circle.

Answer:

Let the radius of the circle be r.
Now,
Circumference = Diameter + 45
$\Rightarrow 2 π r = 2 r + 45 \Rightarrow 2 π r - 2 r = 45 \Rightarrow 2 r (\frac{22}{7} - 1) = 45 \Rightarrow 2 r \times \frac{15}{7} = 45 \Rightarrow r = 10.5 cm$
∴ Circumference = Diameter + 45 = 2(10.5) + 45 = 66 cm
Hence, the circumference of the circle is 66 cm.

Page No 831:

Question 4:

A copper wire when bent in the form of a square encloses an area of 484 cm². The same wire is not bent in the form of a circle. Find the area enclosed by the circle.

Answer:

Area of the circle = 484 cm²
Area of the square = ${Side}^{2}$
$\Rightarrow 484 = {Side}^{2} \Rightarrow 22^{2} = {Side}^{2} \Rightarrow Side = 22 cm$
Perimeter of the square = $4 \times Side$
Perimeter of the square = $4 \times 22$
= 88 cm
Length of the wire = 88 cm
Circumference of the circle = Length of the wire = 88 cm
Now, let the radius of the circle be r cm.
Thus, we have:
$2 πr = 88 \Rightarrow 2 \times \frac{22}{7} \times r = 88 \Rightarrow r = 14$

Area of the circle = ${πr}^{2}$
$= \frac{22}{7} \times 14 \times 14 = 616 {cm}^{2}$

Thus, the area enclosed by the circle is 616 cm².

Page No 831:

Question 5:

A wire when bent in the form of an equilateral triangle encloses an area of $121 \sqrt{3} {cm}^{2}$ . The same wire is bent to form a circle. Find the area enclosed by the circle.

Answer:

$Area of a n equilateral t riangle = \frac{\sqrt{3}}{4} \times {(Side)}^{2} \Rightarrow 121 \sqrt{3} = \frac{\sqrt{3}}{4} \times {(Side)}^{2} \Rightarrow 121 \times 4 = {(Side)}^{2} \Rightarrow Side = 22 cm$

$Perimeter of an equilateral triangle = 3 \times Side$
$= 3 \times 22 = 66 cm$

Length of the wire $= 66 cm$
Now, let the radius of the circle be r cm.
We know:
Circumference of the circle = Length of the wire
$2 π r = 66 \Rightarrow 2 \times \frac{22}{7} \times r = 66 \Rightarrow r = \frac{66 \times 7}{2 \times 22} \Rightarrow r = \frac{21}{2} \Rightarrow r = 10.5$
Thus, we have:
Area of the circle = $π r^{2}$
$= \frac{22}{7} \times 10.5 \times 10.5 = 346.5 sq . cm$
Area enclosed by the circle = 346.5 cm²

Page No 831:

Question 6:

The length of a chain used as the boundary of a semicircular park is 108 m. Find the area of the park.

Answer:

Let the radius of the park be r.
Length of chain = Perimeter of the semicircular park
⇒ 108 = Length of the arc + Diameter
$\Rightarrow 108 = \frac{1}{2} \times 2 π r + 2 r \Rightarrow 108 = r (\frac{22}{7} + 2) \Rightarrow 108 = \frac{36}{7} r \Rightarrow r = 21 m$
Now, Area of park $= \frac{1}{2} π r^{2} = \frac{1}{2} \times \frac{22}{7} \times {(21)}^{2} = 693 m^{2}$
Hence, the area of the park is 693 m² .

Page No 831:

Question 7:

The sum of the radii of two circles is 7 cm, and the difference of their circumferences is 8 cm. Find the circumference of the circles.

Answer:

Let the radii of the two circles be r₁ cm and r₂cm.
Now,
Sum of the radii of the two circles = 7 cm
$r_{1 +} r_{2} = 7 . . . (i)$

Difference of the circumferences of the two circles = 88 cm
$\Rightarrow 2 {πr}_{1} - 2 {πr}_{2} = 8 \Rightarrow 2 π (r_{1} - r_{2}) = 8 \Rightarrow (r_{1} - r_{2}) = \frac{8}{2 π} \Rightarrow r_{1} - r_{2} = \frac{8}{2 \times \frac{22}{7}} \Rightarrow r_{1 -} r_{2} = \frac{8 \times 7}{44}$

$r_{1} - r_{2} = \frac{56}{44} r_{1} - r_{2} = \frac{14}{11} . . . (i i)$

Adding (i) and (ii), we get:
$2 r_{1} = \frac{91}{11} r_{1} = \frac{91}{22}$

∴ Circumference of the first circle = $2 {πr}_{1}$
$= 2 \times \frac{22}{7} \times \frac{91}{22} = 26 cm$
Also,
$r_{1} - r_{2} = \frac{14}{11} \frac{91}{22} - r_{2} = \frac{14}{11} \frac{91}{22} - \frac{14}{11} = r_{2} r_{2} = \frac{63}{22}$

∴ Circumference of the second circle = $2 {πr}_{2}$

$= 2 \times \frac{22}{7} \times \frac{63}{22} = 18 cm$

Therefore, circumferences of the first and second circles are 18 cm and 26 cm, respectively.

Page No 831:

Question 8:

Find the area of a ring whose outer and inner radii are respectively 23 cm and 12 cm.

Answer:

Let r₁ cm and r₂ cm be the radii of the outer and inner boundaries of the ring, respectively.
We have:

$r_{1} = 23 cm r_{2} = 12 cm$

Now,
Area of the outer ring = $π {r_{1}}^{2}$

$= \frac{22}{7} \times 23 \times 23 = 1662.57 {cm}^{2}$

Area of the inner ring = $π {r_{2}}^{2}$

      $= \frac{22}{7} \times 12 \times 12 = 452.57 {cm}^{2}$

Area of the ring = Area of the outer ring $-$ Area of the inner ring
   = 1662.57 $-$ 452.57
       = 1210 ${cm}^{2}$

Page No 831:

Question 9:

A path of 8 m width runs around the outsider of a circular park whose radius is 17 m. Find the area of the path.

Answer:

The radius (r) of the inner circle is 17 m.
The radius (R) of the outer circle is 25 m.          [Includes path, i.e., (17 + 8)]

Area of the path = $π R^{2} - π r^{2}$
           $= π (R^{2} - r^{2}) = \frac{22}{7} (25^{2} - 17^{2}) = \frac{22}{7} \times (25 - 17) (25 + 17) = \frac{22}{7} \times 8 \times 42 = 1056 m^{2}$

∴ Area of the path = 1056 m²

Page No 831:

Question 10:

A race track is in the form of a rig whose inner circumference is 352 m and outer circumference is 396 m. Find the width and the area of the track.

Answer:

Let r m and R m be the radii of the inner and outer tracks.
Now,
Circumference of the outer track = $2 π R$
$\Rightarrow 396 = 2 \times \frac{22}{7} \times R \Rightarrow R = \frac{396 \times 7}{44} \Rightarrow R = 63$

Circumference of the inner track = $2 π r$
$\Rightarrow 352 = 2 \times \frac{22}{7} \times r \Rightarrow r = \frac{352 \times 7}{44} \Rightarrow r = 56$

Width of the track = Radius of the outer track $-$ Radius of the inner track
$= 63 - 56 = 7 m$

Area of the outer circle = $π R^{2}$
    $= \frac{22}{7} \times 63 \times 63 = 12474 m^{2}$

Area of the inner circle = $π R^{2}$
        $= \frac{22}{7} \times 56 \times 56 = 9856 m^{2}$

Area of the track = 12474 $-$ 9856
   = 2618 $m^{2}$

Page No 831:

Question 11:

A sector is cut from a circle of radius 21 cm. The angle of the sector is 150°. Find the length of the arc and the area of the sector.

