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#### Question 1:

Find the circumference and the area of a circle of diameter 35 cm.

Given:
Diameter, d = 35 cm
Thus, we have:

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

Area of the circle = $\pi {r}^{2}$

#### Question 2:

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

Circumference = 39.6 cm
We know:
Circumference of a circle = $2\mathrm{\pi }r$

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

#### Question 3:

The area of a circle is 301.84 cm2. Find its circumference.

Area of the circle = 301.84 cm2
We know:
Area of a circle$=\pi {r}^{2}$

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

#### Question 4:

The circumference of a circle exceeds its diameter by 16.8 cm. Find the circumference of the circle.

Circumference of a circle = $2\mathrm{\pi r}$
Diameter, d = 2r
Thus, we have:

Now,
Circumference of the circle = $2×\frac{22}{7}×3.92$

= 24.64 cm

#### Question 5:

The difference between the circumference and the radius of a circle is 37 cm. Find the area of the circle.

Let the radius of the circle be r cm.
According to the question, we have:

$2\mathrm{\pi }r-r=37\phantom{\rule{0ex}{0ex}}$

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

#### Question 6:

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.

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 7:

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.

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 8:

The length of a chain used as the boundary of a semicircular park is 90 m. Find the area of the park.

Let the radius of the park be r m.
Thus, we have:
$\mathrm{\pi }r+2r=90\phantom{\rule{0ex}{0ex}}⇒r\left(\mathrm{\pi }+2\right)=90$

$⇒r=\frac{90}{\mathrm{\pi }+2}\phantom{\rule{0ex}{0ex}}⇒r=\frac{90}{\frac{22}{7}+2}\phantom{\rule{0ex}{0ex}}⇒r=\frac{90×7}{36}\phantom{\rule{0ex}{0ex}}⇒r=17.5$

Now,
Area of the park = $\frac{1}{2}\mathrm{\pi }{r}^{2}$

​

#### Question 9:

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.

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 10:

The areas of two concentric circles are 962.5 cm2 and 1386 cm2. Find the width of the ring.

Let the radii of the bigger and smaller circles be R cm and r cm, respectively.

Now,
Area of the bigger circle =
Area = $\mathrm{\pi }{R}^{2}$
$⇒1386=\frac{22}{7}×{R}^{2}\phantom{\rule{0ex}{0ex}}⇒\frac{1386×7}{22}={R}^{2}\phantom{\rule{0ex}{0ex}}⇒{R}^{2}=441\phantom{\rule{0ex}{0ex}}⇒R=21$

Area of the smaller circle =
Area = $\mathrm{\pi }{r}^{2}$
$⇒962.5=\frac{22}{7}{r}^{2}\phantom{\rule{0ex}{0ex}}⇒{r}^{2}=\frac{962.5×7}{22}\phantom{\rule{0ex}{0ex}}⇒{r}^{2}=306.25\phantom{\rule{0ex}{0ex}}⇒r=17.5$

∴ Width of the ring = R $-$ r

#### Question 11:

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

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 12:

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.

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 13:

The inner circumference of a circular track is 440 m, and the track is 14 m wide. Calculate the cost of levelling the track at 25 paise/m2. Also, find the cost of fencing the outer boundary of the track at Rs 5 per metre.

Let the radius of the inner circle be r m.
Now,
Inner circumference = 440 m
$⇒2\mathrm{\pi r}=440$

We know that the track is 14 m wide.
∴ Outer radius (R) = (70 + 14) = 84 m

Cost of levelling at 25 paise per square metre
Or,

∴ Outer circumference = $2\mathrm{\pi R}$

Rate of fencing = 5 per metre
∴ Total cost of fencing

#### Question 14:

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.

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 15:

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.

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 16:

In the given figure, AB is a diameter of a circle with centre O and OA = 7 cm. Find the area of the shaded region.

