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Page No 15.10:

Question 19:

The area enclosed between the concentric circles is 770 cm2. If the radius of the outer circle is 21 cm, find the radius of the inner circle.

Answer:

Let the radius of outer and inner two circles be r1  and r2 respectively.
Area enclosed between concentric circles = πr12-πr22
770=227212-r22245=212-r22r22=441-245r22=196r22=142r2=14 cm

Hence, the radius of inner circle is 14 cm.



Page No 15.19:

Question 1:

Find, in terms of π, the length of the arc that subtends an angle of 30° at the centre of a circle of radius 4 cm.

Answer:

The arc length l of a sector of an angle θ in a circle of radius r is given by

It is given that and. Substituting the value of r and θ in above equation,

Page No 15.19:

Question 2:

Find the angle subtended at the centre of a circle of radius 5 cm by an arc of length (5π/3) cm.

Answer:

We know that the arc length l of a sector of an angle θ in a circle of radius r is

It is given that and length . Substituting these value in above equation,

Hence, the angle subtended at the centre of circle is.

Page No 15.19:

Question 3:

An arc of length 20π cm subtends an angle of 144° at the centre of a circle. Find the radius of the circle.

Answer:

We know that the arc length l of a sector of an angle θ in a circle of radius r is

It is given and angle.

Now we substitute the value of l and θ in above formula to find the value of radius r of circle.

Page No 15.19:

Question 4:

An arc of length 15 cm subtends an angle of 45° at the centre of a circle. Find in terms of π, the radius of the circle.

Answer:

We know that the arc length l of a sector of an angle θ in a circle of radius r is

It is given that and angle.

Now we substitute the value of l and θ in above formula to find the value of radius r of circle.

Page No 15.19:

Question 5:

Find the angle subtended at the centre of a circle of radius 'a' by an arc of length (aπ/4) cm.

Answer:

We know that the arc length l of a sector of an angle θ in a circle of radius r is

It is given and radius.

Now we substitute the value of l and r in above formula to find the value of angle θ subtended at the centre of circle.

Page No 15.19:

Question 6:

A sector of a circle of radius 4 cm contains an angle of 30°. Find the area of the sector.

Answer:

We know that the area A of a sector of an angle θ in the circle of radius r is given by

It is given that and angle.

Now we substitute the value of r and θ in above formula,

Page No 15.19:

Question 7:

A sector of a circle of radius 8 cm contains an angle of 135°. Find the area of the sector.

Answer:

We know that the area A of a sector of an angle θ in the circle of radius r is given by

It is given thatand.

Now we substitute the value of r and θ in above formula,

Page No 15.19:

Question 8:

The area of a sector of a circle of radius 2 cm is π cm2. Find the angle contained by the sector.

Answer:

We know that the area A of a sector of an angle θ in the circle of radius r is given by

It is given that and area.

Now we substitute the value of r and A in above formula to find the value of θ,

Page No 15.19:

Question 9:

The area of a sector of a circle of radius 5 cm is 5 π cm2 . Find the angle contained by the sector.

Answer:

We know that the area A of a sector of an angle θ in the circle of radius r is given by

It is given that radius and area.

Now we substitute the value of r and A in above formula to find the value of θ,

Page No 15.19:

Question 10:

AB is a chord of a circle with centre O and radius 4 cm. AB is of length 4 cm. Find the area of the sector of the circle formed by chord AB.

Answer:

We have to find the area of the sector AOB formed by the chord AB.

We have and. So,

Let. Then,

In, we have

Hence,

Now, using the value of and r we will find the area of sector AOB,

Page No 15.19:

Question 11:

In a circle of radius 35 cm, an arc subtends an angle of 72° at the centre. Find the length of the arc and area of the sector.

Answer:

We know that the arc length l and area A of a sector of an angle θ in the circle of radius r is given by and respectively.

It is given that, and.

We will calculate the arc length using the value of r and θ,

Now, we will find the value of area A of the sector

Page No 15.19:

Question 12:

The perimeter of a sector of a circle of radius 5.7 m is 27.2 m. Find the area of the sector.

Answer:

We know that the area A of a sector of circle of radius r and arc length l is given by

Let OAB is the given sector. Then,



arc AB = 15.8 m

So, l = 15.8 m

Now substituting the value of r and l in above formula,

A=12×15.8×5.7=45.03 m2

Page No 15.19:

Question 13:

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

Answer:

We know that the area A of a sector of circle of radius r and arc length l is given by

Let OAB is the given sector. Then,

Perimeter of sector OAB = 27.2 m
OA + OB + arc AB = 27.2 m
5.6 + 5.6 + arc AB = 27.2 m
11.2 + arc AB = 27.2 m

arc AB = 16 m

So, l = 16 m

Now substituting the value of r and l in above formula,

A=12×16×5.6=44.8 m2

Page No 15.19:

Question 14:

A sector is cut-off from a circle of radius 21 cm. The angle of the sector is 120°. Find the length of its arc and the area.

Answer:

We know that the arc length l and area A of a sector of circle at an angle θ of radius r is given by and angle.

Let OAB is the given sector.

It is given that and angle.

Now using the value of r and θ, we will find the value of l and A,

Arc length,

Area of sector,

Page No 15.19:

Question 15:

The minute hand of a clock is 21 cm long. Find the area described by the minute hand on the face of the clock between 7.00 AM and 7.05 AM.

Answer:

We know that the area A of a sector of circle at an angle θ of radius r is given by

.

We have,

Thus,



Page No 15.20:

Question 21:

The length of the minute hand of a clock is 14 cm. Find the area swept by the minute hand in 5 minutes.

Answer:

Angle make by the minute hand in 1 minute = 6
Angle make by the minute hand in 5 minute = 5 ⨯ 6 = 30
Area of the sector having central angle is given by
30°360°π142=112×227142=51.33 cm2
Hence, the area swept by minute hand in 5 minutes is 51.33 cm2

Page No 15.20:

Question 22:

In a circle of radius 21 cm, an arc subtends an angle of 60° at the centre. Find (i) the length of the arc (ii) area of the sector formed by the arc. (Use π = 22/7)

Answer:

Here, we have θ = 60° and r = 21 cm

(i) The length of the arc is given by
60°360°×2π21=16×2×227×21=22 cm


(ii) Area of the sector formed by the arc is given by
60°360°π212=16×227212=231 cm2

Page No 15.20:

Question 16:

The minute hand of a clock is 10 cm long. Find the area of the face of the clock described by the minute hand between 8 AM and 8.25 AM.

Answer:

We know that the area A of a sector of circle at an angle θ of radius r is given by

We have,

∴ Required area

= 150°360°×227×102

= 130.95 cm2

Page No 15.20:

Question 17:

A sector of 56° cut out from a circle contains area 4.4 cm2. Find the radius of the circle.

Answer:

We know that the area A of a sector of circle at an angle θ of radius r is given by

.

It is given that, and angle .

We can find the value of r by substituting these values in above formula,

Page No 15.20:

Question 18:

In a circle of radius 6 cm, a chord of length 10 cm makes an angle of 110° at the centre of the circle. Find:

(i) the circumference of the circle

(ii) the area of the circle

(iii) the length of the arc AB,

(iv) the area of the sector OAB.

Answer:

It is given that the radius of circle, and angle at the centre of circle.

(i) We know that the Circumference C of circle of radius r is,

C=2πr=2×227×6=2647=37.71 cm

(ii) We know that the Area A of circle of radius r is,

(iii) We know that the arc length l of a sector of an angle θ in a circle of radius r is

(iv) We know that the area A of a sector of an angle θ in the circle of radius r is given by

Page No 15.20:

Question 19:

Fig. 15.17, shows a sector of a circle, centre O, containing an angle 0°. Prove that:

(i) Perimeter of the shaded region is tan θ+secθ+πθ180-1

(ii) Area of the shaded region is r22tanθ-πθ180


 

Answer:


It is given that the radius of circle is r and the angle.

In,

It is given that.

(i) We know that the arc length l of a sector of an angle θ in a circle of radius r is

Now we substitute the value of OC, OA and l to find the perimeter of sector AOC,





Hence,

(ii) We know that area A of the sector at an angle θ in the circle of radius r is

Thus

Hence,

Page No 15.20:

Question 20:

Figure 15.18 shows a sector of a circle of radius r cm containing an angle θ°. The area of the sector is A cm2 and perimeter of the sector is 50 cm. Prove that

(i) θ = 360θ25r-1

(ii) A = 25r − r2

Answer:

It is given that the radius of circle is r cm and angle.

(i) We know that the arc length l of a sector of an angle θ in a circle of radius r is

Now we substitute the value of OB, OA and l to find the perimeter of sector AOB,

(ii) We know that area A of the sector at an angle θ in the circle of radius r is

. Thus

Substituting the value of θ,



Page No 15.25:

Question 1:

AB is a chord of a circle with centre O and radius 4 cm. AB is of length 4 cm and divides the circle into two segments. Find the area of the minor segment.

Answer:

We know that the area of minor segment of angle θ in a circle of radius r is,

It is given that the chord AB divides the circle in two segments.

We have and. So,

Let. Then,

In, we have

Hence,

Now using the value of r and θ, we will find the area of minor segment

Page No 15.25:

Question 2:

A chord PQ of length 12 cm subtends an angle of 120° at the centre of a circle. Find the area of the minor segment cut off by the chord PQ.

Answer:

We know that the area of minor segment of angle θ in a circle of radius r is,

It is given that the chord PQ divides the circle in two segments.

We have and. So,

Since,

In, we have

Thus the radius of circle is .

