Rs Aggarwal 2018 Solutions for Class 10 Math Chapter 18 Area Of Circle, Sector And Segment are provided here with simple step-by-step explanations. These solutions for Area Of Circle, Sector And Segment are extremely popular among Class 10 students for Math Area Of Circle, Sector And Segment 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 18 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.

#### Question 1:

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

#### Answer:

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

Now,

Now,
Hence, the circumference of the circle is 44 cm.

#### 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,

Now, Area of quadrant =
Hence, the area of the quadrant of the circle is $\frac{77}{8}$ cm2.

#### 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

Hence, the diameter of the of the circle is 26 cm.

#### 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

Hence, the diameter of the of the circle is 8 cm.

#### 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.

#### 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
Length of arc =
Hence, the length of the arc of the circle is 22 cm.

#### 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

Hence, the diameter of the circle is 10 cm.

#### Question 8:

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

#### Answer:

Let the radius of the circle be r.
​Now,

Now, Area of circle=
Hence, the area of the circle is 16π cm2.

#### 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

Hence, the perimeter of a semicircular protractor is 36 cm.

#### 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

Hence, the radius of the circle is 2 units.

#### 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

Hence, the radius of the required circle is 28 cm.

#### 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

Hence, the radius of the circle is 10 cm.

#### 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
Now, Area of sector =
Hence, the area of the sector of the circle is 9.42 cm2.

#### 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
Length of arc =
Hence, the length of the arc of the circle is 22 cm.

#### 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,
$\frac{c}{C}=\frac{2}{3}\phantom{\rule{0ex}{0ex}}⇒\frac{2\mathrm{\pi }r}{2\mathrm{\pi }R}=\frac{2}{3}\phantom{\rule{0ex}{0ex}}⇒\frac{r}{R}=\frac{2}{3}$
Now, the ratio between their areas is given by
$\frac{a}{A}=\frac{\mathrm{\pi }{r}^{2}}{\mathrm{\pi }{R}^{2}}\phantom{\rule{0ex}{0ex}}={\left(\frac{r}{R}\right)}^{2}\phantom{\rule{0ex}{0ex}}={\left(\frac{2}{3}\right)}^{2}\phantom{\rule{0ex}{0ex}}=\frac{4}{9}$
Hence, the ratio between their areas is 4 : 9.

#### 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,
$\frac{a}{A}=\frac{4}{9}\phantom{\rule{0ex}{0ex}}⇒\frac{\mathrm{\pi }{r}^{2}}{\mathrm{\pi }{R}^{2}}={\left(\frac{2}{3}\right)}^{2}\phantom{\rule{0ex}{0ex}}⇒\frac{r}{R}=\frac{2}{3}$
Now, the ratio between their circumferences is given by
$\frac{c}{C}=\frac{2\mathrm{\pi }r}{2\mathrm{\pi }R}\phantom{\rule{0ex}{0ex}}=\frac{r}{R}\phantom{\rule{0ex}{0ex}}=\frac{2}{3}$
Hence, the ratio between their circumferences is 2 : 3.

#### 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.
$\therefore \sqrt{2}a=2r\phantom{\rule{0ex}{0ex}}⇒a=\sqrt{2}r$
Now,

Hence, the ratio of the areas of the circle and the square is π : 2

#### 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,

We have
Area of sector =
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

#### 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,
$\theta ={30}^{\circ }$

Now,
Length of the arc =$\frac{2\mathrm{\pi r\theta }}{360}$

∴ Length of the pendulum = 16.8 cm

#### 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 = $\frac{360}{60}×20={120}^{\circ }$

We have:

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

#### 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$=\frac{{\mathrm{\pi r}}^{2}\mathrm{\theta }}{360}$

∴ Radius of the circle = 6 cm

#### 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 $=\frac{{\mathrm{\pi r}}^{2}\mathrm{\theta }}{360}$
$⇒69.3=\frac{22}{7}×10.5×10.5×\frac{\theta }{360}\phantom{\rule{0ex}{0ex}}⇒\theta =\frac{69.3×7×360}{22×10.5×10.5}\phantom{\rule{0ex}{0ex}}⇒\theta ={72}^{\circ }$

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

#### 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

Now,
Area of the sector OACBO = $\frac{1}{2}×\mathrm{Radius}×\mathrm{Arc}$

​

#### 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 $=\frac{2\mathrm{\pi r\theta }}{360}$

$⇒44=2×\frac{22}{7}×17.5×\frac{\theta }{360}\phantom{\rule{0ex}{0ex}}⇒\theta =\frac{44×7×360}{44×17.5}\phantom{\rule{0ex}{0ex}}⇒\theta ={144}^{\circ }$
Also,
Area of the sector =$\frac{{\mathrm{\pi r}}^{2}\mathrm{\theta }}{360}$