Answer:

Given:
Radius = 2 cm
Angle of sector = $150^{\circ}$
Now,
$Length of the arc = \frac{2 πr θ}{360} = \frac{2 \times \frac{22}{7} \times 21 \times 150}{360} = 55 cm$

Area of the sector = $\frac{{πr}^{2} θ}{360}$

$= \frac{22}{7} \times 21 \times 21 \times \frac{150}{360} = 577.5 cm$

Page No 831:

Question 12:

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 831:

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:

Length of the arc = 16.5 cm
$θ = 54^{\circ}$
Radius = ?
Circumference=?
We know:
Length of the arc $= \frac{2 πrθ}{360}$
$\Rightarrow 16.5 = \frac{2 \times \frac{22}{7} \times r \times 54}{360} \Rightarrow r = \frac{16.5 \times 360 \times 7}{44 \times 54} \Rightarrow r = 17.5 cm$

$c = 2 πr = 2 \times \frac{22}{7} \times 17.5 = 110 cm$

Circumference = 110 cm

Now,
Area of the circle = ${πr}^{2}$
$= \frac{22}{7} \times 17.5 \times 17.5 = 962.5 {cm}^{2}$

Page No 831:

Question 14:

The radius of a circle with centre O is 7 cm. Two radii OA and OB are drawn at right angles to each other. Find the areas of minor and major
segments.

Answer:

Area of minor segment = Area of sector AOBC − Area of right triangle AOB
$= \frac{90 °}{360 °} π {(OA)}^{2} - \frac{1}{2} \times OA \times OB = \frac{1}{4} \times \frac{22}{7} \times {(7)}^{2} - \frac{1}{2} \times 7 \times 7 = \frac{1}{4} \times \frac{22}{7} \times {(7)}^{2} - \frac{1}{2} \times 7 \times 7 = 38.5 - 24.5 = 14 {cm}^{2}$
Area of major segment APB = Area of circle − Area of minor segment
$= π {(OA)}^{2} - 14 = \frac{22}{7} \times {(7)}^{2} - 14 = 154 - 14 = 140 {cm}^{2}$
Hence, the area of major segment is 140 cm² .

Page No 831:

Question 15:

Find the lengths of the arcs cut off from a circle of radius 12 cm by a chord 12 cm long. Also, find the area of the minor segment.

Answer:

Let AB be the chord. Joining A and B to O, we get an equilateral triangle OAB.
Thus, we have:
$∠ O = ∠ A = ∠ B = 60 °$

Length of the arc ACB:
$2 π \times 12 \times \frac{60}{360} = 4 π = 12.56 cm$

Length of the arc ADB:

$Circumference of the circle - Length of the arc ACB = 2 π \times 12 - 4 π = 20 π cm = 62.80 cm$

Now,
Area of the minor segment:

$Area of the sector - Area of the triangle = [π \times {(12)}^{2} \times \frac{60}{360} - \frac{\sqrt{3}}{4} \times {(12)}^{2}] = 13.08 {cm}^{2}$

Page No 832:

Question 16:

A chord 10 cm long is drawn in a circle whose radius is $5 \sqrt{2}$ cm. Find the areas of both the segments.

Answer:

Let O be the centre of the circle and AB be the chord.

Consider $∆$ OAB.

$OA = OB = 5 \sqrt{2} cm$

${OA}^{2} + {OB}^{2} = 50 + 50 = 100$

Now,
$\sqrt{100} = 10 cm = A B$

Thus, $∆$ OAB is a right isosceles triangle.

Thus, we have:
Area of $∆$ OAB = $\frac{1}{2} \times 5 \sqrt{2} \times 5 \sqrt{2} = 25 {cm}^{2}$

Area of the minor segment = Area of the sector $-$ Area of the triangle
$= (\frac{90}{360} \times π \times {(5 \sqrt{2})}^{2}) - 25 = 14.25 {cm}^{2}$

Area of the major segment = Area of the circle $-$ Area of the minor segment
$= π \times {(5 \sqrt{2})}^{2} - 14.25 = 142.75 {cm}^{2}$

Page No 832:

Question 17:

Find the areas of both the segments of a circle of radius 42 cm with central angle 120°.

Answer:

$Area of the minor sector = \frac{120}{360} \times π \times 42 \times 42 = \frac{1}{3} \times π \times 42 \times 42 = π \times 14 \times 42 = 1848 {cm}^{2}$

Area of the triangle = $\frac{1}{2} R^{2} \sin θ$
Here, R is the measure of the equal sides of the isosceles triangle and θ is the angle enclosed by the equal sides.
Thus, we have:
$\frac{1}{2} \times 42 \times 42 \times \sin (120 °) = 762.93 {cm}^{2}$

Area of the minor segment = Area of the sector $-$ Area of the triangle
= $1848 - 762.93 = 1085.07 {cm}^{2}$

Area of the major segment = Area of the circle $-$ Area of the minor segment
$= (π \times 42 \times 42) - 1085.07 = 5544 - 1085.07 = 4458.93 {cm}^{2}$

Page No 832:

Question 18:

A chord of a circle of radius 30 cm makes an angle of 60° at the centre of the circle. Find the areas of the minor major segments.

Answer:

Let the chord be AB. The ends of the chord are connected to the centre of the circle O to give the triangle OAB.

OAB is an isosceles triangle. The angle at the centre is 60 $°$

Area of the triangle = $\frac{1}{2} {(30)}^{2} \sin 60 ° = 450 \times \frac{\sqrt{3}}{2} = 389.25 {cm}^{2}$

Area of the sector OACBO = $\frac{60}{360} \times π \times 30 \times 30 = 150 π = 471 {cm}^{2}$

Area of the minor segment = Area of the sector $-$ Area of the triangle
= $471 - 389.25 = 81.75 {cm}^{2}$

Area of the major segment = Area of the circle $-$ Area of the minor segment
$= (π \times 30 \times 30) - 81.29 = 2744.71 {cm}^{2}$

Page No 832:

Question 19:

In a circle of radius 10.5 cm, the minor arc is one-fifth of the major arc. Find the area of the sector corresponding to the major arc.

Answer:

Let the length of the major arc be $x$ cm
Radius of the circle = 10.5 cm
∴ Length of the minor arc = $\frac{x}{5} cm$

Circumference = $(x + \frac{x}{5}) = \frac{6 x}{5} cm$

Using the given data, we get:

$\frac{6 x}{5} = 2 \times \frac{22}{7} \times \frac{21}{2} \Rightarrow \frac{6 x}{5} = 66 Or, x = 55$

∴ Area of the sector corresponding to the major arc = $(\frac{1}{2} \times 55 \times \frac{21}{2}) = 288.75 {cm}^{2}$

Page No 832:

Question 20:

The short and long hands of a clock are 4 cm and 6 cm long respectively. Find the sum of distances travelled by their tips in 2 days.

Answer:

In 2 days, the short hand will complete 4 rounds.
Length of the short hand = 4 cm

Distance covered by the short hand = $4 \times 2 π \times 4 = 32 π cm$

In the same 2 days, the long hand will complete 48 rounds.
Length of the long hand = 6 cm
Distance covered by the long hand = $48 \times 2 π \times 6 = 576 π cm$

∴ Total distance covered by both the hands = Distance covered by the short hand + Distance covered by the long hand

$= 32 π + 576 π = 608 π = 608 \times 3.14 = 1909.12 cm$

Page No 832:

Question 21:

Find the area of a quadrant of a circle whose circumference is 88 cm.

Answer:

Let the radius of the circle be r.
Now,
$Circumference = 88 \Rightarrow 2 π r = 88 \Rightarrow r = 14 cm$
Now,
Area of quadrant = $\frac{1}{4} π r^{2} = \frac{1}{4} \times \frac{22}{7} \times {(14)}^{2} = 154 {cm}^{2}$
Hence, the area of the quadrant of the circle is 154 cm².

Page No 832:

Question 22:

A rope by which a cow is tethered is increased from 16 m to 23 m. How much additional ground does it have now graze?

Answer:

r₁ = 16 m
r₂ = 23 m

Amount of additional ground available = Area of the bigger circle $-$ Area of the smaller circle
$= π ({r_{1}}^{2} - {r_{2}}^{2}) = π (23^{2} - 16^{2}) = π (23 + 16) (23 - 16) = 858 m^{2}$

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:

In a rhombus, all sides are congruent to each other.