Now using the value of radius r and angle θ we will find the area of minor segment

Page No 15.25:

Question 3:

A chord of a circle of radius 14 cm makes a right angle at the centre. Find the areas of the minor and major segments of the circle.

Answer:

We know that the area of minor segment of angle θ in a circle of radius r is,

It is given that the chord of the circle of radius makes right angle at the centre.

So,

Substituting the value of r and angle θ in above formula,

Area of minor segment

Hence, area of minor segment is

Page No 15.25:

Question 4:

A chord 10 cm long is drawn in a circle whose radius is 52 cm. Find area of both the segments.

Answer:

We know that the area of minor segment of angle θ in a circle of radius r is,

It is given that the chord AB divides the circle in two segments.

We have and. So,

Let. Then,

In, we have

Hence,

Now using the value of r and θ, we will find the area of minor segment

Page No 15.25:

Question 5:

A chord AB of a circle, of radius 14 cm makes an angle of 60° at the centre of the circle. Find the area of the minor segment of the circle. (Use π = 22/7)

Answer:

We know that the area of minor segment of angle θ in a circle of radius r is,

It is given that,

Substituting these values in above formula



Page No 15.26:

Question 6:

AB is the diameter of a circle, centre O, C is a point on the circumference such that ∠COB = θ. The area of the minor segment cut off by AC is equal to twice the area of the sector BOC. Prove that

sinθ2cosθ2=π12-θ120

Answer:

We know that the area of minor segment of angle θ in a circle of radius r is,

.

It is given that,

So,

Area, A of minor segment cutoff by AC at angle

Now, since and

We know that the area of sector of a circle of radius r at an angle θ is

So, the area of sector BOC,

It is given that,

Page No 15.26:

Question 7:

A chord of a circle subtends an angle of θ at the centre of the circle. The area of the minor segment cut off by the chord is one eighth of the area of the circle. Prove that

8 sinθ2cosθ2+π=πθ45.

Answer:

We know that the area of circle and area of minor segment of angle θ in a circle of radius r is given by, andrespectively.

It is given that,



Page No 15.47:

Question 1:

A plot is in the form of a rectangle ABCD having semi-circle on BC as shown in Fig. 15.64. If AB = 60 m and BC = 28 m, find the area of the plot.

Answer:

It is given that a plot is in form of rectangle ABCD having a semicircle on BC.

Since BC is the diameter of semicircle. Then, radius of semicircle is


                               =308 m2


                                          = 1680 m2

Now,

Area of plot = Area of rectangle + Area of semicircle
                   = 1680 + 308 m2
                   = 1988 m2
 

Page No 15.47:

Question 2:

A play ground has the shape of a rectangle, with two semi-circles on its smaller sides as diameters, added to its outside. If the sides of the rectangle are 36 m and 24.5 m, find the area of the playground. (Take π = 22.7).

Answer:

It is given that a play ground has a shape of rectangle, with two semicircles on its smaller sides as diameter, added to its outside. So,

We have, sides of rectangle l = 36 m and b = 24.5 m.

Since, the diameter of semicircle is. Then,

r=24.52=12.25 m


                              = 235.81 m2


                              = 882 m2

Thus, the area of playground is


                                 = 1353.62 m2

Page No 15.47:

Question 3:

The outer circumference of a circular race-track is 528 m. The track is everywhere 14 m wide. Calculate the cost of levelling the track at the rate of 50 paise per square metre (Use π = 22/7)

Answer:

It is given that the outer circumference C of circular track is 528 m.

We know that the circumference of circle of radius r is

Substituting the value of C,

Thus, the radius of outer circle is.

Since circular race track is 14 m wide everywhere. Then

We know that the area of circle of radius r is

So,

Now,

It is given that,

Page No 15.47:

Question 4:

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

Answer:

It is given that, the quadrants of radius r have been cut from the four corners of a rectangular piece is of length and width.

We have to find the area of remaining part.

We know that,

Now,

Page No 15.47:

Question 5:

Four equal circles, each of radius 5 cm, touch each other as shown in Fig. 15.65. Find the area included between them (Take π = 3.14).

Answer:

It is given that four equal circle touches each other as shown in figure.

Let the side of square is a.

We know that


                                                    =14×78.5 cm2
 


                                     = 21.5 cm2

Page No 15.47:

Question 6:

Four cows are tethered at four corners of a square plot of side 50 m, so that they just cannot reach one another. What area will be left ungrazed? (Fig. 15.66)

Answer:

It is given that four cows are tethered at four corner of square ABCD. We have to find the area of plot that will left ungrazed.

Let the side of square is a.

a = 25 + 25 m = 50 m


Area of square = a2
                        =50×50=2500 m2

Page No 15.47:

Question 7:

A road which is 7 m wide surrounds a circular park whose circumference is 352 m. Find the area of the road.

Answer:

It is given that the circumference C of circular track is 352 m.

We know that the circumference of circle of radius r is

Substituting the value of C,

Thus, the radius of Circular Park is.

Since, 7m wide road surrounds the circular park. Then 

Page No 15.47:

Question 8:

Four equal circles, each of radius a, touch each other. Show that the area between them is 67a2 (Take π = 22/7).

Answer:

It is given that four equal circles of radius a touches each other.

So,

Since circles touches each other, the lines joining their centre make a square ABCD.

The side of square is 2a.

Page No 15.47:

Question 9:

A square water tank has its side equal to 40 m. There are four semi-circular grassy plots all round it. Find the cost of turfing the plot at Rs. 1.25 per square metre (Take π = 3.14).

Answer:

It is given that the side of square.

Since four semicircular grassy plots rounds a square water tank. Then, diameter of semicircular plot is.

So, the radius of semicircle

Now, the total area of plot is sum of area of four semicircular plots.

Since,

 

Page No 15.47:

Question 10:

A rectangular park is 100 m by 50 m. It is surrounding by semi-circular flower beds all round. Find the cost of levelling the semi-circular flower beds at 60 paise per square metre (use π = 3.14).

Answer:

Since four semicircular flower beds rounds the rectangular park. Then, diameters of semicircular plots are and

So, the radius of semicircle at larger side of rectangle

And the radius of semicircle at smaller side of rectangle

Now, the total area of semicircular plot is sum of area of four semicircular plots.

Since,

Page No 15.47:

Question 11:

Prove that the area of a circular path of uniform width h surrounding a circular region of radius r is πh (2r + h).

Answer:

We know that the area of a circle of radius r is

It is given that a circular path of width h surrounds the circle of radius r.

So,

Using the value of radius in above formula,

Hence,



Page No 15.48:

Question 12:

The inside perimeter of a running track (shown in Fig. 15.67) is 400 m. The length of each of the straight portion is 90 m and the ends are semi-circles. If the track is everywhere 14 m wide. find the area of the track. Also find the length of the outer running track.

Answer:

It is given that, and

We know that the circumference C of semicircle of radius be r is

C=πr

The inside perimeter of running track is the sum of twice the length of straight portion and circumferences of semicircles. So,
inside perimeter of running track = 400 m
2l+2πr=400 m
2×90+2×227×r=400 m
r=220×72×22=35 m
Thus, radius of inner semicircle is 35 m.

Now,
radius of outer semi circle r' = 35 + 14 = 49 m

Area of running track = 2×Area of rectangle+2×Area of outer semi circle-2×Area of inner semicircle
            =2×90×14+2×π(49)22-2×π(35)22
            =2520+π×49+3549-35
            =2520+227×84×14
            =2520+3696=6216 m2
Hence, the area of running track = 6216 m2

Now, length L of outer running track is
L = 2 × l + 2πr'
   =2×90+2π×49

   =180+2×227×49
   =180+308=488 m
   
Hence, the length L of outer running track is 488 m.
 

Page No 15.48:

Question 13:

Find the area of Fig. 15.68, in square cm, correct to one place of decimal.  (Take π = 22/7).

Answer:

Let the area of square ABCD be A.

It s given that,

So,

It is given that a semicircle is attached to one side of the square.

We know that the area of semicircle of radius r is

Substituting the value of r,

From the above figure it is seen that a right angle triangle is cutoff from one side of square.

Now, the area of above figure is,

Hence area of given figure is

Page No 15.48:

Question 14:

In Fig. 15.69, AB and CD are two diameters of a circle perpendicular to each other and OD is the diameter of the smaller circle. If OA = 7 cm, find the area of the shaded region.






 

Answer:

It is given that AB and CD are two diameters of a circle perpendicular to each other and OD is the diameter of small circle.

It is given that,

So, radius r of small circle is

We know that the area A of circle of radius r is.

Substituting the value of r in above formula,

Now, let the area of large circle be.

Using the value radius OA,

Hence,

Page No 15.48:

Question 15:

In Fig. 15.70, OACB is a quadrant of a circle with centre O and radius 3.5 cm. If OD = 2 cm, find the area of the (i) quadrant OACB (ii) shaded region.

Answer:

It is given that OACB is a quadrant of circle with centre at O and radius 3.5 cm.

(i) We know that the area of quadrant of circle of radius r is,

Substituting the value of radius,

Hence, the area of OACB is.

(ii) It is given that radius of quadrant of small circle is 2 cm.

Let the area of quadrant of small circle be.

It is clear from the above figure that area of shaded region is the difference of larger quadrant and the smaller one. Hence,

Page No 15.48:

Question 16:

From each of the two opposite corners of a square of side 8 cm, a quadrant of a circle of radius 1.4 cm is cut. Another circle of radius 4.2 cm is also cut from the centre as shown in Fig. 15.71. Find the area of the remaining (Shaded) portion of the square. (Use π = 22/7)

Answer:

It is given that a circle of radius 4.2 cm and two quadrants of radius 1.4 cm are cut from a square of side 8 cm.