#### 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

Hence, the area of the remaining cardboard is 21 cm2

#### 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 $\mathrm{\pi }$ = 3.14.]

#### Answer:

Area of the square ABCD =

Area of the circle = ${\mathrm{\pi r}}^{2}$
Radius = 1 cm

Area of the quadrant of one circle = $\frac{1}{4}{\mathrm{\pi r}}^{2}$

Area of the quadrants of four circles =
Area of the shaded region = Area of the square $-$ Area of the circle $-$ Area of the quadrants of four circles
=

#### 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 =
∴ Area of remaining paper = Area of rectangular sheet −Area of semicircular portion

Hence, the area of the remaining paper is 812 cm2

#### 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

Hence, the area of the shaded region is 10.5 cm2

#### 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

Hence, the area of the shaded region is 77 cm2.

#### 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

Hence, the area of the shaded portion is $\frac{77}{8}$ cm2.

#### 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)

Hence, the area of the shaded region is 42 cm2.

#### 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
Perimeter of the top of the table = Length of the major arc AB + Length of OA + Length of OB

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

#### 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

Hence, the area of the shaded portion is 28 cm2.

#### 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 = $\frac{1}{2}×b×h$
=

Area of the sector AOC  =$\frac{\theta }{360}×\mathrm{\pi }×{r}^{2}$

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

#### 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

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

#### 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

Hence, the required area is 56 cm2 .

#### 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

(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)

#### 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

Hence, the perimeter of shaded region is 31.4 cm.

#### 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

Hence, the area of the shaded region is 228 cm2.

#### 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

Area of the shaded portion = Area of semicircle AQO + Area of semicircle APB

Hence, the area of the shaded portion is 96.25 cm2.

#### 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,

Now,
Area of quadrant =
Hence, the area of the quadrant of the circle is 38.5 cm2.

#### 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 =

Area of the circles =

Area of the shaded region = Area of the square $-$ Area of four circles

#### 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)
Area of the shaded region = Area of circle − Area of rectangle OABC

Hence, the area of the shaded region is 30.57 cm2.

#### 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 =

Perimeter of the square = $4×\mathrm{Side}$
Perimeter of the square = $4×22$
= 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:
​

Area of the circle = ${\mathrm{\pi r}}^{2}$

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

#### 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,

Hence, the area of the square ABCD is 2r2 sq units.

#### 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 $\mathrm{\pi }=\frac{22}{7}$].

#### Answer:

Let the radius of the circle be r.
​Now,

Now,
Area of field =
Cost of ploughing = Rate ⨯ Area of field = 0.5 ⨯ 3850 = Rs 1925
Hence, the cost of ploughing the field is Rs 1925.

#### 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 = $l×b$

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 = $\mathrm{\pi }{r}^{2}$

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

#### 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.

#### 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 =
Area of the smaller circle = ${\mathrm{\pi r}}^{2}$

Radius of the larger circle =
Area of the larger circle = ${\mathrm{\pi r}}^{2}$

∴ Area of the shaded region = Area of the larger circle $-$ Area of the smaller circle
= 2291.14 $-$ 1521.14
= 770 cm2

#### 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

Hence, the area of the shaded region is 6.125 cm2 .

#### 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

Area of major segment APB = Area of circle − Area of minor segment

Hence, the area of major segment is 3500 cm2

#### 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 = $2\mathrm{\pi }r$

Also,
Area of the circle = $\pi {r}^{2}$

#### Question 2:

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

#### Answer:

Let the radius of the circle be r.
​Now,
$\mathrm{Area}=98.56\phantom{\rule{0ex}{0ex}}⇒\mathrm{\pi }{r}^{2}=98.56\phantom{\rule{0ex}{0ex}}⇒\frac{22}{7}×{r}^{2}=98.56\phantom{\rule{0ex}{0ex}}⇒r=5.6$
Now,
Circumference =
Hence, the circumference of the circle is 35.2 cm.

#### 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

∴ Circumference = Diameter + 45 = 2(10.5) + 45 = 66 cm
Hence, the circumference of the circle is 66 cm.