AB is the diameter of the circle.
Here,
OA = 7 cm
OB = 7 cm
OA is the radius of the circle.
∴ OA = OB = OC =OD
Also,
OA is the diameter of the smaller circle.
∴ Radius of the smaller circle = $\frac{\mathrm{OA}}{2}$= 3.5 cm
Area of the smaller circle having diameter AO = ${\mathrm{\pi r}}^{2}$

Area of the triangle $△$CBD = $\frac{1}{2}×b×h$

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

Now,
Area of the shaded region = Area of the semicircle $-$ Area of the triangle $△$CBD
= 77 $-$ 49

Also,
Area of the full-shaded region = 28 + 38.5
= 66.5 sq cm

#### Question 17:

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.

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 18:

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.

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 19:

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.

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 20:

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.

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 21:

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

(​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 22:

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

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:

=

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 23:

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

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 24:

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

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 25:

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

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 26:

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

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 27:

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.

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 28:

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?

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 29:

Find the area of a quadrant of a circle whose circumference is 22 cm.

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

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

Area of the quadrant =

#### Question 30:

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?

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 31:

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.

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 32:

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.

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 33:

Find the area of the shaded region shown in the given figure. The four corners are circle quadrants, and at the centre, there is a circle.

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

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 34:

A rectangular piece is 20 m long and 15 m wide. From its four corners, quadrants of radius 3.5 m have been cut. Find the area of the remaining part.

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

Area of the quadrants of four circles = $4×9.625=38.5$ cm2
Area of the rectangle =

∴ Area of the remaining part = Area of the rectangle $-$ Area of the quadrants of four circles
= 300 $-$ 38.5
= 261.5 ${\mathrm{m}}^{2}$

#### Question 35:

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?

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 36:

In the given figure, AOBC represents a quadrant of a circle of radius 3.5 cm with centre O. Calculate the area of the shaded portion.

Area of the right-angled $∆$AOD = $\frac{1}{2}×b×h$
=

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

Area of the shaded region = Area of the $∆\mathrm{AOD}$ $-$ Area of the sector AOB

#### Question 37:

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

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

#### Question 38:

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

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 39:

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 $\frac{4}{25}{a}^{2}.$

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 40:

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

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 41:

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.

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 42:

A square tank has an area of 1600 m2. There are four semicircular plots around it. Find the cost of turfing the plots at Rs 1.25 per m2.

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

Each piece of land is a semicircle, area =

Total area =

Cost of turfing at Rs 1.25 per sq. m =

#### Question 43:

A lawn is rectangular in the middle, and it has semicircular portions along the shorter sides of the rectangle. The rectangular portion measures 50 m by 35 m. Find the area of the lawn.

Area of the rectangle =
Radius of the circle = 17.5 m
Area of the semicircle = $\frac{1}{2}{\mathrm{\pi r}}^{2}$

∴ Total area of the lawn = Area of the circle + Area of the two semicircles

#### Question 44:

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?

r1 = 16 m
r2 = 23 m

Amount of additional ground available = Area of the bigger circle $-$ Area of the smaller circle

#### Question 45:

In the given figure ∆ABC is right-angled at A, with AB = 6 cm and AC = 8 cm. A circle with centre O has been inscribed inside the triangle. Find the value of r, the radius of the inscribed circle.

Join OC, OA and OB. This gives triangles OAC, OAB and OCB.

Consider $∆$CAB. Using Pythagoras' theorem, we have:

Also,

Further, we have:

#### Question 46:

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.

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 47:

Find the area of the region ABCDEFA shown in the given figure, given that ABDE is a square of side 10 cm, BCD is a semicircle with BD as diameter, EF = 8 cm, AF = 6 cm and ∠AFE = 90°.