Let the side of square be a. Then,

Since the radius of circle is 4.2 cm. So, 

Area of circle=πr2=227×4.2×4.2=55.44 cm2

Now area of quadrant of circle of radius 1.4 cm is,

Area of shaded region = Area of square-Area of circle-2×Area of quadrant=64-55.44-2×1.54=5.48 cm2



Page No 15.49:

Question 18:

In Fig 15.72 (a), OABC is a square of side 7 cm. If OAPC is a quadrant of a circle with centre O, then find the area of the shaded region. (Use π = 22/7)

Answer:

Area of shaded region = Area of square OABC − Area of quadrant OAPC
=Side2-14πr2=72-14×227×7×7=49-38.5=10.5 cm2
Hence, the area of the shaded region is 10.5 cm2

Page No 15.49:

Question 17:

Find the area of the shaded region in Fig. 15.72, if AC = 24 cm, BC = 10 cm and O is the centre of the circle. (Use π = 3.14)

Answer:

It is given a triangle ABC is cut from a circle.

In,

, Since any angle inscribed in semicircle is always right angle.

By applying Pythagoras theorem,

We know that the area A of circle of radius r is

Substituting the value of radius r,

Page No 15.49:

Question 1:

Three circles are placed on a plane in such a way that each circle just touches the other two, each having a radius of 10 cm. Find the area of region enclosed by them.

Answer:

Let there be 3 circles with centre A, B and C respectively such that each circle touches the other two. They all have radius of 10 cm.

We have to find the area enclosed between the circles.

We have,


AB = 10 + 10 cm = 20 cm
BC = 10 + 10 cm = 20 cm
CA = 10 + 10 cm = 20 cm

Henceis an equilateral triangle.

So,,

So, the area enclosed between the circles,

So,

Put the values to get,

Page No 15.49:

Question 2:

The diameters of the front and rear wheels of a tractor are 80 cm and 2 m respectively. Find the number of revolutions that rear wheel will make to cover the distance which the front wheel covers in 1400 revolutions.

Answer:

Let us find the distance covered by front wheel in 1400 revolutions.

We know that distance covered in n revolutions is equal to multiplication of number of revolutions and circumference of wheel.

We have therefore, we will convert it meters.

Now we substitute the values.

Now we will calculate the number of revolutions that rear wheel will make to cover 7040 m.

Therefore, the rear wheel will make revolutions to cover 7040 m.

Page No 15.49:

Question 3:

Find the area of the circle in which a square of area 64 cm2 is inscribed. [Use π = 3.14]

Answer:

We have given area of the square.

Now we will find the diameter of the square.

We know that diagonal of the square is same as the diameter of the circle.

Now we will find the area of the circle as shown below.

Therefore, area of the circle is.

Page No 15.49:

Question 4:

The diameter of a wheel of a bus is 90 cm which makes 315 revolutions per minute. Determine its speed in kilometres per hour. [Use π = 22/7].

Answer:

We have to calculate the speed of the bus. The wheel makes 315 revolutions per minute. We have,

Diameter of the wheel

So distance travelled in one revolution,

So,

So,

Convert the unit of speed as,

Page No 15.49:

Question 5:

In Fig. 15.75, OE = 20 cm. In sector OSFT, square OEFG is inscribed. Find the area  of the shaded region.

Answer:

We have to find the area of the shaded portion. We have, and OEFG is a square.

Use Pythagoras theorem to find OF as,

So, radius of the circle,

Therefore area of the shaded region,

So,

Page No 15.49:

Question 6:

In Fig. 15.76, O is the centre of circle of radius 28 cm. Find the area of minor segment ASB.

Answer:

We have given that radius of the circle is 28 cm and sector angle is.

Now we will find the area of the minor segment ASB.

Therefore, area of the minor segment ASB is.



Page No 15.50:

Question 7:

In Fig. 15.77, PQRS is a square of side 4 cm. Find the area of the shaded square.

Answer:

Area of the shaded region is equal to the area of the square minus area of four sectors with the same radius minus area of the circle.

We know that area of the four sectors with radius is equal to area of one circle.

Therefore, area of the shaded region is

Page No 15.50:

Question 8:

A bucket is raised from a well by means of a rope which is wound round a wheel of diameter 77 cm (Fig .15.78). Given that the bucket ascends in 1 minute 28 seconds with a uniform speed of 1.1 m/s. Calculate the number of complete revolutions the wheel makes in raising the bucket.

Answer:

A bucket is pulled from a well using a rope. We have,

Velocity by which rope is pulled

Diameter of the pulley

Time taken to pull the rope,

So total length of the rope pulled,

Therefore,

So,

Page No 15.50:

Question 9:

A circle is inscribed in an equilateral triangle ABC is side 12 cm, touching its sides (Fig. 15.79). Find the radius of the inscribed circle and the area of the shaded part.

Answer:

We have to find the area of the shaded portion. We havewhich is an equilateral triangle and.

We have O as the incentre and OP, OQ and OR are equal.

So,

Thus,

So area of the shaded region,

Page No 15.50:

Question 10:

In Fig. 15.80, an equilateral triangle ABC of side 6 cm has been inscribed in a circle. Find the area of the shaded region. (Take π = 3.14).

Answer:

We have to find the area of the shaded portion. We havewhich is an equilateral triangle and. Let r be the radius of the circle.

We have O as the circumcentre.

So,

Thus,

So area of the shaded region,

Page No 15.50:

Question 11:

A circular field has a perimeter of 650 m. A square plot having its vertices on the circumference of the field is marked in the field. Calculate the area of the square plot.

Answer:

We have a circular field in which a square field is marked.

Let the radius of the circle be r. We have,

Use Pythagoras theorem to find the side of square as,

So area of the square plot,

Page No 15.50:

Question 12:

In Fig. 15.81, the square ABCD is divided into five equal parts, all having same area. The central part is circular and the lines AE, GC, BF and HD lie along the diagonals AC and BD of the square. If AB = 22 cm, find:

(i) the circumference of the central part.

(ii) the perimeter of the part ABEF.



 

Answer:

We have a square ABCD.

We have,

(i)We have to find the perimeter of the triangle. We have a relation as,

So,

So perimeter of the circular region,

(ii)We have to find the perimeter of ABEF. Let O be the centre of the circular region.

Use Pythagoras theorem to get,

Similarly,

Now length of arc EF,

So, perimeter of ABFE,



Page No 15.51:

Question 13:

In Fig. 15.82, ABCD is a rectangle, having AB = 20 cm and BC = 14 cm. Two sectors of 180° have been cut off. Calculate:

(i) the area of the shaded region.
(ii) the  length of the boundary of the shaded region.

Answer:

(i) We have given two semi-circles and a rectangle.

Area of the shaded region = Area of the rectangle − Area of the two semicircles ……..(1)

Substituting we get,

Therefore, area of shaded region is.

(ii) Now we will find length of the boundary of the shaded region.

Therefore, length of the boundary of the shaded region is.

Page No 15.51:

Question 14:

The diameter of a coin is 1 cm (Fig. 15.83). If four such coins be placed on a table so that the rim of each touches that of the other two, find the area of the shaded region (Take π = 3.1416).

Answer:

Look at the figure carefully shaded region is bounded between four sectors of the circle with same radius and a square of side 1 cm.

Therefore, the area of the shaded region is nothing but the difference the area of the square and area of one circle.

Substituting we get,

Therefore, area of the shaded region is.

Page No 15.51:

Question 15:

From a circular piece of cardboard of radius 3 cm two sectors of 90° have been cut off. Find the perimeter of the remaining portion nearest hundredth centimeters (Take π = 22/7).

Answer:

It is very clear that if we cut two sectors of then we will be left with a semi-circle without diameter.

Look at the figure. If we cut sectors OPA and OPB then we will get curve AB.

Therefore, the required perimeter will be length of the curve AB.

Now we will calculate the perimeter of the remaining curve AB as shown below.

Therefore, perimeter of the remaining portion is.

Page No 15.51:

Question 16:

The area of a sector is one-twelfth that of the complete circle. Find the angle of the sector.

Answer:

We have given that area of the sector is one-twelfth of the area of the circle.

Now we will simplify the above equation as below,

Now we will multiply both sides of the equation by 360.

Therefore, angle of sector is.

Page No 15.51:

Question 17:

The radius of a circle with centre O is 5 cm (Fig . 15.84). Two radii OA and OB are drawn at right angles to each other. Find the areas of the segments made by the chord AB (Take π = 3.14).

Answer:

We have given that the radius of the circle is 5 cm.

First we will find the area of the minor segment AB as given below,

Substituting the values we get,

Now we will calculate the area of the major segment AB as given below,

Therefore, areas of the minor and major segments are respectively.

Page No 15.51:

Question 18:

ABCDEF is a regular hexagon with centre O (Fig. 15.85). If the area of triangle OAB is 9 cm2, find the area of : (i) the hexagon and (ii) the circle in which the haxagon is incribed.

Answer:

We know that a regular hexagon is made up of 6 equilateral triangles.

We have given area of the one of the triangles.

We know that if a regular hexagon is inscribed in the circle, then the radius of the circle is same as the side of the regular hexagon.

We also know that a regular hexagon is made up of 6 equilateral triangles and we have area of one of the equilateral triangle.

Substituting the value of the given equilateral triangle we get,

Now we will find the area of the circle.

Substituting the values we get,

Now we will substitute we get,

Therefore, area of the hexagon and area of the circle are and respectively.

Page No 15.51:

Question 19:

From a thin metallic piece, in the shape of a trapezium ABCD, in which AB || CD and ∠BCD = 90°, a quarter circle BEFC is removed (see Fig. 15.86). Given AB = BC = 3.5 cm and DE = 2 cm, calculate the area of the remaining piece of the metal sheet.