#### 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 = ${\mathrm{Side}}^{2}$

Perimeter of the square = $4×\mathrm{Side}$
Perimeter of the square = $4×22$
= 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:
​

Area of the circle = ${\mathrm{\pi r}}^{2}$

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

#### Question 5:

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

#### Answer:

Length of the wire
Now, let the radius of the circle be r cm.
We know:
Circumference of the circle = Length of the wire

Thus, we have:
Area of the circle =$\mathrm{\pi }{r}^{\mathit{2}}$

Area enclosed by the circle = 346.5 cm2

#### 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

Now, Area of park
Hence, the area of the park is 693 m2 .

#### 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 r2 cm.
Now,
Sum of the radii of the two circles = 7 cm

Difference of the circumferences of the two circles = 88 cm

Adding (i) and (ii), we get:
$2{r}_{1}=\frac{91}{11}\phantom{\rule{0ex}{0ex}}{r}_{1}=\frac{91}{22}$

∴ Circumference of the first circle = $2{\mathrm{\pi r}}_{1}$

Also,
${r}_{1}-{r}_{2}=\frac{14}{11}\phantom{\rule{0ex}{0ex}}\frac{91}{22}-{r}_{2}=\frac{14}{11}\phantom{\rule{0ex}{0ex}}\frac{91}{22}-\frac{14}{11}={r}_{2}\phantom{\rule{0ex}{0ex}}{r}_{2}=\frac{63}{22}$

∴ Circumference of the second circle = $2{\mathrm{\pi r}}_{2}$

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

#### 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:

Now,
Area of the outer ring = $\mathrm{\pi }{{r}_{1}}^{2}$

Area of the inner ring = $\mathrm{\pi }{{r}_{2}}^{2}$

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

#### 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 = $\pi {R}^{2}-\pi {r}^{2}$

∴ Area of the path = 1056 m2

#### 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 = $2\mathrm{\pi }R$
$⇒396=2×\frac{22}{7}×R\phantom{\rule{0ex}{0ex}}⇒R=\frac{396×7}{44}\phantom{\rule{0ex}{0ex}}⇒R=63$

Circumference of the inner track = $2\mathrm{\pi }r$
$⇒352=2×\frac{22}{7}×r\phantom{\rule{0ex}{0ex}}⇒r=\frac{352×7}{44}\phantom{\rule{0ex}{0ex}}⇒r=56$

Width of the track = Radius of the outer track $-$ Radius of the inner track

Area of the outer circle = $\mathrm{\pi }{R}^{2}$

Area of the inner circle = $\mathrm{\pi }{R}^{2}$

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

#### 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,

Area of the sector =$\frac{{\mathrm{\pi r}}^{2}\theta }{360}$

#### 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 $=\frac{{\mathrm{\pi r}}^{2}\mathrm{\theta }}{360}$
$⇒69.3=\frac{22}{7}×10.5×10.5×\frac{\theta }{360}\phantom{\rule{0ex}{0ex}}⇒\theta =\frac{69.3×7×360}{22×10.5×10.5}\phantom{\rule{0ex}{0ex}}⇒\theta ={72}^{\circ }$

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

#### 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
$\theta ={54}^{\circ }$
Radius = ?
Circumference=?
We know:
Length of the arc $=\frac{2\mathrm{\pi r\theta }}{360}$

Circumference = 110 cm

Now,
Area of the circle =${\mathrm{\pi r}}^{2}$

#### 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

Area of major segment APB = Area of circle − Area of minor segment

Hence, the area of major segment is 140 cm2 .

#### 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:
$\angle O=\angle A=\angle B=60°$

Length of the arc ACB:

Length of the arc ADB:

Now,
Area of the minor segment:

#### 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.

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

Now,

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

Thus, we have:
Area of $∆$OAB =

Area of the minor segment = Area of the sector $-$ Area of the triangle

Area of the major segment = Area of the circle $-$ Area of the minor segment

#### Question 17:

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

#### Answer:

Area of the triangle = $\frac{1}{2}{R}^{2}\mathrm{sin}\theta$
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:

Area of the minor segment = Area of the sector $-$ Area of the triangle
=

Area of the major segment = Area of the circle $-$ Area of the minor segment

#### 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 =

Area of the sector OACBO =

Area of the minor segment = Area of the sector $-$ Area of the triangle
=

Area of the major segment = Area of the circle $-$ Area of the minor segment

#### 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 =

Circumference =

Using the given data, we get:

∴ Area of the sector corresponding to the major arc =

#### 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 =

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 =

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

#### 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,

Now,
Area of quadrant =
Hence, the area of the quadrant of the circle is 154 cm2.