Thus, we have:
$O P = P Q = Q R = R O$

Now, consider $∆ Q O P$ .

$O Q = O P (Both are radii .)$

Therefore, $∆ Q O P$ is equilateral.

Similarly, $∆ Q O R$ is also equilateral and $∆ Q O P ≅ ∆ Q O R$ .

$Ar . (Q R O P) = Ar . (∆ Q O P) + A (∆ Q O R) = 2 Ar . [∆ Q O P]$

$Ar . (∆ QOP) = \frac{1}{2} \times 32 \sqrt{3} = 16 \sqrt{3} O r, 16 \sqrt{3} = \frac{\sqrt{3}}{4} s^{2} (where s is the side of the rhombus) O r, s^{2} = 16 \times 4 = 64 \Rightarrow s = 8 cm$

∴ OQ = 8 cm

Hence, the radius of the circle is 8 cm.

Page No 832:

Question 27:

The side of a square is 10 cm. Find (i) the area of the inscribed circle, and (ii) the area of the circumscribed circle.

Answer:

(i) If a circle is inscribed in a square, then the side of the square is equal to the diameter of the circle.
Side of the square = 10 cm
Side = Diameter = 10
∴ Radius = 5 cm
Area of the inscribed circle = ${πr}^{2}$
      $= 3.14 \times 5 \times 5 = 78.5 {cm}^{2}$

(ii) If a circle is circumscribed in a square, then the diagonal of the square is equal to the diameter of the circle.
$Diagonal of the square = \sqrt{2} \times Side of the square = \sqrt{2} \times 10 = 10 \sqrt{2} cm$

Diagonal = Diameter = $10 \sqrt{2} cm$
∴ $r = 5 \sqrt{2}$ cm

Now,
Area of the circumscribed circle = ${πr}^{2}$
                             $= 3.14 \times {(5 \sqrt{2})}^{2} = 3.14 \times 50 = 157 {cm}^{2}$

Page No 832:

Question 28:

If a square is inscribed in a circle, find the ratio of the areas of the circle and the square.

Answer:

If a square is inscribed in a circle, then the diagonals of the square are diameters of the circle.
Let the diagonal of the square be d cm.
Thus, we have:
Radius, r = $\frac{d}{2} cm$

$Area of the circle = π r^{2}$
= $π \frac{d^{2}}{4} {cm}^{2}$

$We know : d = \sqrt{2} \times Side \Rightarrow Side = \frac{d}{\sqrt{2}} cm$

$Area of the square = (Side)^{2} = {(\frac{d}{\sqrt{2}})}^{2} = \frac{d^{2}}{2} {cm}^{2}$

Ratio of the area of the circle to that of the square:
$= \frac{π \frac{d^{2}}{4}}{\frac{d^{2}}{2}} = \frac{π}{2}$
Thus, the ratio of the area of the circle to that of the square is $π : 2$ .

Page No 832:

Question 29:

The area of a circle inscribed in an equilateral triangle is 154 cm². Find the perimeter of the triangle.

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 832:

Question 30:

The radius of the wheel of a vehicle is 42 cm. How many revolutions will it complete in a 19.8-km-long journey?

Answer:

Radius of the wheel = 42 cm
Circumference of the wheel = $2 πr$
$= 2 \times \frac{22}{7} \times 42 = 264 cm = \frac{264}{100} m = 2.64 m$

Distance covered by the wheel in 1 revolution = 2.64 m

Total distance = 19.8 km or 19800 m

∴ Number of revolutions taken by the wheel = $\frac{19800}{2.64} = 7500$

Page No 833:

Question 31:

The wheels of the locomotive of a train are 2.1 m in radius. They make 75 revolutions in one minute.

Answer:

Radius of the wheel = 2.1 m
Circumference of the wheel = $2 πr$
   $= 2 \times \frac{22}{7} \times 2.1 = 13.2 m$
Distance covered by the wheel in 1 revolution = 13.2 m
Distance covered by the wheel in 75 revolutions = $(13.2 \times 75) = (990 \times \frac{1}{100}) m$
    $= (990 \times \frac{1}{1000}) km$
Distance covered by the wheel in 1 minute = Distance covered by the wheel in 75 revolutions = $\frac{990}{1000}$ km

∴ Distance covered by the wheel in 1 hour = $\frac{990}{1000} \times 60$
                = $59.4 km / h$

Page No 833:

Question 32:

The wheels of a car make 2500 revolutions in covering a distance of 4.95 km. Find the diameter of a wheel.

Answer:

Distance = 4.95 km = $4.95 \times 1000 \times 100 cm$
∴ Distance covered by the wheel in 1 revolution $= \frac{Total distance covered}{Number of revolutions}$

   $= \frac{4.95 \times 1000 \times 100}{2500} = 198 cm$

Now,
Circumference of the wheel = 198 cm
$\Rightarrow 2 π r = 198 \Rightarrow 2 \times \frac{22}{7} \times r = 198 \Rightarrow r = \frac{198 \times 7}{44} \Rightarrow r = 31.5 cm$

∴ Diameter of the wheel = 2r
   = 2(31.5)
   = 63 cm

Page No 833:

Question 33:

A boy is cycling in such a way that the wheels of his bicycle are making 140 revolutions per minute. If the diameter of a wheel is 60 cm, calculate the speed (in km/h) at which the boy is cycling.

Answer:

Diameter of the wheel = 60 cm
∴ Radius of the wheel = 30 cm
Circumference of the wheel = $2 πr$
$= 2 \times \frac{22}{7} \times 30 = \frac{1320}{7} cm$
Distance covered by the wheel in 1 revolution = $\frac{1320}{7} cm$
∴ Distance covered by the wheel in 140 revolutions = $(\frac{1320}{7} \times 140 \times \frac{1}{100}) m$
$= (\frac{1320 \times 140}{7 \times 100} \times \frac{1}{1000}) km = \frac{264}{1000} km$

Now,
Distance covered by the wheel in 1 minute = Distance covered by the wheel in 140 revolutions = $\frac{264}{1000} km$

∴ Distance covered by the wheel in 1 hour = $\frac{264}{1000} \times 60 = 15.84 km / h$

Hence, the speed at which the boy is cycling is 15.84 km/h.

Page No 833:

Question 34:

The diameter of the wheels of a bus is 140 cm. How many revolutions per minute do the wheels make when the bus is moving at a speed of 72.6 km per hour?

Answer:

Diameter of the wheel = 140 cm
Radius = 70 cm
Circumference = $2 πr$
     $= 2 \times \frac{22}{7} \times 70 = 440 cm$
Speed of the wheel = 72.6 km per hour
Distance covered by the wheel in 1 minute = $\frac{72.6 \times 1000 \times 100}{60}$ = 121000 cm
Number of revolutions made by the wheel in 1 minute = $\frac{Total distance covered}{Circumference}$

                                                                                       $= \frac{121000}{440} = 275$

Hence, the wheel makes 275 revolutions per minute.

Page No 833:

Question 35:

The diameters of the front and rear wheels of a tractor are 80 cm and 2 m respectively. Find the number of revolutions that a rear wheel makes to cover the distance which the front wheel covers is 800 revolutions.