Answer:

We have given a trapezium. We are asked to find the area of the shaded region.

We can find the area of the remaining part that is area of the shaded region as shown below.

……..(1)

Now we find the value of CD.

………(Since, CE is radius of the sector, therefore, CE = 3.5)

Substituting the values of CD and in equation (1),

Therefore, area of the remaining part is.



Page No 15.52:

Question 20:

Find the area enclosed between two concentric circles of radii 3.5 cm and 7 cm. A third concentric circle is drawn outside the 7 cm circle, such that the area enclosed between it and the 7 cm circle is same as that between the two inner circles. Find the radius of the third circle correct to one decimal place.

Answer:

We know the length of OA and OB.

Now we will calculate the area between the circles of radii 3.5 cm and 7 cm as shown below,

Substituting the values of, and we get,

Therefore, area enclosed by the two circles with radii 3.5 cm and 7 cm is.

Now we will draw another circle such that radius of the circle will.

Now we will find.

We have given that area enclosed by the circle with radius and is equal to the area of the enclosed by the circle with radius and.

Substituting the values we get,

Adding 49 both sides of the equation we get,

Now we will take the square root.

Therefore, radius of the third circle is.

Page No 15.52:

Question 21:

A path of 4 m width runs round a semi-circular grassy plot whose circumference is 163 37 m Find:

(i) the area of the path

(ii) the cost of gravelling the path at the rate of Rs 1.50 per square metre

(iii)  the cost of turfing the plot at the rate of 45 paise per m2.

Answer:

We have given AB = 4m and circumference of semicircle with radius OA as.

We are asked to find the area between the two semi-circles.

For that we will first find OA.

πr=16337

Now we will substitute

    227×r=16337r=11447×722r=52OA=52 m

Now we will find OB.

OB=52+4OB=56  m

Now we will find the area between two semi-circles as given below,
Area=π×56×562-π×52×522             =π21568-1352 m2             =π2×216             =227×2×216             =339.43 m2

Therefore, area of the path is 339.43 m2.

Now we will find the cost of gravelling the path.

Cost=339.43×1.50        =Rs 509.14

Therefore, cost of gravelling the path is Rs 509.14.

Now we will find the cost of turfing the plot. For that we will find the area of the plot.

Area of the plot=12πr2                           =12×227522                           =12×227×52×52                           =4249.14Cost of turfing the plot= 4249.14×0.45                                      =Rs 1912.11

Therefore, cost of the turfing the plot is Rs 1912.11.
Disclaimer: Due to some error in the question, we get the different answers.

Page No 15.52:

Question 22:

Fig. 15.87, shows the cross-section of railway tunnel. The radius OA of the circular part is 2 m. If ∠AOB = 90°, calculate:

(i) the height of the tunnel

(ii) the perimeter of the cross-section

(iii) the area of the cross-section.



figure

 

Answer:

We have a cross section of a railway tunnel. is a right angled isosceles triangle, right angled at O. let OM be perpendicular to AB.

(i) We have to find the height of the tunnel. We have,

Use Pythagoras theorem into get,

Let the height of the tunnel be h. So,

Thus,

Therefore,

h=2+2 m

(ii)

Perimeter of cross-section is,


=3π+22 m

(iii)


3π+2 m2

Page No 15.52:

Question 23:

In Fig. 15.88, ABC is a right-angled triangle, ∠B = 90°, AB = 28 cm and BC = 21 cm. With AC as diameter a semicircle is drawn and with BC as radius a quarter circle is drawn. Find the area of the shaded region correct to two decimal places.

Answer:

We have given two semi-circles and one circle.

Area of the shaded region = area of semicircle with diameter AC + area of right angled triangle ABC − area of sector

First we will find the hypotenuse of right angled triangle ABC.

Substituting we get,

Therefore, area of shaded region is.

Page No 15.52:

Question 24:

In Fig. 15.89, there are three semicircles, A, B and C having diameter 3 cm each, and another semicircle E having a circle D with diameter 4.5 cm are shown. Calculate:

(i) the area of the shaded region

(ii)  the cost of painting the shaded region at the rate of 25 paise per cm2 , to the nearest rupee.

Answer:

(i) Area of the shaded region can be calculated as shown below,

Area of the shaded region = Area of the semi-circle with diameter of 9 cm − area of 2 semi-circles with radius 3cm − area of the circle with centre D + area of semi-circle with radius 3 cm

Substituting we get,

Therefore, area of the shaded region is.

Now we will find the cost of painting the shaded region at the rate of 25 paise per cm2.

paise

Therefore, cost of painting the shaded region to the nearest rupee is.



Page No 15.53:

Question 25:

In Fig. 15.90, O is the centre of a circular arc and AOB is a straight line. Find the perimeter and the area of the shaded region correct to one decimal place. (Take π = 3.142)

Answer:

(i) Let us find the perimeter of the shaded region.

Therefore, perimeter of the shaded region is .

Now we will find the area of the shaded region can be calculated as shown below,

Area of the shaded region = Area of the semi-circle − area of the right angled triangle

First, we will find the length of AB as shown below,

Substituting we get,

Therefore, area of the shaded region is.

Page No 15.53:

Question 26:

In Fig. 15.91, ABCD is a square of side 2a, Find the ratio between

(i) the circumferences

(ii) the areas of the in circle and the circum-circle of the square. 

Answer:

We have a square ABCD having. From the given diagram we can observe that,

Radius of incircle

Radius of circumcircle

(i) We have to find the ratio of the circumferences of the two circles. So the required ratio is,

(ii) We have to find the ratio of the areas of the two circles. So the required ratio is,

Page No 15.53:

Question 27:

In Fig. 15.92, two circles with centres A and B touch each other at the point C. If AC = 8 cm and AB = 3 cm, find the area of the shaded region.

Answer:

Area of the shaded region can be calculated as shown below,

Area of the shaded region = Area of circle with radius AC − area of circle with radius BC

We have given radius of the outer circle that is 8 cm but we don’t know the radius of the inner circle.

We can calculate the radius of the inner circle as shown below,

Substituting we get,

Therefore, area of the shaded region is.

Page No 15.53:

Question 28:

In Fig. 15.93, ABCD is a rectangle with AB = 14 cm and BC = 7 cm. Taking DC, BC and AD as diameters, three semi-circles are drawn as shown in the figure. Find the area of the shaded region.

Answer:

Area of the shaded region can be calculated as shown below,

Area of the shaded region = Area of rectangle − area of the semi-circle with diameter DC triangle + 2 area of two semicircles with diameters AD and BC

Substituting we get,

Therefore, area of the shaded region is.



Page No 15.54:

Question 29:

In Fig. 15.94, ABC is an equilateral triangle of side 8 cm. A, B and C are the centres of circular arcs of radius 4 cm. Find the area of the shaded region correct upto 2 decimal places. (Take π =3.142 and 3 = 1.732).

Answer:

Area of the shaded region can be calculated as shown below,

Area of the shaded region = Area of equilateral triangle − 3 area of circular arc

Substituting and we get,

Therefore, area of the shaded region is.

Page No 15.54:

Question 30:

In Fig. 15.95, the boundary of the shaded region consists of four semi-circular arcs, the smallest two being equal. If the diameter of the largest is 14 cm and of the smallest is 3.5 cm, find

(i) the length of the boundary.
(ii) the area of the shaded region.

Answer:

(i) We will first find the length of the boundary.

Length of the boundary perimeter of semi-circle with diameter AB + boundary of semi-circle with diameter 7 cm

Therefore, length of the boundary is.

Now we will find the area of the shaded region as shown below,

Area of the shaded region = Area of the semi-circle with AB as a diameter − area of the semi-circle with radius AE − area of the semi-circle with radius BC + area of the semi-circle with diameter 7 cm.

Therefore, area of the shaded region is.

Page No 15.54:

Question 31:

Fig. 15.96, shows a kite in which BCD is the shape of a quadrant of a circle of radius 42 cm. ABCD is a square and Δ CEF is an isosceles right angled triangle whose equal sides are 6 cm long. Find the area of the shaded region.

Answer:

We will find the area of the shaded region as shown below,

Area of the shaded region = area of quadrant + area of isosceles triangle ……..(1)

Substituting we get,

Therefore, area of the shaded region is.



Page No 15.55:

Question 34:

In Fig. 15.99 a square OABC is inscribed in a quadrant OPBQ of a circle. If OA = 21 cm, find the area of the shaded region.

 

Answer:

Construction: Join OB



In right triangle AOB
OB2 = OA2 + AB2
= 212 + 212
= 441 + 441
= 882
∴ OB2 = 882
Area of the shaded region = Area of quadrant OPBQ − Area of Square OABC
=14πOB2-OA2=14×227×882-441=693-441=252 cm2
14π(OB)2(OA)2=14×3.14×800400=628400=228 cm2
Hence, the area of the shaded region is 252 cm2.

Page No 15.55:

Question 35:

Two circular pieces of equal radii and maximum area, touching each other are cut out from a rectangular card board of dimensions 14 cm × 7 cm. Find the area of the remaining card board. (Use π = 22/7).

Answer:

We know that we can cut two circular pieces of equal radii and maximum area from the rectangular cardboard whose diameter is equal to the width of the rectangular cardboard.
∴ Radii of two circuar pieces = Half of the width of the rectangular cardboard = 3.5 cm
Now,
Area of remaining cardboard = Area of rectangular cardboard − 2 ⨯ Area of circular piece having radius 3.5 cm
=14×7-2227×3.5×3.5=98-77=21 cm2
=14×72(227×3.5×3.5)=9877=21 cm2
Hence, the area of the remaining cardboard is 21 cm2

Page No 15.55:

Question 32:

In Fig. 15.97, AB = 36 cm and M is mid-point of AB. Semi-circles are drawn on AB, AM and MB as diameters. A circle with centre C touches all the three circles. Find the area of the shaded region.