#### 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

#### 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

Total area of the field =

Area left ungrazed = Area of the field $-$ Area of the grazed field
=

#### 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 =$\frac{\sqrt{3}}{4}×\left(\mathrm{Side}{\right)}^{2}$

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$°$

=$\frac{\theta }{360}×\mathrm{\pi }{r}^{\mathit{2}}$

Area of the field the horse cannot graze = Area of the equilateral triangle $-$ Area of the field the horse can graze

#### 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 =

Area that each cow grazes =  $\frac{1}{4}×\mathrm{\pi }×{r}^{2}$

Total area grazed =

Now,
Area left ungrazed = Area of the square $-$ Grazed area
=

#### 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:
$OP=PQ=QR=RO$

Now, consider $∆QOP$.

Therefore, $∆QOP$ is equilateral.

Similarly, $∆QOR$ is also equilateral and .

OQ = 8 cm

Hence, the radius of the circle is 8 cm.

#### 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 = ${\mathrm{\pi r}}^{2}$

(ii) If a circle is circumscribed in a square, then the diagonal of the square is equal to the diameter of the circle.

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

Now,
Area of the circumscribed circle = ${\mathrm{\pi r}}^{2}$

#### 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 =

=

Ratio of the area of the circle to that of the square:
$=\frac{\pi \frac{{d}^{2}}{4}}{\frac{{d}^{2}}{2}}\phantom{\rule{0ex}{0ex}}=\frac{\mathrm{\pi }}{2}$
Thus, the ratio of the area of the circle to that of the square is $\mathrm{\pi }:2$.

#### 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 ${\mathrm{cm}}^{2}$
We know:
Area of the circle =$\pi {r}^{2}$
$⇒154=\frac{22}{7}{r}^{2}\phantom{\rule{0ex}{0ex}}⇒\frac{154×7}{22}={r}^{2}\phantom{\rule{0ex}{0ex}}⇒{r}^{2}=49\phantom{\rule{0ex}{0ex}}⇒r=7$
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:

$⇒h=3r\phantom{\rule{0ex}{0ex}}⇒h=21$

Let each side of the triangle be a cm.

∴ Perimeter of the triangle = 3a
​

#### 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 = $2\mathrm{\pi r}$

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 = $\frac{19800}{2.64}=7500$

#### 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 = $2\mathrm{\pi r}$

Distance covered by the wheel in 1 revolution = 13.2 m
Distance covered by the wheel in 75 revolutions =

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 = $\frac{990}{1000}×60$
=

#### 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 =
∴ Distance covered by the wheel in 1 revolution

Now,
Circumference of the wheel = 198 cm
$⇒2\pi r=198\phantom{\rule{0ex}{0ex}}⇒2×\frac{22}{7}×r=198\phantom{\rule{0ex}{0ex}}⇒r=\frac{198×7}{44}\phantom{\rule{0ex}{0ex}}⇒r=31.5\mathrm{cm}$

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

#### 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 = $2\mathrm{\pi r}$

Distance covered by the wheel in 1 revolution =
∴ Distance covered by the wheel in 140 revolutions =

Now,
Distance covered by the wheel in 1 minute = Distance covered by the wheel in 140 revolutions =

∴ Distance covered by the wheel in 1 hour =

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

#### 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 = $2\mathrm{\pi r}$

Speed of the wheel = 72.6 km per hour
Distance covered by the wheel in 1 minute = $\frac{72.6×1000×100}{60}$ = 121000 cm
​Number of revolutions made by the wheel in 1 minute =

$=\frac{121000}{440}\phantom{\rule{0ex}{0ex}}=275$

Hence, the wheel makes 275 revolutions per minute.

#### 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 =

Circumference of the front wheel =

Distance covered by the front wheel in 800 revolutions =
Radius of the rear wheel = 1 m
Circumference of the rear wheel =

∴ Required number of revolutions =
$=\frac{640\mathrm{\pi }}{2\mathrm{\pi }}\phantom{\rule{0ex}{0ex}}=320$

#### 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 = $\frac{1}{4}{\mathrm{\pi r}}^{2}$

Area of the quadrants of four circles =

Now,
Area of the square = ${\left(\mathrm{Side}\right)}^{2}$

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

#### 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 = ${\mathrm{Side}}^{2}$
Area of the square =
Area of the quadrant of one circle = $\frac{1}{4}{\mathrm{\pi r}}^{2}$

Area of the quadrants of four circles =
Area of the shaded portion = Area of the square $-$ Area of the quadrants of four circles