Answer:

Radius of the front wheel = $40 cm = \frac{2}{5} m$

Circumference of the front wheel = $(2 π \times \frac{2}{5}) m = \frac{4 π}{5} m$

Distance covered by the front wheel in 800 revolutions = $(\frac{4 π}{5} \times 800) m = (640 π) m$
Radius of the rear wheel = 1 m
Circumference of the rear wheel = $(2 π \times 1) = 2 π m$

∴ Required number of revolutions = $\frac{Distance covered by the front wheel in 800 revolutions}{Circumference of the rear wheel}$
$= \frac{640 π}{2 π} = 320$

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 833:

Question 41:

In the given figure, ABCD is a trapezium of area 24.5 cm² , If AD || BC, ∠DAB = 90°, AD = 10 cm, BC = 4 cm and ABE is quadrant of a circle, then find the area of the shaded region. [CBSE 2014]

Answer:

Area of trapezium = $\frac{1}{2} (AD + BC) \times AB$
$\Rightarrow 24.5 = \frac{1}{2} (10 + 4) \times AB \Rightarrow AB = 3.5 cm$
Area of shaded region = Area of trapezium ABCD − Area of quadrant ABE
$= 24.5 - \frac{1}{4} π {(AB)}^{2} = 24.5 - \frac{1}{4} \times \frac{22}{7} {(3.5)}^{2} = 24.5 - 9.625 = 14.875 {cm}^{2}$
Hence, the area of shaded region is 14.875 cm²

Page No 834:

Question 42:

ABCD is a field in the shape of a trapezium, AD || BC, ∠ABC = 90° and ∠ADC = 60°. Four sectors are formed with centres A, B, C and D, as shown in the figure. The radius of each sector is 14 m. Find the following: [CBSE 2013C]
(i) total area of the four sectors,
(ii) area of the remaining portion, given that AD = 55 m, BC = 45 m and AB = 30 m.

Answer:

(i) Area of fours sector = Area of sector having central angle 60° + Area of sector having central angle 90° + Area of sector having central angle 90° + Area of sector having central angle 120°
$= \frac{60 °}{360 °} π {(14)}^{2} + \frac{90 °}{360 °} π {(14)}^{2} + \frac{90 °}{360 °} π {(14)}^{2} + \frac{120 °}{360 °} π {(14)}^{2} = (\frac{60 °}{360 °} + \frac{90 °}{360 °} + \frac{90 °}{360 °} + \frac{120 °}{360 °}) π {(14)}^{2} = (\frac{360 °}{360 °}) π {(14)}^{2} = 616 m^{2}$

(ii) Area of the remaining portion = Area of trapezium ABCD − Area of four quadrants
$= \frac{1}{2} (AD + BC) \times AB - 616 = \frac{1}{2} (55 + 45) \times 30 - 616 = 1500 - 616 = 884 m^{2}$

Page No 834:

Question 43:

Find the area of the shaded region in the given figure, where a circular arc of radius 6 cm has been drawn with vertex of an equilateral triangle of side 12 cm as centre and a sector of circle of radius 6 cm with centre B is made. [Use $\sqrt{3} = 1.73, π = 3.14$ ]
[CBSE 2014]

Answer:

In equilateral traingle all the angles are of 60°
∴ ∠ABO = ∠AOB = 60°
Area of the shaded region = (Area of triangle AOB − Area of sector having central angle 60°) + Area of sector having central angle (360° − 60°)
$= \frac{\sqrt{3}}{4} {(AB)}^{2} - \frac{60 °}{360 °} π {(6)}^{2} + \frac{300 °}{360 °} π {(6)}^{2} = \frac{1.73}{4} {(12)}^{2} - \frac{1}{6} \times 3.14 {(6)}^{2} + \frac{5}{6} \times 3.14 {(6)}^{2} = 62.28 - 18.84 + 94.2 = 137.64 {cm}^{2}$
Hence, the area of shaded region is 137.64 cm²

Page No 834:

Question 44:

In the given figure, ABCD is a rectangle with AB = 80 cm and BC = 70 cm, ∠AED = 90° and DE = 42 cm. A semicircle is drawn, taking BC as diameter. Find the area of the shaded region. [CBSE 2014]

Answer:

We know that the opposite sides of a rectangle are equal
AD = BC = 70 cm
In right triangle AED
AE² = AD² − DE²
= (70)² − (42)²
= 4900 − 1764
= 3136
∴ AE² = 3136
⇒ AE = 56
= Area of the shaded region = Area of rectangle − (Area of triangle AED + Area of semicircle)
$= AB \times BC - [\frac{1}{2} \times AE \times DE + \frac{1}{2} π {(\frac{BC}{2})}^{2}] = 80 \times 70 - [\frac{1}{2} \times 56 \times 42 + \frac{1}{2} \times \frac{22}{7} {(\frac{70}{2})}^{2}] = 5600 - 3101 = 2499 {cm}^{2}$
Hence, the area of shaded region is 2499 cm²

Page No 834:

Question 45:

In the given figure, from a rectangular region ABCD with AB = 20 cm, a right triangle AED with AE = 9 cm and DE = 12 cm, is cut off. On the other end, taking BC as diameter, a semicircle is added on outside' the region. Find the area of the shaded region. [Use π = 3.14] [CBSE 2014]

Answer:

In right triangle AED
AD² = AE² + DE²
= (9)² + (12)²
= 81 + 144
= 225
∴ AD² = 225
⇒ AD = 15 cm
We know that the opposite sides of a rectangle are equal
AD = BC = 15 cm
= Area of the shaded region = Area of rectangle − Area of triangle AED + Area of semicircle
$= AB \times BC - \frac{1}{2} \times AE \times DE + \frac{1}{2} π {(\frac{BC}{2})}^{2} = 20 \times 15 - \frac{1}{2} \times 9 \times 12 + \frac{1}{2} \times 3.14 {(\frac{15}{2})}^{2} = 300 - 54 + 88.31 = 334.31 {cm}^{2}$
Hence, the area of shaded region is 334.31 cm²

Page No 834:

Question 46:

In the given figure, O is the centre of the circle with AC = 24 cm, AB = 7 cm and ∠BOD = 90°. Find the area of shaded region.
[Use π = 3.14] [CBSE 2012]

Answer:

In right triangle ABC
BC² = AB² + AC²
= (7)² + (24)²
= 49 + 576
= 625
∴ BC² = 625
⇒ BC = 25
Now, ∠COD + ∠BOD = 180° (Linear pair angles)
⇒∠COD = 180° − 90° = 90°
Now, Area of the shaded region = Area of sector having central angle (360° − 90°) − Area of triangle ABC
$= \frac{270 °}{360 °} π {(\frac{BC}{2})}^{2} - \frac{1}{2} AB \times AC = \frac{3}{4} \times 3.14 {(\frac{25}{2})}^{2} - \frac{1}{2} \times 7 \times 24 = 367.97 - 84 = 283.97 {cm}^{2}$
Hence, the area of shaded region is 283.97 cm²

Page No 835:

Question 47:

In the given figure, a circle is inscribed in an equilateral triangle ABC of side 12 cm. Find the radius of inscribed circle and the area of the shaded region.
[Use $\sqrt{3} = 1.73, π = 3.14$ ] [CBSE 2014]

Answer:

We can find the radius of the incircle by using the formula
$r = \frac{2 \times Area of triangle}{Perimeter of triangle} = \frac{2 \times \frac{\sqrt{3}}{4} \times {(12)}^{2}}{3 \times 12} = 2 \sqrt{3} cm$
Now, area of shaded region = Area of triangle − Area of circle
$= \frac{\sqrt{3}}{4} \times {(12)}^{2} - 3.14 \times {(2 \sqrt{3})}^{2} = 62.28 - 37.68 = 24.6 {cm}^{2}$
Hence, the area of shaded region is 24.6 cm²

Page No 835:

Question 48:

On a circular table cover of radius 42 cm, a design is formed by a girl leaving an equilateral triangle ABC in the middle, as shown in the figure. Find the covered area of the design. [Use $\sqrt{3} = 1.73, π = \frac{22}{7}$ ] [CBSE 2013C]

Answer:

Construction: Join AO and extend it to D on BC.

Radius of the circle, r = 42 cm
∠OCD= 30°

$\cos 30 ° = \frac{DC}{OC} \Rightarrow \frac{\sqrt{3}}{2} = \frac{DC}{42} \Rightarrow DC = 21 \sqrt{3} \Rightarrow BC = 2 \times DC = 42 \sqrt{3} = 72.66 cm \sin 30 ° = \frac{OD}{OC} \Rightarrow \frac{1}{2} = \frac{OD}{42} \Rightarrow OD = 21 cm Now, AD = AO + OD = 42 + 21 = 63 cm$

Area of shaded region = Area of circle − Area of triangle ABC
$= π {(OA)}^{2} - \frac{1}{2} \times AD \times AB = \frac{22}{7} {(42)}^{2} - \frac{1}{2} \times 63 \times 72.66 = 5544 - 2288.79 = 3255.21 {cm}^{2}$

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.