Answer:

We have given two semi-circles and one circle.

Area of the shaded region = area of semicircle with diameter AB − area of two semicircles with diameters AM and MB - area of circle ……..(1)

Let us calculate the area of the semi-circle with AB as a diameter.

Now we will find the area of the semi-circle with AM as a diameter.

Area of the semi-circle with MB as a diameter is same as the area of the semi-circle with diameter with AM as a diameter.

Now we will find the area of the circle with centre C.

We know that radius of the circle is one sixth of AB.

Now we will substitute all these values in equation (1).

Therefore, area of shaded region is.

Page No 15.55:

Question 33:

In Fig. 15.98, ABC is a right angled triangle in which ∠A = 90°, AB = 21 cm and AC = 28 cm. Semi-circles are described on AB, BC and AC as diameters. Find the area of the shaded region.

Answer:

We have given three semi-circles and one right angled triangle.

     ……..(1)

Let us calculate the area of the semi-circle with AB as a diameter.

Now we will find the area of the semi-circle with AC as a diameter.

Now we will find the length of BC.

In right angled triangle ABC, we will use Pythagoras theorem,

Now we will calculate the area of the right angled triangle ABC.

Now we will find the area of the semi-circle with BC as a diameter.

Now we will substitute all these values in equation (1).

Therefore, area of shaded region is.



Page No 15.57:

Question 1:

What is the ratio of the areas of a circle and an equilateral triangle whose diameter and a side are respectively equal?

Answer:

We are given that diameter and side of an equilateral triangle are equal.

Let d and a are the diameter and side of circle and equilateral triangle respectively.

Therefore d = a

We know that area of the circle

Area of the equilateral triangle

Now we will find the ratio of the areas of circle and equilateral triangle.

We know that radius is half of the diameter of the circle.

Now we will substitute in the above equation,

Therefore, ratio of the areas of circle and equilateral triangle is.



Page No 15.58:

Question 2:

If the circumference of two circles are in the ratio  2 : 3, what is the ratio of their areas?

Answer:

We are given ratio of circumferences of two circles. If and are circumferences of two circles such that

…… (1)

Simplifying equation (1) we get,

Let and are the areas of the respective circles and we are asked to find their ratio.

…… (2)

We know that substituting this value in equation (2) we get,

Therefore, ratio of their areas is.

Page No 15.58:

Question 3:

Write the area of the sector of a circle whose radius is r and length of the arc is l.

Answer:

We know that area of the sector of the circle of radius r

Length of the arc

But we have given that length of the arc

…… (1)

Area of the sector

Now we will adjust 2 in the following way,

Area of the sector

Area of the sector

From equation (1) we will substitute

Area of the sector

Area of the sector

Therefore, area of the sector =.

Page No 15.58:

Question 4:

What is the length (in terms of π)  of the arc that subtends an angle of 36° at the centre of a circle of radius 5 cm?

Answer:

We have

We have to find the length of the arc.

Substituting the values we get,

……….(1)

Now we will simplify the equation (1) as below,

Therefore, length of the arc is

Page No 15.58:

Question 5:

What is the angle subtended at the centre of a circle of radius 6 cm by an arc of length 3 π cm?

Answer:

We have

We will find the angle subtended at the centre of a circle.

Substituting the values we get,

…… (1)

Now we will simplify the equation (1) as below,

Therefore, angle subtended at the centre of the circle is.

Page No 15.58:

Question 6:

What is the area of a sector of a circle of radius 5 cm formed by an arc of length 3.5 cm?

Answer:

We have

Now we will find the area of the sector.

Substituting the values we get,

…… (1)

Now we will simplify the equation (1) as below,

Therefore, area of the sector is.

Page No 15.58:

Question 7:

In a circle of radius 10 cm, an arc subtends an angle of 108° at the centre. what is the area of the sector in terms of π?

Answer:

We have given the radius of the circle and angle subtended at the centre of the circle.

Now we will find the area of the sector.

Substituting the values we get,

…… (1)

Now we will simplify the equation (1) as below,

Therefore, area of the sector is.

Page No 15.58:

Question 8:

If a square is inscribed in a circle, what is the ratio of the areas of the circle and the square?

Answer:

We have the following situation

Let BD be the diameter and diagonal of the circle and the square respectively.

We know that area of the circle

Area of the square

As we know that diagonal of the square is the diameter of the square.

…… (1)

Substituting in equation (1) we get,

Now we will find the ratio of the areas of circle and square.

Now we will simplify the above equation as below,

Therefore, ratio of areas of circle and square is.

Page No 15.58:

Question 9:

Write the formula for the area of a sector of angle θ (in degrees) of a circle of radius r.

Answer:

Let r be the radius of the circle and angle θ subtended at the centre of the circle.

Area of the sector of the circle

Therefore, area of the sector is.

Page No 15.58:

Question 10:

Write the formula for the area of a segment in a circle of radius r given that the sector angle is θ (in degrees).

Answer:

In this figure, centre of the circle is O, radius OA = r and

We are going to find the area of the segment AXB.

…… (1)

We know that

We also know that

Substituting these values in equation (1) we get,

Therefore, area of the segment is.

Page No 15.58:

Question 11:

If the adjoining figure is a sector of a circle of radius 10.5 cm, what is the perimeter of the sector? (Take π=22/7)

Answer:

Given figure is a quadrant of a circle. We have given radius of sector that is 10.5 cm. Arc AB subtended an angle of 60° at the centre of the circle.

Perimeter of the sector

Substituting the values we get,

Perimeter of the sector …… (1)

Now we will simplify equation (1) as shown below,

Now we will substitute.

Therefore, perimeter of the given sector is.



Page No 15.59:

Question 12:

If the diameter of a semi-circular protractor is 14 cm, then find its perimeter.

Answer:

Let AB be the diameter of the semi-circular protractor.

We know that perimeter of the semicircle ……(1)

We have given the diameter of the protractor.

Therefore, radius of the protractor

So, radius of the protractor

 

Substituting the value of r in equation (1) we get,

Substituting we get,

Therefore, perimeter of the semi-circular protractor is.

Page No 15.59:

Question 13:

An arc subtends an angle of 90° at the centre of the circle of the radius 14 cm. Write the area of minor sector thus formed in terms of π.

Answer:

We have given an angle subtended by an arc at the centre of the circle and radius of the circle.

Now we will find the area of the minor sector.

Area of the minor sector =

Substituting the values we get,

Area of the minor sector = …… (1)

Now we will simplify the equation (1) as below,

Area of the minor sector =

Area of the minor sector =

Area of the minor sector = π × 7 × 7

Area of the minor sector = 49π

Therefore, area of the minor sector is.

Page No 15.59:

Question 1:

If the circumference and the area of a circle are numerically equal, then diameter of the circle is

(a) π2

(b) 2π

(c) 2

(d) 4

Answer:

We have given that circumference and area of a circle are numerically equal.

Let it be x.

Let r be the radius of the circle, therefore, circumference of the circle is and area of the circle will be.

Therefore, from the given condition we have,

………(1)

………(2)

Therefore, from equation (1) get . Now we will substitute this value in equation (2) we get,

Simplifying further we get,

Cancelling x we get,

Now we will cancel

………(3)

Now we will multiply both sides of the equation (3) by we get,

We can rewrite this equation as given below,

Comparing equation (4) with equation (1) we get r = 2.

Therefore, radius of the circle is 2. We know that diameter of the circle is twice the radius of the circle.

Therefore, diameter of the circle is.

Hence, option (d) is correct.

Page No 15.59:

Question 2:

If the difference between the circumference and radius of a circle is 37 cm, then using π = 227, the circumference (in cm) of the circle is

(a) 154

(b) 44

(c) 14

(d) 7

Answer:

We know that circumference; C of the circle with radius r is equal to.

We have given difference between circumference and radius of the circle that is 37 cm.

Substituting we get,

Dividing both sides of the equation by, we get,

Therefore, circumference of the circle will be 
2πr=2×227×7=44 cm2

Hence, the correct choice is (b).

Page No 15.59:

Question 3:

A wire can be bent in the form of a circle of radius 56 cm. If it is bent in the form fo a square, then its area will be

(a) 3520 cm2

(b) 6400 cm2

(c) 7744 cm2

(d) 8800 cm2

Answer:

We have given that a wire is bent in the form of circle of radius 56 cm. If we bent the same wire in the form of square of side a cm, the perimeter of the wire will not change.

perimeter of the circle = perimeter of the square

We know that r = 56 cm.

Now we will substitute the value of r in the equation,

………(1)

Dividing both sides of the equation by 4 we get,

Now we obtained side of the square. Now we can calculate the area of the square as given below.

Hence, the area of the square is .

Therefore, the correct answer is (c).

Page No 15.59:

Question 4:

If a wire is bent into the shape of a square, then the area of the square is 81 cm2 . When wire is bent into a semi-circular shape, then the  area of the semi-circle will be

(a) 22 cm2

(b) 44 cm2

(c) 77 cm2

(d) 154 cm2

Answer:

We have given that a wire is bent in the form of square of side a cm such that the area of the square is . If we bent the same wire in the form of a semicircle with radius r cm, the perimeter of the wire will not change.

perimeter of the square = perimeter of semi circle

………(1)

We know that area of the square = .

Now we will substitute the value of a in the equation (1),

Now we will substitute.

Multiplying both sides of the equation by 7 we get,

Now we will divide both sides of the equation by 36 we get, r = 7

Therefore, radius of the semi circle is 7cm.