#### Question 38:

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

Area occupied by the four sectors

Area between the circles = Area of the square $-$ Area of the four sectors

#### 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 =$\frac{\sqrt{3}}{4}×{\mathrm{Side}}^{2}$

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

#### 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 = $\frac{\sqrt{3}}{4}×2a×2a=\sqrt{3}{a}^{2}$

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

Area of the region between the circles =
$=\left(\sqrt{3}-\frac{11}{7}\right){a}^{2}\phantom{\rule{0ex}{0ex}}=\left(1.73-1.57\right){a}^{2}\phantom{\rule{0ex}{0ex}}=0.16{a}^{2}\phantom{\rule{0ex}{0ex}}=\frac{4}{25}{a}^{2}$

#### 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 = $\frac{1}{2}\left(\mathrm{AD}+\mathrm{BC}\right)×\mathrm{AB}$

Area of shaded region = Area of trapezium ABCD − Area of quadrant ABE

Hence, the area of shaded region is 14.875 cm2

#### 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°

(ii) Area of the remaining portion = Area of trapezium ABCD − Area of four quadrants

#### 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 ]
[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°)

Hence, the area of shaded region is 137.64 cm2

#### 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)

Hence, the area of shaded region is 2499 cm2

#### 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

Hence, the area of shaded region is 334.31 cm2

#### 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

Hence, the area of shaded region is 283.97 cm2

#### 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 ]     [CBSE 2014]

#### Answer:

We can find the radius of the incircle by using the formula

Now, area of shaded region = Area of triangle − Area of circle

Hence, the area of shaded region is 24.6 cm2

#### 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 ] [CBSE 2013C]

#### Answer:

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

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

Area of shaded region = Area of circle − Area of triangle ABC

#### 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 = $\frac{1}{4}\left(2\mathrm{\pi }r\right)+2r$

Area of quadrant =
Hence, the area of quadrant is 38.5 cm2

#### 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

Hence, the area of minor segment is 28.5 cm2

#### 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

Cost of levelling the road = Area of the road ⨯ Rate
= 6594 ⨯ 20
= Rs 131880

#### 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:

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

Hence, the required area is 7.77 cm2

#### 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, HQFI and ELDF. 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 = $B×H$

Area of parallelogram FGHI = $B\mathit{×}H$

Area of the square = ${\mathrm{Side}}^{2}$
=

In $∆$ELF, we have:

Area of $△$DEF = $\frac{1}{2}×B×H$

Area of the semicircle =$\frac{1}{2}{\mathrm{\pi r}}^{2}$

∴ 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

#### 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° =

Now,  Area of sector having central angle 90° : Area of sector having central angle 120° : Area of sector having central angle 150°
$=\frac{90°}{360°}\mathrm{\pi }{\left(6\right)}^{2}:\frac{120°}{360°}\mathrm{\pi }{\left(6\right)}^{2}:\frac{150°}{360°}\mathrm{\pi }{\left(6\right)}^{2}\phantom{\rule{0ex}{0ex}}=\frac{1}{4}:\frac{1}{3}:\frac{5}{12}\phantom{\rule{0ex}{0ex}}=3:4:5$

#### 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.

#### 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:

OR = OQ = 12.5 cm
Now,
Area of the circle = ${\mathrm{\pi r}}^{2}$

Area of the semicircle =
Area of the triangle
Thus, we have:

#### 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:

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

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,

Because the circle is an incircle, AE and AD are tangents to the circle.

Also,
$\angle A=90°$
Therefore, AEOD is a square.
Thus, we can say that $AE=EO=OD=AD=r$.

Area of the shaded part = Area of the triangle $-$ Area of the circle

#### 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:

#### 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) = $2\mathrm{\pi r}$
We know:
Perimeter of a semicircular arc = $\mathrm{\pi r}$
Now,
For the arc PTS, radius is 6 cm.
∴ Circumference of the semicircle PTS =

For the arc QES, radius is 4 cm.
​∴ Circumference of the semicircle QES =

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

Now,
Perimeter of the shaded region = $6\mathrm{\pi }+4\mathrm{\pi }+2\mathrm{\pi }$
$=12\mathrm{\pi cm}$

Area of the semicircle PBQ = $\frac{1}{2}{\mathrm{\pi r}}^{2}$

Area of the semicircle PTS = $\frac{1}{2}{\mathrm{\pi r}}^{2}$

Area of the semicircle QES = $\frac{1}{2}{\mathrm{\pi r}}^{2}$

Area of the shaded region = Area of the semicircle PBQ + Area of the semicircle PTS $-$ Area of the semicircle QES