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

Page No 836:

Question 56:

In the given figure, PQ = 24, PR = 7 cm and O is the centre of the circle. Find the area of the shaded region.

Answer:

In the right $∆$ RPQ, we have:
$R Q = \sqrt{R P^{2} + P Q^{2}} = \sqrt{7^{2} + 24^{2}} = \sqrt{49 + 576} = 25 cm$

OR = OQ = 12.5 cm
Now,
Area of the circle = ${πr}^{2}$
$= 3.14 \times 12.5 \times 12.5 = 490.625 sq . cm$

Area of the semicircle = $\frac{490.625}{2} = 245.31 sq . cm$
Area of the triangle $= \frac{1}{2} \times b \times h = \frac{1}{2} \times 7 \times 24 = 84 sq . cm$
Thus, we have:

$Area of the shaded part = Area of the semicircle - Area of the triangle = 245.31 - 84 = 161.31 {cm}^{2}$

Page No 836:

Question 57:

In the given figure, ∆ABC is right-angled at A. Find the area of the shaded region if AB = 6 cm, BC = 10 cm and O is the centre of the incircle of ∆ABC.

Answer:

Using Pythagoras' theorem for triangle ABC, we have:

$C A^{2} + A B^{2} = B C^{2}$

$C A = \sqrt{B C^{2} - A B^{2}} = \sqrt{100 - 36} = \sqrt{64} = 8 cm$

Now, we must find the radius of the incircle. Draw OE, OD and OF perpendicular to AC, AB and BC, respectively.

Consider quadrilateral AEOD.
Here,
$E O = O D (Both are radii .)$

Because the circle is an incircle, AE and AD are tangents to the circle.
$∠ A E O = ∠ A D O = 90 °$

Also,
$∠ A = 90 °$
Therefore, AEOD is a square.
Thus, we can say that $A E = E O = O D = A D = r$ .

$C E = C F = 8 - r B F = B D = 6 - r C F + B F = 10 \Rightarrow (8 - r) + (6 - r) = 10 \Rightarrow 14 - 2 r = 10 \Rightarrow r = 2 cm$

Area of the shaded part = Area of the triangle $-$ Area of the circle
$= \{\frac{1}{2} \times 6 \times 8\} - \{π \times 2 \times 2\} = 24 - 12.56 = 11.44 {cm}^{2}$

Page No 836:

Question 58:

In the given figure, ∆ABC is right-angled at A. Semicircles are drawn on AB, AC and BC as diameters. It is given that AB = 3 cm and AC = 4 cm. Find the area of the shaded region.

Answer:

In triangle $∆$ ABC, we have:
$B C = \sqrt{A B^{2} + A C^{2}} = \sqrt{9 + 16} = \sqrt{25} = 5 cm$

$Ar (shaded part) = Ar (∆ A B C) + Ar (semicircle A P B) + Ar (semicircle A Q C) - Ar (semicircle B A C) = (\frac{1}{2} \times 3 \times 4) + (\frac{1}{2} π \times 1.5 \times 1.5) + (\frac{1}{2} π \times 2 \times 2) - (\frac{1}{2} π \times 2.5 \times 2.5) = \{6 + \frac{1}{2} π (4 + \frac{9}{4} - \frac{25}{4})\} = 6 + 0 = 6 {cm}^{2}$

Page No 836:

Question 59:

PQRS is a diameter of a circle of radius 6 cm. The lengths PQ, QR and RS are equal. Semicircles are drawn with PQ and QS as diameters, as shown in the given figure. If PS = 12 cm, find the perimeter and area of the shaded region.

Answer:

Perimeter (circumference of the circle) = $2 πr$
We know:
Perimeter of a semicircular arc = $πr$
Now,
For the arc PTS, radius is 6 cm.
∴ Circumference of the semicircle PTS = $πr = 6 π cm$

For the arc QES, radius is 4 cm.
∴ Circumference of the semicircle QES = $πr = 4 π cm$

For the arc PBQ, radius is 2 cm.
∴ Circumference of the semicircle PBQ = $πr = 2 π cm$

Now,
Perimeter of the shaded region = $6 π + 4 π + 2 π$
   $= 12 πcm$
   $= 12 \times 3.14 = 37.68 cm$

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

          $= \frac{1}{2} \times 3.14 \times 2 \times 2 = 6.28 {cm}^{2}$

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

$= \frac{1}{2} \times 3.14 \times 6 \times 6 = 56.52 {cm}^{2}$

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

      $= \frac{1}{2} \times 3.14 \times 4 \times 4 = 25.12 {cm}^{2}$

Area of the shaded region = Area of the semicircle PBQ + Area of the semicircle PTS $-$ Area of the semicircle QES
   $= 6.28 + 56.52 - 25.12 = 37.68 {cm}^{2}$

Page No 836:

Question 60:

The inside perimeter of a running track shown in the figure is 400 m. The length of each of the straight portions is 90 m, and the ends are semicircles. If the track is 14 m wide everywhere, find the area of the track. Also, find the length of the outer boundary of the track.

Answer:

Length of the inner curved portion $(400 - 2 \times 90) = 220 m$
∴ Length of each inner curved path = $\frac{220}{2}$ = 110 m
Thus, we have:
$\Rightarrow πr = 110 \Rightarrow \frac{22}{7} r = 110 \Rightarrow r = \frac{110 \times 7}{22} \Rightarrow r = 35 m$

Inner radius = 35 m
Outer radius = (35 + 14) = 49 m
Area of track = {Area of the two rectangles [each $(90 \times 14) m$ ] + Area of the circular ring with R = 49 m and r = 35 m)}
$= (2 \times 90 \times 14) + \frac{22}{7} \times [{(49)}^{2} - {(35)}^{2}] = 2520 + \frac{22}{7} \times (2401 - 1225) = 2520 + \frac{22}{7} \times 1176 = 2520 + 3696 = 6216 m^{2}$

Length of the outer boundary of the track
$= (2 \times 90 + 2 \times \frac{22}{7} \times 49) = 488 m$

Therefore, the length of the outer boundary of the track is 488 m and the area of the track is 6216 sq. m.

Page No 846:

Question 1:

The area of a circle is 38.5 cm². The circumference of the circle is
(a) 6.2 cm
(b) 12.1 cm
(c) 11 cm
(d) 22 cm

Answer:

(d) 22 cm
Let the radius be r cm.
We know:
Area of a circle $= π r^{2} {cm}^{2}$
Thus, we have:
$π r^{2} = 38.5$
$\Rightarrow \frac{22}{7} \times r^{2} = 38.5 \Rightarrow r^{2} = (38.5 \times \frac{7}{22}) \Rightarrow r^{2} = (\frac{385}{10} \times \frac{7}{22}) \Rightarrow r^{2} = \frac{49}{4} \Rightarrow r = \frac{7}{2}$
Now,
Circumference of the circle $= 2 π r$
$= 2 \times \frac{22}{7} \times r = (2 \times \frac{22}{7} \times \frac{7}{2}) = 22 cm$

Page No 846:

Question 2:

The area of a circle is 49 π cm². Its circumference is
(a) 7 π cm
(b) 14 π cm
(c) 21 π cm
(d) 28 π cm

Answer:

(b) 14π cm
Let the radius be r cm.
We know:
Area of a circle $= π r^{2}$
Thus, we have:
$π r^{2} = 49 π \Rightarrow r^{2} = 49 \Rightarrow r = \sqrt{49} \Rightarrow r = 7$

Now,
Circumference of the circle $= 2 πr$
$= (2 \times \frac{22}{7} \times 7) cm = 14 π cm$

Page No 846:

Question 3:

The difference between the circumference and radius of a circle is 37 cm. The area of the circle is
(a) 111 cm²
(b) 184 cm²
(c) 154 cm²
(d) 259 cm²