Now we will find the area of the semicircle.

Area of the semicircle

Therefore, area of the semicircle is .

Hence the correct answer is option (c).

Page No 15.59:

Question 5:

A circular park has a path of uniform width around it. The difference between the outer and inner circumferences of the circular path is 132 m. Its width is

(a) 20 m

(b) 21 m

(c) 22 m

(d) 24 m

Answer:

Let OA = r be the radius of the inner circle and OB = r be the radius of the outer circle.

Therefore, circumference of the inner circle and circumference of the outer circle

Here we have to find the width of the circular park that is we have to find .

We have given the difference between the circumferences of outer circle and inner circle.

Substituting we get,

Now we will multiply both sides of the equation by .

Therefore, the width is.

Hence the correct answer is option (b).

Page No 15.59:

Question 6:

The radius of a wheel is 0.25 m. The number of revolutions it will make to travel a distance of 11 km will be

(a) 2800  

(b) 4000

(c) 5500

(d) 7000

Answer:

We have given the radius of the wheel that is 0.25 cm.

We know that distance covered by the wheel in one revolution.

Distance covered by the wheel in one revolution is equal to the circumference of the wheel.

………(1)

Distance moved is given as 11 km so we will first convert it to m.

11 km = 11000 m

Now we will substitute the values in equation (1),

Now we will substitute.

Simplifying equation (1) we get,

Therefore, it will make revolutions to travel a distance of 11 km.

Hence, the correct answer is option (d).

Page No 15.59:

Question 7:

The ratio of the outer and inner perimeters of a circular path is 23 : 22. If the path is 5 metres wide, the diameter of the inner circle is

(a) 55 m

(b) 110 m

(c) 220 m

(d) 230 m

Answer:

Let OA = r be the radius of the inner circle and OB = r be the radius of the outer circle.

Therefore, circumference of the inner circle and circumference of the outer circle

Here we have to find the diameter of the inner circle.

We have given the ratio of outer and inner perimeters of a circular path.

Simplifying the above equation we get,

……..(1)

We have also given that the path is 5 meters wide, that is we have given

……..(2)

We are asked to find the diameter of the inner circle hence, we will eliminate r using equations (1) and (2) for that we will multiply equation (2) by 22.

……..(3)

Subtracting equation (1) from equation (3) we get,

Therefore, radius of the inner circle is 110 meters.

Therefore, diameter of the inner circle meters

Therefore, diameter of the inner circle is .

Hence, the correct answer is option (c).

Page No 15.59:

Question 8:

The circumference of a circle is 100 m. The side of a square inscribed in the circle is

(a) 502

(b) 50π

(c) 502π

(d) 1002π

Answer:

We have given the circumference of the circle that is 100 cm. If d is the diameter of the circle, then its circumference will be.

We obtained diameter of the circle. Look at the figure, diameter of the circle is also the diagonal of the square ABCD.

We know that if we have diagonal of the circle we can calculate the side of the square, using the formula given below,

Substituting the value of diagonal we get,

Therefore, side of the inscribed square is.

Hence, the correct answer is option (d).

Page No 15.59:

Question 9:

The area of the incircle of an equilateral triangle of side 42 cm is

(a) 223 cm2

(b) 231 cm2

(c) 462 cm2

(d) 924 cm2

Answer:

Let ABC be the equilateral triangle such that AB = BC = CA = 42 cm. Also, let O be the centre and r be the radius of its incircle.

AB, BC and CA are tangents to the circle at M, N and P.

∴ OM = ON = OP = r

Area of ΔABC = Area (ΔOAB) + Area (ΔOBC) + Area (ΔOCA)

Area of the circle = 

Hence, the correct answer is option (c).

Page No 15.59:

Question 10:

The area of incircle of an equilateral triangle is 154 cm2 . The perimeter of the triangle is

(a) 71.5 cm

(b) 71.7 cm

(c) 72.3 cm

(d) 72.7 cm

Answer:

Area of incircle of equilateral triangle is 154

We have to find the perimeter of the triangle. So we will use area to get,

As triangle is equilateral so,

So,

So,

Therefore perimeter of the triangle is,

Therefore the answer is (d).

Page No 15.59:

Question 11:

The area of the largest triangle that can be inscribed in a semi-circle of radius r, is

(a) r2

(b) 2r2

(c) r3

(d) 2r3

Answer:

The triangle with the largest area will be symmetrical as shown in the figure.

Let the radius of the circle be r.

Hence,

Therefore the answer is (a).



Page No 15.60:

Question 12:

The perimeter of a triangle is 30 cm and the circumference of its incircle is 88 cm. The area of the triangle is

(a) 70 cm2

(b) 140 cm2

(c) 210 cm2

(d) 420 cm2

Answer:

We have to find the area of the given triangle.

Perimeter of triangle is 30 cm.

Let the radius of the circle be r.

We have,

Therefore,

Therefore the answer is (c).

Page No 15.60:

Question 13:

The area of a circle is 220 cm2. The area of ta square inscribed in it is

(a) 49 cm2

(b) 70 cm2

(c) 140 cm2

(d) 150 cm2

Answer:

Let BD be the diameter and diagonal of the circle and the square respectively.

We know that area of the circle

Area of the circle

Multiplying both sides of the equation by 7 we get,

Dividing both sides of the equation by 22 we get,

As we know that diagonal of the square is the diameter of the square.

………(1)

Substituting in equation (1) we get,

Area of the square

Therefore, area of the square is.

Hence, the correct answer is option (c).

Page No 15.60:

Question 14:

If the circumference of a circle increases from 4π to 8π, then its area is

(a) halved

(b) doubled

(c) tripled

(d) quadrupled

Answer:

Let the circumference

Therefore, area of the circle when radius of the circle is 2 can be calculated as below,

……..(1)

Now when circumference is, then the radius of the circle is calculated as below,

Therefore, area of the circle when radius of the circle is 2 can be calculated as below,

……..(2)

Therefore, from equation (1) and (2) we can say that its area is.

Hence, the correct answer is option (d).

Page No 15.60:

Question 15:

If the radius of a circle is diminished by 10%, then its area is diminished by

(a) 10%

(b) 19%

(c) 20%

(d) 36%

Answer:

Let x be the initial radius of the circle.

Therefore, its area is ……..(1)

It is given that the radius is diminished by 10%, therefore, its new radius is calculated as shown below,

Now we will find the percentage decreased in the area.

Therefore, its area is diminished by.

Hence, the correct answer is option (b).

Page No 15.60:

Question 16:

If the area of a square is same as the area of a circle, then the ratio of their perimeters, in terms of π, is

(a) π : 3

(b) 2 : π

(c) 3 : π

(d) π : 2

Answer:

We have given that area of a circle of radius r is equal to the area of a square of side a.

We have to find the ratio of the perimeters of circle and square.

………(1)

Now we will substitute in equation (1).

Therefore, ratio of their perimeters is.

Hence, the correct answer is (b).

Page No 15.60:

Question 17:

The area of the largest triangle that can be inscribed in a semi-circle of radius r is

(a) 2r

(b) r2

(c) r

(d) r

Answer:

The triangle with the largest area will be symmetrical as shown in the figure.

Let the radius of the circle be r.

Hence,

Therefore the answer is (b).

Page No 15.60:

Question 18:

The ratio of the areas of a circle and an equilateral triangle whose diameter and a side are respectively equal, is

(a) π:2

(b) π:3

(c) 3:π

(d) 2:π

Answer:

We are given that diameter and side of an equilateral triangle are equal.

Let d and a are the diameter and side of circle and equilateral triangle respectively.

We know that area of the circle

Area of the equilateral triangle

Now we will find the ratio of the areas of circle and equilateral triangle.

We know that radius is half of the diameter of the circle.

Now we will substitute in the above equation,

Therefore, ratio of the areas of circle and equilateral triangle is.

Hence, the correct answer is option (b).

Page No 15.60:

Question 19:

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  r12+r22
(a) >r2

(b) =r2

(c) <r2

(d) None of these

Answer:

We have given area of the circle of radius r1 plus area of the circle of radius r2 is equal to the area of the circle of radius r.

Therefore, we have,

Cancelling, we get

Therefore, .

Hence, the correct answer is option (b).

Page No 15.60:

Question 20:

If the perimeter of a semi-circular protractor is 36 cm, then its diameter is

(a) 10 cm

(b) 12 cm

(c) 14 cm

(d) 16 cm

Answer:

We know that perimeter of a semi-circle of radius r ………(1)

We have given the perimeter of the semi-circle and we are asked to find the diameter of the semi-circle.

Therefore, substituting the perimeter of the semi-circle in equation (1) we get,

Multiplying both sides of the equation by 2 we get,

Substituting we get,

Now we will multiply both sides of the equation by 7.

Adding like terms we get,

Dividing both sides of the equation 72 we get,

Therefore, radius of the semi circle is 7cm.

Now we will find the diameter.

Hence, diameter of the semi-circle is.

Therefore, the correct answer is (c).

Page No 15.60:

Question 21:

The perimeter of the sector OAB shown in Fig. 15.106,  is

(a) 643 cm

(b) 26 cm

(c) 645 cm

(d) 19 cm

Answer:

We know that perimeter of a sector of radius r ………(1)

We have given sector angle and radius of the sector and we are asked to find perimeter of the sector OAB.

Therefore, substituting the corresponding values of the sector angle and radius in equation (1) we get,

………(2)

We will simplify equation (2) as shown below,

Substituting we get,

Now we will make the denominator same.

Therefore, perimeter of the sector is.

Hence, the correct answer is option (a).