#### 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
∴ Length of each inner curved path = $\frac{220}{2}$ = 110 m
​Thus, we have:

Inner radius = 35 m
Outer radius = (35 + 14) = 49 m
Area of track = {Area of the two rectangles [each ] + Area of the circular ring with R = 49 m and r = 35 m)}

​Length of the outer boundary of the track

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

#### 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
Thus, we have:
$\mathrm{\pi }{r}^{2}=38.5$
$⇒\frac{22}{7}×{r}^{2}=38.5\phantom{\rule{0ex}{0ex}}⇒{r}^{2}=\left(38.5×\frac{7}{22}\right)\phantom{\rule{0ex}{0ex}}⇒{r}^{2}=\left(\frac{385}{10}×\frac{7}{22}\right)\phantom{\rule{0ex}{0ex}}⇒{r}^{2}=\frac{49}{4}\phantom{\rule{0ex}{0ex}}⇒r=\frac{7}{2}$
Now,
Circumference of the circle$=2\mathrm{\pi }r$

#### 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$=\mathrm{\pi }{r}^{2}\phantom{\rule{0ex}{0ex}}$
Thus, we have:
$\mathrm{\pi }{r}^{2}=49\mathrm{\pi }\phantom{\rule{0ex}{0ex}}⇒{r}^{2}=49\phantom{\rule{0ex}{0ex}}⇒r=\sqrt{49}\phantom{\rule{0ex}{0ex}}⇒r=7$

Now,
Circumference of the circle$=2\mathrm{\pi r}$

#### 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$=2\mathrm{\pi r}$
Thus, we have:

Radius = 7 cm
Now,
Area of the circle$=\mathrm{\pi }{r}^{2}$

#### 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
Thus, we have:
$2\mathrm{\pi }r=242$
$⇒2×\frac{22}{7}×r=242\phantom{\rule{0ex}{0ex}}⇒\frac{44}{7}×r=242\phantom{\rule{0ex}{0ex}}⇒r=\left(242×\frac{7}{44}\right)\phantom{\rule{0ex}{0ex}}⇒r=\frac{77}{2}$

∴ Area of the circle$=\mathrm{\pi }{r}^{2}$

#### 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$=\mathrm{\pi }×{\left(\frac{d}{2}\right)}^{2}$
$=\frac{\mathrm{\pi }{d}^{2}}{4}$
New diameter
$=\left(\frac{140}{100}×d\right)\phantom{\rule{0ex}{0ex}}=\frac{7d}{5}$
Now,
New radius$=\frac{7d}{5×2}$
$=\frac{7d}{10}$
New area$=\mathrm{\pi }×{\left(\frac{7d}{10}\right)}^{2}$
$=\frac{49\mathrm{\pi }{d}^{2}}{10}$
Increase in the area$=\left(\frac{49\mathrm{\pi }{d}^{2}}{10}-\frac{\mathrm{\pi }{d}^{2}}{4}\right)$
$=\frac{24\mathrm{\pi }{d}^{2}}{100}\phantom{\rule{0ex}{0ex}}=\frac{6\mathrm{\pi }{d}^{2}}{25}$
We have:
Increase in the area$=\left(\frac{6\mathrm{\pi }{d}^{2}}{25}×\frac{4}{\mathrm{\pi }{d}^{2}}×100\right)%$
= 96%

#### 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$=\mathrm{\pi }{r}^{2}$
Also,
New radius
$=\left(\frac{70}{100}×r\right)\phantom{\rule{0ex}{0ex}}=\frac{7r}{10}$
New area$=\mathrm{\pi }×{\left(\frac{7r}{10}\right)}^{2}$
$=\frac{49\mathrm{\pi }{r}^{2}}{100}$
Decrease in the area$=\left(\mathrm{\pi }{r}^{2}-\frac{49\mathrm{\pi }{r}^{2}}{100}\right)$
$=\frac{59\mathrm{\pi }{r}^{2}}{100}$
Thus, we have:
Decrease in the area$=\left(\frac{59\mathrm{\pi }{r}^{2}}{100}×\frac{1}{\mathrm{\pi }{r}^{2}}×100\right)%$
=51%

#### 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) $\sqrt{\mathrm{\pi }}:2$

#### Answer:

(d) $\sqrt{\mathrm{\pi }}:2$
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$=\mathrm{\pi }{r}^{2}$
Because the area of the square is the same as the area of the circle, we have:
${a}^{2}=\mathrm{\pi }{r}^{2}\phantom{\rule{0ex}{0ex}}⇒\frac{{r}^{2}}{{a}^{2}}=\frac{1}{\mathrm{\pi }}\phantom{\rule{0ex}{0ex}}⇒\frac{r}{a}=\frac{1}{\sqrt{\mathrm{\pi }}}$
∴ Ratio of their perimeters