Join AE.
Now, AEDB is a square.
Area of the square = ${\mathrm{Side}}^{2}$ =

Area of semi-circle = $\frac{1}{2}{\mathrm{\pi r}}^{2}$=

Area of $∆$EFA =

Area of the region ABCDEFA = Area of the square + Area of the semicircle $-$ Area of $∆$EFA
= 100 + 39.25 $-$ 24
= 115.25 sq. cm

#### Question 48:

In the given figure, ABCD is a square of side 14 cm. Find the area of the shaded region in the given figure.

Area of the square =

Area of the circles =

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

#### Question 49:

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

We know:
Perimeter (circumference of a circle) = $2\mathrm{\pi r}$
Perimeter of a semicircular arc = $\mathrm{\pi r}$

Now,
For the arc ADC, radius is 2.1 cm.
∴ Perimeter of the arc ADC =

For the arc AEB, radius is 1.4 cm.
∴ ​Perimeter of the arc AEB =

For the arc BFC, radius is 0.7 cm.
∴ Perimeter of the arc BFC =

Thus, we have:
Perimeter of the shaded region = $2.1\mathrm{\pi }+1.4\mathrm{\pi }+0.7\mathrm{\pi }$

#### Question 50:

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.

In triangle $∆$ABC, we have:

#### Question 51:

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

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 52:

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.

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

The hexagon is made up of six triangles.

#### Question 53:

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.

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.

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 54:

The area of an equilateral triangle is . Taking each angular point as centre, circle as 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.

Let the side of the equilateral triangle be $a$ cm.

Thus, we have:

$\frac{\sqrt{3}}{4}{a}^{2}=49\sqrt{3}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}⇒{a}^{2}=196\phantom{\rule{0ex}{0ex}}⇒a=14$

The radius of each circle is 7 cm. The angle at the vertex of each triangle is $60°$.

Area of the sector with angle $60°$ and radius 7 cm:

There are three such sectors.
Total area =

Area not included in the circles = Area of the triangle $-$ Area of the three sectors

#### Question 55:

In the given figure, ∆ABC is a right-angled triangle with ∠B = 90°, AB = 48 cm and BC = 14 cm. With AC as diameter a semicircle is drawn and with BC as radius, A quadrant of a circle is drawn. Find the area of the shaded region.

Consider the triangle ABC.

Now,

#### Question 56:

Calculate the area other than the area common between two quadrants of circles of radius 16 cm each, which is shown as the shaded region in the given figure.

#### Question 57:

In a circular table cover of radius 70 cm, a design is formed leaving an equilateral ∆ABC in the middle, as shown in the figure. Find the total area of the design.

$∆$ABC is equilateral.
Thus, we have:

$\angle A=\angle B=\angle C=60°$

OB bisects $\angle B$. Therefore, $\angle OBD=30°$.

$∆$OBD is a right-angled triangle.

We have:

Now,
$BC=2BD=35\sqrt{3}×2=70\sqrt{3}=AB=AC$

Area of the circle =

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

#### Question 58:

Find the area of the sector of a circle of radius 14 cm with central angle 45°.

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

#### Question 59:

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.

Given:
Radius = 2 cm
Angle of sector = ${150}^{\circ }$
Now,

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

#### Question 60:

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.

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 61:

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

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 62:

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

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 63:

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

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 64:

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.

Length of the arc = 16.5 cm
$\theta ={54}^{\circ }$
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 65:

The circumference of a circle is 88 cm. Find the area of the sector whose central angle is 72°.

Given:
Circumference of the circle = 88 cm
$\theta ={72}^{\circ }$
Area of the sector = ?

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

#### Question 66:

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

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 67:

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

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 68:

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.

#### Question 69:

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.

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 70:

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.

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 71:

The radius of a circle with centre O is 6 cm. Two radii OA and OB are drawn at right angles to each other. Find the areas of the minor and major segments.

The triangle OAB is a right isosceles triangle.

Area of triangle OAB =

Now,
Area of the minor segment:

Area of the major segment:

#### Question 72:

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

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 73:

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

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 74:

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.

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 75:

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.

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 76:

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.