Page No 15.60:

Question 22:

If the perimeter of a sector of a circle of radius 6.5 cm is 29 cm, then its area is

(a) 58 cm2

(b) 52 cm2

(c) 25 cm2

(d) 56 cm2

Answer:

We know that perimeter of a sector of radius r ………(1)

We have given perimeter of the sector and radius of the sector and we are asked to find the area of the sector. For that we have to find the sector angle.

Therefore, substituting the corresponding values of perimeter and radius in equation (1) we get,

………(2)

We will simplify equation (2) as shown below,

Dividing both sides of the equation by , we get,

Subtracting 1 from both sides of the equation we get,

………(3)

We know that area of the sector

From equation (3), we get

Area of the sector

Substituting we get,

Area of the sector

Therefore, area of the sector is .

Hence, the correct answer is option (b).

Page No 15.60:

Question 23:

If the area of a sector of a circle bounded by an arc of length 5π cm is equal to 20π cm2, then its radius is

(a) 12 cm

(b)16 cm

(c) 8 cm

(d) 10 cm

Answer:

We have given length of the arc and area of the sector bounded by that arc and we are asked to find the radius of the circle.

We know that area of the sector .

Now we will substitute the values.

……..(1)

……..(2)

Now we will divide equation (1) by equation (2),

Now we will cancel the like terms.

Therefore, radius of the circle is.

Hence, the correct answer is option (c).

Page No 15.60:

Question 24:

The area of the circle that can be inscribed in a square of side 10 cm is

(a) 40 π cm2

(b) 30 π cm2

(c) 100 π cm2

(d) 25 π cm2

Answer:

We know that ABCD is a square of length 10 cm. A circle is inscribed in the square therefore, all the sides of the square are become tangents of the circle.

By, the tangent property, we have

If we join PR then it will be the diameter of the circle of 10 cm.

Therefore, radius of the circle = 5cm

Therefore, area of the circle is.

Hence, the correct answer is option (d).



Page No 15.61:

Question 25:

If the difference between the circumference and radius of a circle is 37 cm, then its area is

(a) 154 cm2

(b) 160 cm2

(c) 200 cm2

(d) 150 cm2

Answer:

We have given the difference between circumference and radius of the circle.

Let C be the circumference, r be the radius and A be the area of the circle.

Therefore, from the given condition we have

Now we will substitute.

Now we will substitute the value of r in.

Now we will substitute.

Therefore, area of the circle is.

Hence, the correct answer is option (a).

Page No 15.61:

Question 26:

The area of a circular path of uniform width h surrounding a circular region of radius r is

(a) π(2r+h)r

(b) π(2r+h)h

(c) π(h+r)r

(d) π(h+r)h

Answer:

We have

Therefore, radius of the outer circle will be.

Now we will find the area between the two circles.

Cancelling we get,

Therefore, area of the circle is.

Hence, the correct answer is option (b).

Page No 15.61:

Question 27:

If AB is a chord of length 53 cm of a circle with centre O and radius 5 cm, then area of sector OAB is

(a) 3π8cm2

(b) 8π3cm2

(c) 25πcm2

(d) 25π3cm2

Answer:

We have to find the area of the sector OAB.

We have,

So,

Hence,

Therefore area of the sector,

So answer is (d)

Page No 15.61:

Question 28:

The area of a circle whose area and circumference are numerically equal, is

(a) 2π sq. units

(b) 4 π sq. units

(c) 6π sq. units

(d) 8π sq. units

Answer:

We have given that circumference and area of a circle are numerically equal.

Let it be x.

Let r be the radius of the circle, therefore, circumference of the circle is and area of the circle will be.

Therefore, from the given condition we have,

………(1)

………(2)

Therefore, from equation (1) get . Now we will substitute this value in equation (2) we get,

Simplifying further we get,

Cancelling x we get,

Now we will cancel

………(3)

Now we will multiply both sides of the equation (3) by we get,

Therefore, area of the circle is.

Hence, option (b) is correct.

Page No 15.61:

Question 29:

If diameter of a circle is increased by 40%, then its area increase by

(a) 96%

(b) 40%

(c) 80%

(d) 48%

Answer:

If d is the original diameter of the circle, then the original radius is.

If diameter of the circle increases by 40%, then new diameter of the circle is calculated as shown below,

That is new diameter

So, new area will be.

Now we will calculate the change in area.

Therefore, its area is increased by.

Hence, the correct answer is option (a).

Page No 15.61:

Question 30:

In Fig. 15.107, the shaded area is

(a) 50 (π−2) cm2

(b) 25 (π−2) cm2

(c) 25 (π+2) cm2

(d) 5 (π−2) cm2

Answer:

Area of the shaded region is-

So the answer is (b).

Page No 15.61:

Question 31:

In Fig. 15.108, the area of the segment PAQ is

(a) a24π+2

(b) a24π-2

(c) a24π-1

(d) a24π+1

Answer:

We have to find area of segment PAQ.

We know that.

Substituting the values we get,

Substituting and we get,

Now we will make the denominator same.

Therefore, area of the segment PAQ is.

Hence, the correct answer is option (b).



Page No 15.62:

Question 32:

In Fig. 15.109, the area of segment ACB is

(a) π3-32r2

(b) π3+32r2

(c) π3-23r2

(d) None of these

Figure

Answer:

We have to find area of segment ACB.

We know that.

Substituting the values we get,

Substituting and we get,

Therefore, area of the segment ACB is.

Hence, the correct answer is option (d).

Page No 15.62:

Question 33:

If the area of a sector of a circle bounded by an arc of length 5π  cm is equal to 20π cm2, then the radius of the circle

(a) 12 cm

(b) 16 cm

(c) 8 cm

(d) 10 cm

Answer:

We have given length of the arc and area of the sector bounded by that arc and we are asked to find the radius of the circle.

If l is the length of the arc, A is the area of the arc and r is the radius of the circle, then we know the expression of the area of the sector in terms of the length of the arc and radius of the circle.

Now we will substitute the corresponding values of length of the arc and area of the sector.

Multiplying both sides of the equation by 2 we get,

Dividing both sides of the equation by we get,

Therefore, radius of the circle is .

Hence, the correct answer is option (c).

Page No 15.62:

Question 34:

In Fig. 15.110, the ratio of the areas of two sectors S1 and S2 is

(a) 5 : 2

(b) 3 : 5

(c) 5 : 3

(d) 4 : 5

figure

Answer:

Area of the sector,

Area of the sector,

Now we will take the ratio,

Now we will simplify the ratio as below,

Substituting the values we get,

Therefore, ratio of the areas of the two sectors is .

Hence, the correct answer is option (d).

Page No 15.62:

Question 35:

If the area of a sector of a circle is 518 of the area of the circle, then the sector angle is equal to

(a) 60°

(b) 90°

(c) 100°

(d) 120°

Answer:

We have given that area of the sector is of the area of the circle.

Therefore, area of the sector area of the circle

Now we will simplify the equation as below,

Now we will multiply both sides of the equation by 360,

Therefore, sector angle is.

Hence, the correct answer is option (c).

Page No 15.62:

Question 36:

If he area of a sector of a circle is 720 of the area of the circle, then the sector angle is equal to

(a) 110°

(b) 130°

(c) 100°

(d) 126°

Answer:

We have given that area of the sector is of the area of the circle.

Therefore, area of the sector area of the circle

Now we will simplify the equation as below,

Now we will multiply both sides of the equation by 360,

Therefore, sector angle is.

Hence, the correct answer is option (d).

Page No 15.62:

Question 37:

In Fig. 15.111, If ABC is an equilateral triangle, then shaded area is equal to

(a) π3-34r2

(b) π3-32r2

(c) π3+34r2

(d) π3+3r2

Figure

Answer:

We have given that ABC is an equilateral triangle.

As we know that,.

Area of the shaded region = area of the segment BC.

Let

Substituting the values we get,

Substituting and we get,

Therefore, area of the shaded region is . Hence, the correct answer is option (a).



Page No 15.63:

Question 38:

In Fig. 15.112, the area of the shaded region is

(a) 3π cm2
 
(b) 6π cm2

(c) 9π cm2

(d) 7π cm2

Figure

Answer:

In the figure,

and,

Therefore, area of the shaded region is.

Hence, the correct answer is option (a).

Page No 15.63:

Question 39:

If the perimeter of a circle is equal to that of a square, then the ratio of their areas is

(a) 13 : 22

(b) 14 : 11

(c) 22 : 13

(d) 11 : 14

Answer:

We have given that perimeter of circle of radius r is equal to square of side a.

perimeter of the circle = perimeter of the square

Now we will substitute value of r in the following equation

Substituting we get,

Hence, ratio of the areas of the circle and square is.

Therefore, the correct answer is (b).

Page No 15.63:

Question 40:

The radius of a circle is 20 cm. It is divided into four parts of equal area by drawing three concentric circles inside it. Then, the radius of the largest of three concentric circles drawn is

(a) 105 cm

(b) 103 cm

(c) 105 cm

(d) 102 cm

Answer:

The circle can be divided into four parts of equal area by drawing three concentric circles inside it as,

It is given that OB = 20 cm. Let OA = x.

Since the circle is divided into four parts of equal area by the three concentric circles, we have,

Area of the fourth region = Area of the given circle

Therefore, the correct answer is (b).

 

Page No 15.63:

Question 41:

The area of a sector whose perimeter is four times its radius r units, is

(a) r24 sq. units

(b) 2r2  sq. units

(c) r2 sq.units

(d) r22 sq. units

Answer:

We know that perimeter of the sector .

We have given that perimeter of the sector is four times the radius.

Subtracting 2r from both sides of the equation,

Dividing both sides of the equation 2r we get,

……….(1)

Let us find the area of the sector.