#### 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:
$2\pi r=2\pi {r}_{1}+2\pi {r}_{2}$
$⇒2\pi r=\left(2\pi ×18\right)+\left(2\pi ×10\right)\phantom{\rule{0ex}{0ex}}⇒2\pi r=2\pi ×\left(18+10\right)\phantom{\rule{0ex}{0ex}}⇒2\pi r=\left(2\pi ×28\right)\phantom{\rule{0ex}{0ex}}⇒2\pi r=\left(2×\frac{22}{7}×28\right)$

#### 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:
$\pi {r}^{2}=\pi {{r}_{1}}^{2}+\pi {{r}_{2}}^{2}$

∴ Diameter of the new circle
= 50 cm

#### 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

Hence, the correct answer is option (b).

#### 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}
(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  is equal to the area of a circle with radius R, we have:
$\mathrm{\pi }{{R}_{1}}^{2}+\mathrm{\pi }{{R}_{2}}^{2}=\mathrm{\pi }{R}^{2}\phantom{\rule{0ex}{0ex}}⇒\mathrm{\pi }\left({{\mathrm{R}}_{1}}^{2}+{{\mathrm{R}}_{2}}^{2}\right)=\mathrm{\pi }{\mathrm{R}}^{2}\phantom{\rule{0ex}{0ex}}⇒{{\mathrm{R}}_{1}}^{2}+{{\mathrm{R}}_{2}}^{2}={\mathrm{R}}^{2}$

#### 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}
(d) none of these

#### Answer:

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

$2\mathrm{\pi }{R}_{1}+2\mathrm{\pi }{R}_{2}=2\mathrm{\pi }R\phantom{\rule{0ex}{0ex}}⇒2\mathrm{\pi }\left({R}_{1}+{R}_{2}\right)=2\mathrm{\pi }R\phantom{\rule{0ex}{0ex}}⇒{R}_{1}+{R}_{2}=R$

#### 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$=2\mathrm{\pi }r\phantom{\rule{0ex}{0ex}}$
Now,
Let a be the side of the square.
We know:
Perimeter of the square = 4a
Now,
$2\mathrm{\pi }r=4a\phantom{\rule{0ex}{0ex}}⇒r=\frac{4a}{2\mathrm{\pi }}$
∴ Area of the circle$=\mathrm{\pi }{r}^{2}$
$=\mathrm{\pi }×{\left(\frac{4a}{2\mathrm{\pi }}\right)}^{2}\phantom{\rule{0ex}{0ex}}=\mathrm{\pi }×\frac{16{a}^{\mathit{2}}}{4{\mathrm{\pi }}^{2}}\phantom{\rule{0ex}{0ex}}=\frac{4{a}^{2}}{\mathrm{\pi }}\phantom{\rule{0ex}{0ex}}=\frac{4×7{a}^{2}}{22}\phantom{\rule{0ex}{0ex}}=\frac{14{a}^{2}}{11}$

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

#### 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$=\mathrm{\pi }\left({R}^{2}-{r}^{2}\right)$

#### 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:
${\mathrm{\pi R}}^{2}=1386$

Also,

∴ Width of the ring$=\left(R-r\right)$

#### 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,
$\frac{c}{C}=\frac{3}{4}\phantom{\rule{0ex}{0ex}}⇒\frac{2\mathrm{\pi }r}{2\mathrm{\pi }R}=\frac{3}{4}\phantom{\rule{0ex}{0ex}}⇒\frac{r}{R}=\frac{3}{4}$
Now, the ratio between their areas is given by
$\frac{a}{A}=\frac{\mathrm{\pi }{r}^{2}}{\mathrm{\pi }{R}^{2}}\phantom{\rule{0ex}{0ex}}={\left(\frac{r}{R}\right)}^{2}\phantom{\rule{0ex}{0ex}}={\left(\frac{3}{4}\right)}^{2}\phantom{\rule{0ex}{0ex}}=\frac{9}{16}$
Hence, the correct answer is option (c).