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 1:

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

(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 2:

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

(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 3:

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

(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 4:

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

(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 5:

The circumferences of two circles are in the ratio 2 : 3. The ratio between their areas is
(a) 2 : 3
(b) 4 : 9
(c) 9 : 4
(d) none of these

(b) 4:9
Let cm be the radii of the two circles.
Thus, we have:
Perimeter of the first circle$=2\mathrm{\pi }{r}_{1}\phantom{\rule{0ex}{0ex}}$
And,
Perimeter of the second circle$=2\mathrm{\pi }{r}_{2}$
Now,
$\frac{2\mathrm{\pi }{r}_{1}}{2\mathrm{\pi }{r}_{2}}=\frac{2}{3}\phantom{\rule{0ex}{0ex}}⇒\frac{{r}_{1}}{{r}_{2}}=\frac{2}{3}$
Also,
Area of the first circle$=\mathrm{\pi }{{r}_{1}}^{2}$
And,
Area of the second circle$=\mathrm{\pi }{{r}_{2}}^{2}$
Thus, we have:
$\frac{\mathrm{\pi }{{r}_{1}}^{2}}{\mathrm{\pi }{{r}_{2}}^{2}}=\frac{{{r}_{1}}^{2}}{{{r}_{2}}^{2}}$

Hence, the ratio of the areas of the two circles is 4:9.

#### Question 6:

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

(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 7:

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

(d) None of these
Let r be the original radius.
Thus, we have:
Original area$=\mathrm{\pi }{r}^{2}$
Also,
$=\left(\frac{70}{100}×r\right)\phantom{\rule{0ex}{0ex}}=\frac{7r}{10}$
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 8:

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

(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 9:

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

(a) 2:3
Let  be the radii of the two circles.
Now,
Area of the first circle$=\mathrm{\pi }{{r}_{1}}^{2}$
And,
Area of the second circle$=\mathrm{\pi }{{r}_{2}}^{2}$
Thus, we have:
$\frac{\mathrm{\pi }{{r}_{1}}^{2}}{\mathrm{\pi }{{r}_{2}}^{2}}==\frac{4}{9}\phantom{\rule{0ex}{0ex}}⇒\frac{{{r}_{1}}^{2}}{{{r}_{2}}^{2}}=\frac{4}{9}$
Also,
Perimeter of the first circle$=2\mathrm{\pi }{r}_{1}\phantom{\rule{0ex}{0ex}}$
And,
Perimeter of the second circle $=2\mathrm{\pi }{r}_{2}$
Thus, we have:

=2:3

Hence, the ratio of their circumferences is 2:3.

#### Question 10:

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

(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 11:

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

(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 12:

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

(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 13:

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

(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 14:

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

(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 15:

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) ${R}_{1}^{2}+{R}_{2}^{2}={R}^{2}$

(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 16:

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

(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 17:

If the perimeter of a square is equal to the circumference of a circle, then the ratio of their areas  is.
(a) 14 : 11
(b) 11 : 14
(c) 22 : 7
(d) 7 : 22

(b) 11:14
Let P be the perimeter of the square.
Now,
Each side of the square$=\frac{P}{4}$
Let r be the radius of the circle.
We know:
Circumference of the circle$=2\mathrm{\pi }r$
Now,
$2\mathrm{\pi }r=P\phantom{\rule{0ex}{0ex}}⇒r=\frac{P}{4\mathrm{\pi }}$
∴ Area of the square$={\left(\frac{P}{4}\right)}^{2}$
$=\frac{{P}^{2}}{16}$