Substituting we get,

Hence, area of the sector is.

Therefore, the correct answer is (c).

Page No 15.63:

Question 42:

If a chord of a circle of  radius 28 cm makes an angle of 90 ° at the centre, then the area of the major segment is

(a) 392 cm2

(b) 1456 cm2

(c) 1848 cm2

(d) 2240 cm2

Answer:

Area of major segment,

So the answer is (d)

Page No 15.63:

Question 43:

If area of a circle inscribed in an equilateral triangle is 48π square units, then perimeter of the triangle is

(a) 173 units

(b) 36 units

(c) 72 units

(d) 483 units

Answer:

Let the circle of radius r be inscribed in an equilateral triangle of side a.

Area of the circle is given as 48π.

⇒ πr2 = 48π

r2 = 48

Now, it is clear that ONBC. So, ON is the height of ΔOBC corresponding to BC.

Area of ΔABC = Area of ΔOBC + Area of ΔOCA + Area of ΔOAB = 3 × Area of ΔOBC

Thus, perimeter of the equilateral triangle = 3 × 24 units = 72 units

So the answer is (c).

Page No 15.63:

Question 44:

The hour hand of a clock is 6 cm long. The area swept by it between 11.20 am and 11.55 am is

(a) 2.75 cm2

(b) 5.5 cm2

(c) 11 cm2

(d) 10 cm2

Answer:

Hour hand movesin one minute.

So, area,

So the answer is (b)

Page No 15.63:

Question 45:

ABCD is a square of side 4 cm. If E is a point in the interior of the square such that ΔCED is equilateral, then area of Δ ACE is

(a) 23-1 cm2

(b) 43-1 cm2

(c) 63-1 cm2

(d) 83-1 cm2

Answer:

We have the following diagram.

Since is equilateral,

Therefore,

Now,

Since AC is diagonal of sqr.ABCD

Therefore,

Therefore we get,

Now, in , draw a perpendicular EM to the base AC.

So in ,

Therefore,

Now in ,

So the answer is (b)



Page No 15.64:

Question 46:

If the area of a circle is equal to the sum of the areas of two circles of diameters 10 cm and 24 cm, then diameter of the large circle (in cm) is

(a) 34

(b) 26

(c) 17

(d) 14

Answer:

Let the diameter of the larger circle be d
Now, Area of larger circle = Area of circle having diameter 10 cm + Area of circle having diameter 24 cm
πd22=π1022+π2422d22=52+122

d22=25+144d22=132d2=13d=26 cm
Hence, the correct answer is option (b).

Page No 15.64:

Question 47:

If π is taken as 227, the distance (in metres) covered by a wheel of diameter 35 cm, in one revolution, is

(a) 2.2

(b) 1.1

(c) 9.625

(d) 96.25

Answer:

The disstance covered by the wheel in one revolution is equal to the circumference of the wheel.
Circumference=πr2=2273522=2273522=962.5 cm
The distance covered by the wheel in one revolution (in m) is given by
962.5100=9.625 m
Hence, the correct answer is option (c).



Page No 15.9:

Question 1:

Find the circumference and area of a circle of radius 4.2 cm.

Answer:

We know that the circumference C and area A of a circle of radius r are given by and respectively.

Here,

So substituting the value of r in above formulas,

Circumference of the circle

Area of the circle

Page No 15.9:

Question 2:

Find the circumference of a circle whose area is 301.84 cm2.

Answer:

Let r cm be the radius of the circle. Then

Area of a circle is

We know that the Circumference of circle of radius r is

So substituting the value of r in above formula

Page No 15.9:

Question 3:

Find the area of a circle whose circumference is 44 cm.

Answer:

Let r be the radius of the circle. Then

Circumference of the circle

We know that the area of a circle of radius r is

Substituting the value of r in above formula

Page No 15.9:

Question 4:

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

Answer:

Let the radius of a circle be r cm, then diameter of circle is 2r cm and Circumference iscm.

It is given that the circumference exceeds the diameter of circle by 16.8 cm.

So,

Now the circumference is

Page No 15.9:

Question 5:

A horse is tied to a pole with 28 m long string. Find the area where the horse can graze. (Take π = 22/7).

Answer:

We know that the horse is tied to a pole with 28 m long string. So the horse can graze the area of a circle of radius 28 m.

Area of circle is

Hence the horse can graze area.

Page No 15.9:

Question 6:

A steel wire when bent in the form of a square encloses an area of 121 cm2. If the same wire is bent in the form of a circle, find the area of the circle.

Answer:

Let a cm be the side of square. Then area of square is

We have,

Let the radius of circle be r cm. Then,

Now, we will calculate area of circle.

Page No 15.9:

Question 7:

A horse is placed for grazing inside a rectangular field 40 m by 36 m and is tethered to one corner by a rope 14 m long. Over how much area can it graze? (Take π = 22/7)

Answer:

It is given that a horse is tethered to one corner of a rectangular field (40 m × 36 m) by a 14 m long rope.

Let r m be the radius of a circle. Then area A of circle is

A=πr2 m2=227×14×14 m2=616 m2

Since the horse can graze inside the rectangular field only, the required area is quadrant of circle. So,

Required area=A4=6164=154 m2

Hence the horse can graze 154 m2 area.

Page No 15.9:

Question 8:

A sheet of paper is in the form of a rectangle ABCD in which AB = 40 cm and AD = 28 cm. A semi-circular portion with BC as diameter is cut off. Find the area of the remaining paper.

Answer:

The length and width of rectangle ABCD is given by andrespectively.

Now, we will find the area of rectangle.

It is given that a semicircular portion with BC as diameter is cutoff from rectangle. So,

Now,

Substituting the value of r,

The area A of remaining paper is

Thus, the area of remaining paper is.

Page No 15.9:

Question 9:

The circumference of two circles are in the ratio 2 : 3. Find the ratio of their areas.

Answer:

Let the radius of two circles beandrespectively. Then their circumferences are and respectively and their areas are and respectively.

It is given that,

Now we will calculate the ratio of their areas,

Substituting the value of ,

Hence the ratio of their Areas is.

Page No 15.9:

Question 10:

The side of a square is 10 cm. Find the area of circumscribed and inscribed circles.

Answer:

It is given that the side of square is 10 cm.

So, the diameter of circle inscribed the square is 10 cm.

We know that the area A of circle inscribed the square is

Substituting the value of radius of inscribed circle,

Hence the area of circle inscribed the square is

Now we will find the diameter of circle circumscribed the square.

So,

We know that the area of circle inscribed the square is

Substituting the value of radius,

Hence the area of circle circumscribed the square is .

Page No 15.9:

Question 11:

The sum of the radii of two circles is 140 cm and the difference of their circumferences is 88 cm. Find the diameters of the circles.

Answer:

Let the radius of two circles beandrespectively. Then their circumferences are and respectively and their areas are and respectively.

It is given that the sum of the radii of two circles is 140 cm and difference of their circumferences is 88 cm. So,

……(A)

……(B)

Now, solving (A) and (B)

Thus diameters of circles are,

Page No 15.9:

Question 12:

The area of a circle inscribed in an equilateral triangle is 154 cm2. Find the perimeter of the triangle. [Use π = 22/7 and 3 = 1.73]

Answer:

It is given that the area A of circle inscribed in an equilateral triangle is 154 cm2.

We know that the area A of circle inscribed in an equilateral triangle is

Now, we will find the value of r.

Substituting the value of area,

Let the height of triangle be h. Then

If a is the side of triangle, then

Substituting the value of h,

Hence perimeter of triangle is .

Page No 15.9:

Question 13:

A field is in the form of a circle. A fence is to be erected around the field. The cost fencing would be Rs. 2640 at the rate of Rs. 12 per metre. The, the field is to be thoroughly ploughed at the cost of Re. 0.50 per m2. What is the amount required to plough the field? [Take π = 22/7].

Answer:

We know that the circumference C of a circle of radius r is

It is given that cost of fencing around the circular field would be Rs.2640 at the rate of Rs.12 per meter. So,

We know that the area A of circle of radius r,

Substituting the value of r

Since,

Hence, amount required to plough the field is.

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Question 14:

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

Answer:

Let ABCD be the square inscribed in a circle of radius r.

Here, OA = OB = r.

∴ OA2 + OB2 = AB2

r2 + r2 = AB2

⇒ 2r2 = AB2

Now, area of square ABCD

Now we will find the ratio of area of the circle and the square.

Hence, the ratio of area of the circle and square is.

Page No 15.9:

Question 15:

A park is in the form of a rectangle 120 m × 100 m. At the centre of the park there is a circular lawn, The area of park excluding lawn is 8700 m2. Find the radius of the circular lawn. (Use π = 22/7).

Answer:

Let the radius of circular lawn be r. Then,

It is given that

Hence, radius of circular lawn is.

Page No 15.9:

Question 16:

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

Answer:

Let the radius of circles be , andrespectively. Then their areas are, and respectively.

It is given that,

We have, and

Substituting the values of ,

Hence, the radius of circle is.

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Question 17:

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

Answer:

Let the radius of circles be, andrespectively. Then their circumferences are, and respectively.

It is given that,

We have, and

Substituting the values of,

Hence the radius of the circle is.

We know that the area A of circle is

Substituting the value of r

Hence the area of the circle is.

Page No 15.9:

Question 18:

A car travels 1 kilometre distance in which each wheel makes 450 complete revolutions. Find the radius of the its wheels.

Answer:

Let the radius of wheel be r. Thus, circumference C of the wheel

Since car travels 1 km distance in which wheel makes 450 complete revolutions. Then

We know that,

 

Hence the radius of wheel is.



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