Answer:

(c) 154 cm²
Let the radius be r cm.
We know:
Circumference of the circle $= 2 πr$
Thus, we have:
$2 πr - r = 37 \Rightarrow r (2 π - 1) = 37 \Rightarrow r (2 \times \frac{22}{7} - 1) = 37 \Rightarrow r (\frac{37}{7}) = 37 \Rightarrow r = (37 \times \frac{7}{37}) \Rightarrow r = 7 cm$

Radius = 7 cm
Now,
Area of the circle $= π r^{2}$
$= (\frac{22}{7} \times 7 \times 7) {cm}^{2} = 154 {cm}^{2}$

Page No 847:

Question 4:

The perimeter of a circular field is 242 m. The area of the field is
(a) 9317 m²
(b) 18634 m²
(c) 4658.5 m²
(d) none of these

Answer:

(c) 4658.5 m²
Let the radius be r m.
We know:
Perimeter of a circle $= 2 π r$
Thus, we have:
$2 π r = 242$
$\Rightarrow 2 \times \frac{22}{7} \times r = 242 \Rightarrow \frac{44}{7} \times r = 242 \Rightarrow r = (242 \times \frac{7}{44}) \Rightarrow r = \frac{77}{2}$

∴ Area of the circle $= π r^{2}$
$= (\frac{22}{7} \times \frac{77}{2} \times \frac{77}{2}) m^{2} = 4658.5 m^{2}$

Page No 847:

Question 5:

On increasing the diameter of a circle by 40%, its area will be increased by
(a) 40%
(b) 80%
(c) 96%
(d) 82%

Answer:

(c) 96%
Let d be the original diameter.
Radius $= \frac{d}{2}$
Thus, we have:
Original area $= π \times {(\frac{d}{2})}^{2}$
$= \frac{π d^{2}}{4}$
New diameter $= 140 % of d$
     $= (\frac{140}{100} \times d) = \frac{7 d}{5}$
Now,
New radius $= \frac{7 d}{5 \times 2}$
   $= \frac{7 d}{10}$
New area $= π \times {(\frac{7 d}{10})}^{2}$
    $= \frac{49 π d^{2}}{10}$
Increase in the area $= (\frac{49 π d^{2}}{10} - \frac{π d^{2}}{4})$
$= \frac{24 π d^{2}}{100} = \frac{6 π d^{2}}{25}$
We have:
Increase in the area $= (\frac{6 π d^{2}}{25} \times \frac{4}{π d^{2}} \times 100) %$
= 96%

Page No 847:

Question 6:

On decreasing the radius of a circle by 30%, its area is decreased by
(a) 30%
(b) 60%
(c) 45%
(d) none of these

Answer:

(d) None of these
Let r be the original radius.
Thus, we have:
Original area $= π r^{2}$
Also,
New radius $= 70 % of r$
$= (\frac{70}{100} \times r) = \frac{7 r}{10}$
New area $= π \times {(\frac{7 r}{10})}^{2}$
$= \frac{49 π r^{2}}{100}$
Decrease in the area $= (π r^{2} - \frac{49 π r^{2}}{100})$
$= \frac{59 π r^{2}}{100}$
Thus, we have:
Decrease in the area $= (\frac{59 π r^{2}}{100} \times \frac{1}{π r^{2}} \times 100) %$
=51%

Page No 847:

Question 7:

The area of a square is the same as the area of a square. Their perimeters are in the ratio
(a) 1 : 1
(b) 2 : π
(c) π : 2
(d) $\sqrt{π} : 2$

Answer:

(d) $\sqrt{π} : 2$
Let a be the side of the square.
We know:
Area of a square $= a^{2}$
Let r be the radius of the circle.
We know:
Area of a circle $= π r^{2}$
Because the area of the square is the same as the area of the circle, we have:
$a^{2} = π r^{2} \Rightarrow \frac{r^{2}}{a^{2}} = \frac{1}{π} \Rightarrow \frac{r}{a} = \frac{1}{\sqrt{π}}$
∴ Ratio of their perimeters $= \frac{2 π r}{4 a} [Because perimeter of the circle is 2 π r and perimeter of the square is 4 a]$

$= \frac{π r}{2 a} = \frac{π}{2} \times \frac{r}{a} = \frac{π}{2} \times \frac{1}{\sqrt{π}} [Since \frac{r}{a} = \frac{1}{\sqrt{π}}] = \frac{\sqrt{π}}{2} = \sqrt{π} : 2$

Page No 847:

Question 8:

The circumference of a circle is equal to the sum of the circumference of two circles having diameters 36 cm and 20 cm. The radius of the new circle is
(a) 16 cm
(b) 28 cm
(c) 42 cm
(d) 56 cm

Answer:

(b) 28 cm
Let r cm be the radius of the new circle.
We know:
Circumference of the new circle = Circumference of the circle with diameter 36 cm + Circumference of the circle with diameter 20 cm
Thus, we have:
$2 π r = 2 π r_{1} + 2 π r_{2}$
$\Rightarrow 2 π r = (2 π \times 18) + (2 π \times 10) \Rightarrow 2 π r = 2 π \times (18 + 10) \Rightarrow 2 π r = (2 π \times 28) \Rightarrow 2 π r = (2 \times \frac{22}{7} \times 28)$
$\Rightarrow 2 π r = 176 \Rightarrow 2 \times \frac{22}{7} \times r = 176 \Rightarrow r = (176 \times \frac{7}{44}) \Rightarrow r = 28 cm$

Page No 847:

Question 9:

The area of circle is equal to the sum of the areas of two circles of radii 24 cm and 7 cm. The diameter of the new circle is
(a) 25 cm
(b) 31 cm
(c) 50 cm
(d) 62 cm

Answer:

(c) 50 cm
Let r cm be the radius of the new circle.
Now,
Area of the new circle = Area of the circle with radius 24 cm + Area of the circle with radius 7 cm
Thus, we have:
$π r^{2} = π {r_{1}}^{2} + π {r_{2}}^{2}$
$\Rightarrow π r^{2} = [π \times {(24)}^{2} + π \times {(7)}^{2}] {cm}^{2} \Rightarrow π r^{2} = [π \times 576 + π \times 49] {cm}^{2} \Rightarrow π r^{2} = π \times (576 + 49) {cm}^{2} \Rightarrow r^{2} = 625 π {cm}^{2} \Rightarrow r^{2} = 625 \Rightarrow r = 25$

∴ Diameter of the new circle $= (25 \times 2) cm$
= 50 cm

Page No 847:

Question 10:

If the perimeter of a square is equal to the circumference of a circle then the ratio of their areas is
(a) 4 : π
(b) π : 4
(c) π : 7
(d) 7 : π

Answer:

Let the side of the square be a and the radius of the circle be r.
Now, Perimeter of circle = Circumference of the circle
$\Rightarrow 4 a = 2 π r \Rightarrow \frac{a}{r} = \frac{π}{2} Now, \frac{Area of square}{Area of circle} = \frac{a^{2}}{π r^{2}} = \frac{1}{π} \times {(\frac{a}{r})}^{2} = \frac{1}{π} \times \frac{π^{2}}{4} = \frac{π}{4}$
Hence, the correct answer is option (b).