#### 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,
$\frac{a}{A}=\frac{9}{4}\phantom{\rule{0ex}{0ex}}⇒\frac{\mathrm{\pi }{r}^{2}}{\mathrm{\pi }{R}^{2}}={\left(\frac{3}{2}\right)}^{2}\phantom{\rule{0ex}{0ex}}⇒\frac{r}{R}=\frac{3}{2}$
Now, the ratio between their circumferences is given by
$\frac{c}{C}=\frac{2\mathrm{\pi }r}{2\mathrm{\pi }R}\phantom{\rule{0ex}{0ex}}=\frac{r}{R}\phantom{\rule{0ex}{0ex}}=\frac{3}{2}$
Hence, the correct answer is option (a)

#### 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$=2\mathrm{\pi }r$

Number of revolutions taken to cover 11 km$=\left(11×1000×\frac{7}{11}\right)$
= 7000

#### 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$=\mathrm{\pi }d$

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

#### 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
= 88 m
We have:
Circumference of the wheel = 88 m
Now, let the diameter of the wheel be d m.
Thus, we have:

#### Question 21:

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

#### Answer:

(d) $\frac{\mathrm{\pi }{R}^{2}\theta }{360}$

#### Question 22:

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

#### Answer:

(b) $\frac{2\mathrm{\pi }R\theta }{360}$

#### 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
Hence, the correct answer is option (a).

#### 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

Hence, the correct answer is option (c).

#### 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
Length of arc =
Hence, the correct answer is option (b)

#### 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°

Area of minor segment = Area of sector OAPB − Area of triangle AOB

Hence, the correct answer is option (a).

#### 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.

Hence, the radius of the circle is .
Now,
Area of the shaded region = Area of the quadrant $-$ Area of the square OABC

#### 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$=\mathrm{\pi }×d$

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

#### Question 3:

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

#### Answer:

(d) $\frac{x}{360}×\mathrm{\pi }{r}^{2}$

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

#### 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.

∴ Radius of the circle
=5 cm
Now,
Area of the shaded region = Area of the circle with radius 5 cm $-$ Area of the rectangle ABCD

#### 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:

Also,
Area of the circle$=\mathrm{\pi }{r}^{2}$

#### 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:

∴ Length of the arc ACB$=\frac{2\mathrm{\pi }r}{360}$

#### 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$=360°\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$
Angle described by the minute hand in 35 minutes$=\left(\frac{360}{60}×35\right)°$
$=210°$
Now,

∴ Required area swept by the minute hand in 35 minutes = Area of the sector with

#### 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:

Now,
Area of the sector OACBO

#### 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 $\theta$ be the angle.
We have:

Area of the sector$=\frac{\mathrm{\pi }{r}^{2}\theta }{360}$

#### 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 $-$ (Area of the sector with )

#### Question 11:

A wire when bent in the form of an equilateral triangle encloses an area of $121\sqrt{3}{\mathrm{cm}}^{2}$. 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:

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$=2\mathrm{\pi }r$

Also,
Area of the circle$=\mathrm{\pi }{r}^{2}$

#### 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$=\mathrm{\pi }×d$

Distance covered in 1 second
= 1320 cm
Distance covered in 1 hour

#### 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

(ii) Area of the shaded region = Area of the quadrant OACB $-$ Area of $∆AOD$

#### 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:

=14 cm
Now,
Area of the shaded region = (Area of the square ABCD$-$ 4(Area of the sector where )

#### 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 $OD\perp BC\phantom{\rule{0ex}{0ex}}$.
Because $∆ABC$ is equilateral, $\angle A=\angle B=\angle C=60°$.
Thus, we have:

Also,

∴ Area of the shaded region = (Area of the circle) $-$ (Area of  $∆ABC$)

#### 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$=360°\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$
Angle described by the minute hand in 35 minutes$=\left(\frac{360}{60}×35\right)°$
$=210°$
Now,

∴ Required area described by the minute hand in 35 minutes = Area of the sector where

#### 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:

Width of the track$=\left(R-r\right)$

Area of the track$=\mathrm{\pi }\left({\mathrm{R}}^{2}-{\mathrm{r}}^{2}\right)$

#### 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 $\angle AOB=60°$.
Area of the sector OACBO $=\frac{\mathrm{\pi }{r}^{2}\theta }{360}$

Area of $∆OAB$

Area of the minor segment = (Area of the sector OACBO$-$ (Area of $∆OAB$)

Area of the major segment = (Area of the circle) $-$ (Area of the minor segment)

#### 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:

= 25 m
Area left ungrazed = (Area of the square) $-$ 4(Area of the sector where )

#### 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$={a}^{2}\phantom{\rule{0ex}{0ex}}$
Thus, we have:

Area of the plots = 4(Area of the semicircle of radius 20 m)

∴ Cost of turfing the plots at
= Rs 31400

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