Also,
Area of the circle $=\mathrm{\pi }{r}^{2}$
$=\mathrm{\pi }×{\left(\frac{P}{2\mathrm{\pi }}\right)}^{2}\phantom{\rule{0ex}{0ex}}=\mathrm{\pi }×\frac{{P}^{\mathit{2}}}{4{\mathrm{\pi }}^{2}}\phantom{\rule{0ex}{0ex}}=\frac{{P}^{\mathit{2}}}{4\mathrm{\pi }}$
∴ Required ratio$=\frac{{P}^{\mathit{2}}}{16}×\frac{4\mathrm{\pi }}{{P}^{\mathit{2}}}$
$=\frac{\mathrm{\pi }}{4}\phantom{\rule{0ex}{0ex}}=\left(\frac{22}{7×4}\right)\phantom{\rule{0ex}{0ex}}=\frac{11}{14}$

=11:14

#### Question 18:

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

(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 19:

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

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

#### Question 20:

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

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

#### Question 21:

A chord of a circle of radius 28 cm subtends an angle of 45° at the centre of the circle. The area of the minor segment is
(a) 30.256 cm2
(b) 30.356 cm2
(c) 30.456 cm2
(d) 30.856 cm2

(d) 30.856 cm2
Let r be the radius of the circle and $\theta$ be the angle.
∴ Area of the minor segment $=\left(\frac{\mathrm{\pi }{r}^{2}\theta }{360}-\frac{1}{2}r\right)$ sin θ cm2

#### Question 22:

A chord of a circle subtends an angle of 60° at the centre of the circle. If the length of the chord is 10 cm, then the area of the major segment is
(a) 305 cm2
(b) 295 cm2
(c) 310 cm2
(d) 335 cm2

(a) 305 cm2
Let AB be the chord of a circle with centre O.
Now,
OA = OB = AB = 10 cm
Thus, we have:

Area of the minor segment$=\left(\frac{\mathrm{\pi }{r}^{2}\theta }{360}-\frac{1}{2}r\right)$2sin θ cm2

Area of the major segment

#### Question 23:

The perimeter of a sector of a circle with central angle 90° is 25 cm. The area of the minor segment of the circle is
(a) 14 cm2
(b) 16 cm2
(c) 18 cm2
(d) 24 cm2

(a) 14 cm2
Let r be the radius of the circle and $\theta$ be the angle.
Now,
Perimeter of the sector$=\left(2r+\frac{2\mathrm{\pi }r\theta }{360}\right)$
$=2r+2×\frac{22}{7}×r×\frac{90}{360}\phantom{\rule{0ex}{0ex}}=\left(2r+\frac{11r}{7}\right)\phantom{\rule{0ex}{0ex}}=\frac{25r}{7}$
Also,

Area of the minor segment$=\left(\frac{\mathrm{\pi }{r}^{2}\theta }{360}-\frac{1}{2}r\right)$2sin θ cm2

#### Question 24:

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

(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 25:

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

(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 26:

Match the following columns:

 Column I Column II (a) The circumference of a circle is 44 cm. The area of this circle is .........cm2. (p) 1936 (b) A wire is looped in the form of a circle of radius 28 cm. It is bent into a square. The area of the square is .......cm2 (q) 10 (c) The radii of two circles are 9 cm and 19 cm respectively. The radius of the circle whose circumference is equal to the sum of the circumferences of the given circles is .........cm2 (r) 154 (d) The radii of two circles are 8 cm are 6 cm respectively. The radius of the circle having its area of the given circles is .........cm2 (s) 28

(a) Let r be the radius of the circle.
Now,
Circumference of the circle$=2\mathrm{\pi }r$
We have:

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

∴ $\left(\mathrm{a}\right)⇒\left(\mathrm{r}\right)$

(b) Let r be the radius of the circle.
Length of the circle = Circumference of the circle
​
Perimeter of the square = Length of the wire
∴ Side of the square
= 44 cm
Area of the square

$\left(b\right)⇒\left(p\right)$

(c) Let r be the radius of the circle whose circumference is equal to the sum of the circumferences of the given circles.
Thus, we have:

∴ $\left(c\right)⇒\left(s\right)$

(d) Let r be the radius of the circle with area equal to the sum of the areas of the given circles.
Thus, we have:

$\left(d\right)⇒\left(q\right)$

#### Question 27:

Match the following columns:

 Column I Column II (a) In a circle of radius 6 cm, the angle of a sector is 60°. The area of the sector is ..........cm2 (p) $\frac{25\mathrm{\pi }}{6}$ (b) In a circle with centre O and radius 5 cm, AB is a chord of length $5\sqrt{3}$ cm. The area of sector OAB is .........cm2 (q) 44.8 (c) A chord of a circle of radius 14 cm subtends a right angle at the centre. The area of the sector is ........cm2 (r) 154 (d) The perimeter of the sector of a circle of radius 5.6 cm is 27.2 cm. The area of the sector is ..........cm2 (s) $18\frac{6}{7}$

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

∴ $\left(a\right)⇒\left(s\right)$

(b) Draw OD such that $OD\perp AB$.
Now,
$DB=\frac{1}{2}AB$

From the right $∆ODB$, we have:

∴ Area of $∆AOB=\frac{1}{2}×AB×OD$

Also,
Area of

We have:

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

$\left(b\right)⇒\left(p\right)$

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

∴ $\left(c\right)⇒\left(r\right)$

(d) Let O be the centre of the circle of radius 5.6 cm and OACB be its sector with perimeter 27.2 cm.
Now,

Area of the sector OACBO

∴ $\left(\mathrm{d}\right)⇒\left(\mathrm{q}\right)$

#### Question 28:

Match the following columns:

 Column I Column II (a) If the perimeter of a semi- circular protractor is 66 cm, then its radius is.........cm. (p) 35 (b) Each wheel of a car makes 450 complete revolution in covering 0.99 km. The radius of each wheel is ........m. (q) 32 (c) A bicycle wheel makes 5000 revolutions in covering 11 km. The diameter of its wheel is .........cm2 (r) $12\frac{5}{6}$ (d) The given figure is a sector of a circle of radius 10.5 cm. Perimeter of the sector is ............cm2 figure (s) 70

(a)
Let the radius of the protractor be r cm.
Then, perimeter$=\left(\mathrm{\pi }r+2r\right)$

Therefore,

Hence, $\left(\mathrm{a}\right)⇒\left(r\right)$

(b)
Distance covered in 1 revolution

Let r be the radius of the wheel. Then,
Circumference of the wheel $=2\mathrm{\pi }r$

Hence, $\left(\mathrm{b}\right)⇒\left(\mathrm{p}\right)$

(c)
Distance covered in 1 revolution

Let the diameter of the wheel be d m.
Then,

Hence, $\left(c\right)⇒\left(s\right)$

(d)
Let r be the radius of the arc. Then,
Arc length$=\frac{2\mathrm{\pi }r\theta }{360}$

Therefore, Perimeter = OA+OB+arc AB
= (10.5 + 10.5 +arc AB) cm
= 32 cm
Hence, $\left(\mathrm{d}\right)⇒\left(\mathrm{q}\right)$

#### Question 29:

Assertion (A)
The area of the quadrant of a circle having a circumference of 22 cm is $9\frac{5}{8}{\mathrm{cm}}^{2}.$

Reason (R)
The area of a sector of a circle of radius r with central angle x° is $\left(\frac{x×\mathrm{\pi }{r}^{2}}{360}\right).$

(a) Both Assertion (A) and Reason (R) are true and Reason (R) is a correct explanation of Assertion (A).
(b) Both Assertion (A) Reason (R) true but Reason (R) is not a correct explanation of Assertion (A).
(c) Assertion (A) is true and Reason (R) is false.
(d) Assertion (A) is false and Reason (R) is true.

(a) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A).

Assertion (A):
Let r be the radius of the circle.
Now,
Circumference of the circle $=2\mathrm{\pi }r\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$
We have:

Area of the quadrant$=\frac{90°}{360°}\mathrm{\pi }{r}^{2}$

Hence, assertion (A) is true.