Page No 847:

Question 11:

If the sum of the areas of two circles with radii R₁ and R₂ is equal to the area of a circle of radius R, then
(a) $R_{1} + R_{2} = R$
(b) $R_{1} + R_{2} < R$
(c) $R_{1}^{2} + R_{2}^{2} < R^{2}$
(d) $R_{1}^{2} + R_{2}^{2} = R^{2}$

Answer:

(d) $R_{1}^{2} + R_{2}^{2} = R^{2}$
Because the sum of the areas of two circles with radii $R_{1} and R_{2}$ is equal to the area of a circle with radius R, we have:
$π {R_{1}}^{2} + π {R_{2}}^{2} = π R^{2} \Rightarrow π ({R_{1}}^{2} + {R_{2}}^{2}) = π R^{2} \Rightarrow {R_{1}}^{2} + {R_{2}}^{2} = R^{2}$

Page No 847:

Question 12:

If the sum of the circumferences of two circles with radii R₁ and R₂ is equal to the circumference of a circle of radius R, then
(a) $R_{1} + R_{2} = R$
(b) $R_{1} + R_{2} > R$
(c) $R_{1} + R_{2} < R$
(d) none of these

Answer:

(a) $R_{1} + R_{2} = R$
Because the sum of the circumferences of two circles with radii $R_{1} and R_{2}$ is equal to the circumference of a circle with radius R, we have:

$2 π R_{1} + 2 π R_{2} = 2 π R \Rightarrow 2 π (R_{1} + R_{2}) = 2 π R \Rightarrow R_{1} + R_{2} = R$

Page No 847:

Question 13:

If the circumference of a circle and the perimeter of a square are equal, then
(a) area of the circle = area of the square
(b) (area of the circle) > (area of the square)
(c) (area of the circle) < (area of the square)
(d) none of these

Answer:

(b) Area of the circle > Area of the square
Let r be the radius of the circle.
We know:
Circumference of the circle $= 2 π r$
Now,
Let a be the side of the square.
We know:
Perimeter of the square = 4a
Now,
$2 π r = 4 a \Rightarrow r = \frac{4 a}{2 π}$
∴ Area of the circle $= π r^{2}$
$= π \times {(\frac{4 a}{2 π})}^{2} = π \times \frac{16 a^{2}}{4 π^{2}} = \frac{4 a^{2}}{π} = \frac{4 \times 7 a^{2}}{22} = \frac{14 a^{2}}{11}$

Also,
Area of the square $= a^{2}$
Clearly, $\frac{14 a^{2}}{11} > a$ ².
∴ Area of the circle > Area of the square

Page No 847:

Question 14:

The radii of two concentric circles are 19 cm and 16 cm respectively. The area of the ring enclosed by these circles is
(a) 320 cm²
(b) 330 cm²
(c) 332 cm²
(d) 340 cm²

Answer:

(b) 330 cm²
Let:
R = 19 cm and r = 16 cm
Thus, we have:
Area of the ring $= π (R^{2} - r^{2})$
$= π (R + r) (R - r) = |\frac{22}{7} \times (19 + 16) \times (19 - 16)| {cm}^{2} = (\frac{22}{7} \times 35 \times 3) {cm}^{2} = 330 {cm}^{2}$

Page No 848:

Question 15:

The areas of two concentric circles are 1386 cm² and 962.5 cm². The width of the ring is
(a) 2.8 cm
(b) 3.5 cm
(c) 4.2 cm
(d) 3.8 cm

Answer:

(b) 3.5 cm
Let r cm and R cm be the radii of two concentric circles.
Thus, we have:
${πR}^{2} = 1386$
$\Rightarrow \frac{22}{7} \times R^{2} = 1386 \Rightarrow R^{2} = (1386 \times \frac{7}{22}) {cm}^{2} \Rightarrow R^{2} = 441 {cm}^{2} \Rightarrow R = 21 cm$

Also,
${πr}^{2} = 962.2 \Rightarrow \frac{22}{7} \times r^{2} = 962.2 \Rightarrow r^{2} = (962.2 \times \frac{7}{22}) {cm}^{2} \Rightarrow r^{2} = (\frac{962.2}{10} \times \frac{7}{22}) {cm}^{2} \Rightarrow r^{2} = \frac{1225}{4} {cm}^{2} \Rightarrow r = \frac{35}{2} cm$
∴ Width of the ring $= (R - r)$
$= (21 - \frac{35}{2}) cm = \frac{7}{2} cm = 3.5 cm$

Page No 848:

Question 16:

The circumferences of two circles are in the ratio 3 : 4. The ratio of their areas is
(a) 3: 4
(b) 4 : 3
(c) 9 : 16
(d) 16: 9

Answer:

Let the the radii of the two circles be r and R, the circumferences of the circles be c and C and the areas of the two circles be a and A.
Now,
$\frac{c}{C} = \frac{3}{4} \Rightarrow \frac{2 π r}{2 π R} = \frac{3}{4} \Rightarrow \frac{r}{R} = \frac{3}{4}$
Now, the ratio between their areas is given by
$\frac{a}{A} = \frac{π r^{2}}{π R^{2}} = {(\frac{r}{R})}^{2} = {(\frac{3}{4})}^{2} = \frac{9}{16}$
Hence, the correct answer is option (c).

Page No 848:

Question 17:

The areas of two circles are in the ratio 9 : 4. The ratio of their circumferences is
(a) 3 : 2
(b) 4 : 9
(c) 2 : 3
(d) 81 : 16

Answer:

Let the the radii of the two circles be r and R, the circumferences of the circles be c and C and the areas of the two circles be a and A.
Now,
$\frac{a}{A} = \frac{9}{4} \Rightarrow \frac{π r^{2}}{π R^{2}} = {(\frac{3}{2})}^{2} \Rightarrow \frac{r}{R} = \frac{3}{2}$
Now, the ratio between their circumferences is given by
$\frac{c}{C} = \frac{2 π r}{2 π R} = \frac{r}{R} = \frac{3}{2}$
Hence, the correct answer is option (a)

Page No 848:

Question 18:

The radius of a wheel is 0.25 m. How many revolutions will it make in covering 11 km?
(a) 2800
(b) 4000
(c) 5500
(d) 7000

Answer:

(d) 7000
Distance covered in 1 revolution $= 2 π r$
$= (2 \times \frac{22}{7} \times 0.25) m = (2 \times \frac{22}{7} \times \frac{25}{100}) m = \frac{11}{7} m$
Number of revolutions taken to cover 11 km $= (11 \times 1000 \times \frac{7}{11})$
= 7000

Page No 848:

Question 19:

The diameter of a wheel is 40 cm. How many revolutions will it make in covering 176 m?
(a) 140
(b) 150
(c) 160
(d) 166

Answer:

(a) 140
Distance covered by the wheel in 1 revolution $= π d$
$= (\frac{22}{7} \times 40) cm = \frac{880}{7} cm = \frac{880}{7 \times 100} m$

Number of revolutions required to cover 176 m $= (\frac{176}{\frac{880}{7 \times 100}})$
$= (176 \times 100 \times \frac{7}{880})$
=140

Page No 848:

Question 20:

In making 1000 revolutions, a wheel covers 88 km. The diameter of the wheel is
(a) 14 m
(b) 24 m
(c) 28 m
(d) 40 m

Answer:

(c) 28 m
Distance covered by the wheel in 1 revolution $= (\frac{88 \times 1000}{1000}) m$
= 88 m
We have:
Circumference of the wheel = 88 m
Now, let the diameter of the wheel be d m.
Thus, we have:

$π d = 88 \Rightarrow \frac{22}{7} \times d = 88 \Rightarrow d = (88 \times \frac{7}{22}) \Rightarrow d = 28 m$

Page No 848:

Question 21:

The area of the sector of angle $θ °$ of a circle with radius R is
(a) $\frac{2 π R θ}{180}$
(b) $\frac{π R^{2} θ}{180}$
(c) $\frac{2 π R θ}{360}$
(d) $\frac{π R^{2} θ}{360}$

Answer:

(d) $\frac{π R^{2} θ}{360}$

Page No 848:

Question 22:

The length of an arc of the sector of angle $θ °$ of a circle with radius R is
(a) $\frac{2 π R θ}{180}$
(b) $\frac{2 π R θ}{360}$
(c) $\frac{π R^{2} θ}{180}$
(d) $\frac{π R^{2} θ}{360}$

Answer:

(b) $\frac{2 π R θ}{360}$

Page No 848:

Question 23:

The length of the minute hand of a clock is 21 cm. The area swept by the minute hand in 10 minutes is [CBSE 2012]
(a) 231 cm²
(b) 210 cm²
(c) 126 cm²
(d) 252 cm²

Answer:

Angle subtends by the minute hand in 1 minute = 6^∘
∴ Angle subtends by the minute hand in 10 minutes = 60^∘
Now,
Area of the sector $= \frac{θ}{360 °} π r^{2} = \frac{60 °}{360 °} \times \frac{22}{7} {(21)}^{2} = 231 {cm}^{2}$
Hence, the correct answer is option (a).