Reason (R):
The given statement is true.

Assertion (A) is true and reason (R) is the correct explanation of assertion (A).

#### Question 30:

Assertion (A)
An arc of a circle of length 5π cm bounds a sector whose area is 20π cm2. Then, the radius of the circle is 4 cm.

Reason (R)
A chord of a circle of radius 12 cm subtends an angle of 60° at the centre of the circle. The area of the minor segment of the circle is 13.08 cm2.

(a) Both Assertion (A) and Reason (R) are true and Reason (R) is a correct explanation of Assertion (A).
(b) Both Assertion (A) Reason (R) true but Reason (R) is not a correct explanation of Assertion (A).
(c) Assertion (A) is true and Reason (R) is false.
(d) Assertion (A) is false and Reason (R) is true.

(d) Assertion (A) is false and reason (R) is true.
Assertion (A):
Let r be the radius of the circle.
We have:
Arc length$=\frac{\mathrm{\pi }r\theta }{180}$
Now,

Area of the sector$=\frac{{\mathrm{\pi r}}^{2}\mathrm{\theta }}{360}$
Thus, we have:
$\frac{\mathrm{\pi }{r}^{2}\theta }{360}=20\mathrm{\pi }⇒\frac{{\mathrm{r}}^{2}\mathrm{\theta }}{360}=20$

$⇒{r}^{2}\mathrm{\theta }=\left(20×360\right)$

Now,
$\frac{{r}^{2}\mathrm{\theta }}{r\mathrm{\theta }}=\frac{\left(20×360\right)}{\left(5×180\right)}$
r = 8 cm

Hence, assertion (A) is false.

Reason (R):
Let r be the radius of the circle.
Now,
Area of the minor segment$=\left(\frac{\mathrm{\pi }{r}^{2}\theta }{360}-\frac{1}{2}r\right)$2sin θ

Hence, reason (R) is true.

#### Question 31:

Assertion (A)
If the circumferences of two circles are in the ratio 2 : 3, then the ratio of their areas is 4 : 9.

Reason (R)
The circumference of a circle of radius r is 2πr.

(a) Both Assertion (A) and Reason (R) are true and Reason (R) is a correct explanation of Assertion (A).
(b) Both Assertion (A) Reason (R) true but Reason (R) is not a correct explanation of Assertion (A).
(c) Assertion (A) is true and Reason (R) is false.
(d) Assertion (A) is false and Reason (R) is true.

(b) Both assertion (A) and reason (R) are true, but reason (R) is not the correct explanation of assertion (A).

Assertion (A):
Let  be the radii of two circles.
Now,
Circumference of the first circle$=2\mathrm{\pi }{r}_{1}\phantom{\rule{0ex}{0ex}}$
Circumference of the second circle$=2\mathrm{\pi }{r}_{2}$
Thus, we have:
$\frac{2\mathrm{\pi }{r}_{1}}{2\mathrm{\pi }{r}_{2}}=\frac{2}{3}\phantom{\rule{0ex}{0ex}}⇒\frac{{r}_{1}}{{r}_{2}}=\frac{2}{3}$
Also,
Area of the first circle$=\mathrm{\pi }{{r}_{1}}^{2}$
Area of the second circle$=\mathrm{\pi }{{r}_{2}}^{2}$
Thus, we have:
$\frac{\mathrm{\pi }{{r}_{1}}^{2}}{\mathrm{\pi }{{r}_{2}}^{2}}=\frac{{{r}_{1}}^{2}}{{{r}_{2}}^{2}}$

Hence, the ratio of their areas is 4:9.
Hence, assertion (A) is true.

Reason (R):
The given statement is true.
Hence, both assertion (A) and reason (R) are true, but reason (R) is not the correct explanation of assertion (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

(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

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

(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

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.

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.

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.

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.

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.

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.

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?

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.

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.

(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.

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.

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.

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.

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.

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?

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.

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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