Page No 848:

Question 24:

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 7:

The minute hand of a clock is 12 cm long. Find the area swept by in it 35 minutes.

Answer:

Angle described by the minute hand in 60 minutes $= 360 °$
Angle described by the minute hand in 35 minutes $= (\frac{360}{60} \times 35) °$
$= 210 °$
Now,
$r = 12 cm and θ = 210 °$
∴ Required area swept by the minute hand in 35 minutes = Area of the sector with $r = 12 cm and θ = 210 °$
$= \frac{{πr}^{2} θ}{360} = (\frac{22}{7} \times 12 \times 12 \times \frac{210}{360}) {cm}^{2} = 264 {cm}^{2}$

Page No 852:

Question 8:

The perimeter of a sector of a circle of radius 5.6 cm is 27.2 cm. Find the area of the sector.

Answer:

Let O be the centre of the circle with radius 5.6 cm and OACB be its sector with perimeter 27.2 cm.
Thus, we have:
$O A + O B + a r c A B = 27.2 \Rightarrow 5.6 + 5.6 + a r c A B = 27.2 \Rightarrow a r c A B = 16 cm$
Now,
Area of the sector OACBO $= (\frac{1}{2} \times Radius \times l) square units$
$= (\frac{1}{2} 5.6 \times 16) {cm}^{2} = 44.8 {cm}^{2}$

Page No 852:

Question 9:

A chord of a circle of radius 14 cm a makes a right angle at the centre. Find the area of the sector.

Answer:

Let r cm be the radius of the circle and $θ$ be the angle.
We have:
$r = 14 cm and θ = 90 °$
Area of the sector $= \frac{π r^{2} θ}{360}$
$= (\frac{22}{7} \times 14 \times 14 \times \frac{90}{360}) {cm}^{2} = 154 {cm}^{2}$

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:

Area of the shaded region = (Area of the sector with $r = 7 cm and θ = 30 °$ ) $-$ (Area of the sector with $r = 3.5 cm and θ = 30 °$ )
$= |(\frac{22}{7} \times 7 \times 7 \times \frac{30}{360}) - (\frac{22}{7} \times 3.5 \times 3.5 \times \frac{30}{360})| {cm}^{2} = (\frac{77}{6} - \frac{77}{24}) {cm}^{2} = \frac{77}{8} {cm}^{2} = 9.625 {cm}^{2}$

Page No 853:

Question 11:

A wire when bent in the form of an equilateral triangle encloses an area of $121 \sqrt{3} {cm}^{2}$ . If the same wire is bent into the form of a circle, what will be the area of the circle?

Answer:

Let a cm be the side of the equilateral triangle.
Now,
Area of the equilateral triangle $= \frac{\sqrt{3}}{4} a^{2}$
We have:
$\frac{\sqrt{3}}{4} a^{2} = 121 \sqrt{3} \Rightarrow \frac{a^{2}}{4} = 121 \Rightarrow a^{2} = 484 \Rightarrow a = 22$

Perimeter of the triangle = Circumference of the circle
Perimeter of the triangle = (22 + 22 + 22) cm
= 66 cm
Now, let r cm be the radius of the circle.
We know:
Circumference of the circle $= 2 π r$
$\Rightarrow 2 \times \frac{22}{7} \times r = 66 \Rightarrow r = (66 \times \frac{7}{44}) cm \Rightarrow r = \frac{21}{2} cm$
Also,
Area of the circle $= π r^{2}$
$= (\frac{22}{7} \times \frac{21}{2} \times \frac{21}{2}) {cm}^{2} = \frac{693}{2} {cm}^{2} = 346.5 {cm}^{2}$

Page No 853:

Question 12:

The wheel of a cart is making 5 revolutions per second. If the diameter of the wheel is 84 cm, find its speed in km per hour.

Answer:

Distance covered in 1 revolution $= π \times d$
      $= (\frac{22}{7} \times 84) cm = 264 cm$
Distance covered in 1 second $= (5 \times 264) cm$
     = 1320 cm
Distance covered in 1 hour $= (60 \times 60 \times 1320) cm$

   $= 4752000 cm = (\frac{4752000}{1000 \times 100}) km = 47.52 km$

Page No 853:

Question 13:

OACB is a quadrant of a circle with centre O and its radius is 3.5 cm. If OD = 2 cm. find the area of (i) quadrant OACB (ii) 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 $∆ A B C$ )
$= |(3.14 \times 4 \times 4) - (\frac{\sqrt{3}}{4} \times 4 \sqrt{3} \times 4 \sqrt{3})| {cm}^{2} = |50.24 - (12 \times 1.73)| {cm}^{2} = (50.24 - 20.76) {cm}^{2} = 29.48 {cm}^{2}$

Page No 853:

Question 16:

The minute hand of a clock is 12 cm long. Find the area of the face of the clock described by the minute hand in 35 minutes.

Answer:

Angle described by the minute hand in 60 minutes $= 360 °$
Angle described by the minute hand in 35 minutes $= (\frac{360}{60} \times 35) °$
$= 210 °$
Now,
$r = 12 cm and θ = 210 °$
∴ Required area described by the minute hand in 35 minutes = Area of the sector where $r = 12 cm and θ = 210 °$
$= \frac{{πr}^{2} θ}{360} = (\frac{22}{7} \times 12 \times 12 \times \frac{210}{360}) {cm}^{2} = 264 {cm}^{2}$

Page No 853:

Question 17:

A racetrack is in the form of a ring whose inner circumference is 352 m and outer circumference is 396 m. Find the width and the area of the track.

Answer:

Let r m and R m be the inner and outer boundaries, respectively.
Thus, we have:

$2 π r = 352 \Rightarrow r = \frac{352}{2 π} Also, 2 πR = 396 \Rightarrow R = \frac{396}{2 π}$

Width of the track $= (R - r)$
$= (\frac{396}{2 π} - \frac{352}{2 π}) m = \frac{1}{2 π} (396 - 352) m = (\frac{1}{2} \times \frac{7}{22} \times 44) m = 7 m$

Area of the track $= π (R^{2} - r^{2})$
$= π (R + r) (R - r) = [π (\frac{396}{2 π} + \frac{352}{2 π}) \times (\frac{396}{2 π} - \frac{352}{2 π})] m^{2} = (π \times \frac{748}{2 π} \times 7) m^{2} = \frac{748}{2} \times 7 m^{2} = 2618 m^{2}$

Page No 853:

Question 18:

A chord of a circle of radius 30 cm makes an angle of 60° at the centre of the circle. Find the area of the minor and major segments.

Answer:

Let AB be the chord of a circle with centre O and radius 30 cm such that $∠ A O B = 60 °$ .
Area of the sector OACBO $= \frac{π r^{2} θ}{360}$
     $= (3.14 \times 30 \times 30 \times \frac{60}{360}) {cm}^{2} = 471 {cm}^{2}$
Area of $∆ O A B$ $= \frac{1}{2} r^{2} \sin θ$
   $= (\frac{1}{2} \times 30 \times 30 \times \sin 60 °) {cm}^{2} = (225 \times 1.732) {cm}^{2} = 389.7 {cm}^{2}$

Area of the minor segment = (Area of the sector OACBO) $-$ (Area of $∆ O A B$ )
            $= (471 - 389.7) {cm}^{2} = 81.3 {cm}^{2}$

Area of the major segment = (Area of the circle) $-$ (Area of the minor segment)
            $= |(3.14 \times 30 \times 30) - 81.3| {cm}^{2} = 2744.7 {cm}^{2}$

Page No 854:

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 $r = 25 m and θ = 90 °$ )

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

Page No 854:

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

View NCERT Solutions for all chapters of Class 10

Login or Create a free account