RD SHARMA SOLUTION CHAPTER- 15 Areas Related to Circles | CLASS 10TH MATHEMATICS-EDUGROWN

Table of Contents

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Find the radius of a circle whose circumference is equal to the sum of the circumference of two circles of radii 15 cm and 18 cm.Solution 9

Question 10

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.Solution 10

Question 11

The radii of two circles are 19 cm and 9 cm respectively. Find the radius and area of the circle which has its circumference equal to the sum of the circumferences of the two circles.Solution 11

Area of a circle = πr² = (22/7) × 28 × 28 = 2464 cm²Question 12

The area of a circular playground is 22176 m². Find the cost of fencing this ground at the rate of Rs. 50 per metre.Solution 12

Question 13

Solution 13

Question 14

Solution 14

Question 15

Solution 15

Question 16

Solution 16

Question 17

A park is in the form of a rectangle 120 m x 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 = 22/7).Solution 17

Question 18

Solution 18

Question 19

Solution 19

Question 20

An archery target has three regions formed by three concentric circles as shown in figure. If the diameters of the concentric circles are in the ratio 1 : 2 : 3, then find the ratio of the areas of three regions.

Solution 20

Question 21

The wheel of a motor cycle is of radius 35 cm. How many revolutions per minute must the wheel make so as to keep a speed of 66 km/hr?Solution 21

Question 22

A circular pond is 17.5 m in diameter. It is surrounded by a 2 m wide path. Find the cost of constructing the path at the rate of Rs.25 per m².Solution 22

Question 23

A circular park is surrounded by a road 21 m wide. If the radius of the park is 105 m, find the area of the road.Solution 23

Question 24

A square of diagonal 8 cm is inscribed in a circle. Find the area of the region lying inside the circle and outside the square.Solution 24

Question 25

Solution 25

Question 26

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.

Solution 26

Question 27

A path of width 3.5 m runs around a semi-circular grassy plot whose perimeter is 72 m. Find the area of the path. (Use π = 22/7)Solution 27

Question 28

A circular pond is of diameter 17.5 m. It is surrounded by a 2 m wide path. Find the cost of constructing the path at the rate of Rs. 25 per square meter (π = 3.14)Solution 28

table attributes columnalign left end attributes row cell text Area of the surrounding path = end text pi left parenthesis R squared minus r squared right parenthesis end cell row cell R equals left parenthesis 17.5 divided by 2 right parenthesis plus 2 equals 8.75 plus 2 equals 10.75 text m is the outer radius. end text end cell row cell r equals 17.5 divided by 2 equals 8.75 text m is the inner radius. end text end cell row cell text Area of the path = end text pi left parenthesis R squared minus r squared right parenthesis equals pi left parenthesis 10.75 squared minus 8.75 squared right parenthesis end cell row cell equals 3.14 cross times 39 equals 122.46 text m end text to the power of text 2 end text end exponent end cell row cell text Total cost = 122.46 end text cross times text 25 = Rs. 3061.50 end text end cell end table

Question 29

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 = 22/7).Solution 29

Question 30

Solution 30

Question 31

Prove that the area of a circular path of uniform width h surrounding a circular region of radius r is h (2r + h).Solution 31

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

The area of a sector of a circle of radius 5 cm is 5 cm². Find the angle contained by the sector.Solution 9

Question 10

Find the area of the sector of a circle of radius 5 cm, if the corresponding arc length is 3.5 cm.Solution 10

Question 11

Solution 11

Question 12

The perimeter of a scetor of a circle of radius 5.7 m is 27.2 m. Find the area of the sector.Solution 12

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.Solution 13

Question 14

Solution 14

Question 15

Solution 15

Question 16

Solution 16

Question 17

A sector of 56^o cut out from a circle contains area 4.4 cm². Find the radius of the circle.Solution 17

Question 18

Area of a sector of central angle 200° of a circle is 770 cm². Find the length of the corresponding arc of this sector.Solution 18

Question 19

The length of minute hand of a clock is 5 cm. Find the area swept by the minute hand during the time period 6:05 am and 6:40 am.Solution 19

Question 20

The length of the minute hand of a clock is 14 cm. Find the area swept by the minute hand in 5 minutes.Solution 20

table attributes columnalign left end attributes row cell text A minute hand covers 360 end text to the power of text o end text end exponent text in 60 minutes. end text end cell row cell text Hence , the minute hand covers end text fraction numerator text 360 end text over denominator text 60 end text end fraction equals 6 degree text in one minute. end text end cell row cell text The minute hand covers 6 end text cross times text 5 = 30 end text to the power of text o end text end exponent text in 5 minutes. end text end cell row cell text The length of minute hand is 14 cm . end text end cell row cell text Area swept = end text text 30 end text to the power of text o end text end exponent over text 360 end text to the power of text o end text end exponent cross times pi R squared equals 1 over 6 cross times 22 over 7 cross times 14 squared equals 102.67 text cm end text to the power of text 2 end text end exponent end cell end table

*Answer does not match with textbook answer.Question 21

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

(i) the length of arc

(ii) area of the sector formed by the arc. (use π = 22/7)Solution 21

table attributes columnalign left end attributes row cell left parenthesis text i end text right parenthesis text Length of an arc end text equals fraction numerator theta over denominator 360 degree end fraction cross times 2 pi R end cell row cell equals fraction numerator 60 degree over denominator 360 degree end fraction cross times 2 cross times 22 over 7 cross times 21 end cell row cell equals 1 over 6 cross times 2 cross times 22 over 7 cross times 21 equals 22 text cm end text end cell end table

table attributes columnalign left end attributes row cell left parenthesis i i right parenthesis text Area of the sector = end text fraction numerator theta over denominator 360 degree end fraction cross times pi R squared end cell row cell equals fraction numerator 60 degree over denominator 360 degree end fraction cross times pi cross times 21 squared equals 1 over 6 cross times 22 over 7 cross times 21 squared equals 231 text cm end text to the power of text 2 end text end exponent end cell end table

Question 22

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)Solution 22

*Note: Answer given in the book is incorrect.Question 23

The area of a sector is one-twelfth that of the complete circle. Find the angle of the sector.Solution 23

Question 24

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.Solution 24

Question 25

Solution 25

Question 26

Solution 26

Question 27

Solution 27

Question 1

Solution 1

Question 2

A chord PQ of length 12 cm subtends an angle of 120^o at the centre of a circle. Find the area of the minor segment cut off by the chord PQ.Solution 2

Question 3

Solution 3

Question 4

A chord 10 cm long is drawn in a circle whose radius is cm. Find area of both the segments. (Take = 3.14).Solution 4

Question 5

Solution 5

Question 6

Find the area of the minor segment of a circle of radius 14 cm, when the angle of the corresponding sector is 60°.Solution 6

Question 7

A chord of a circle of radius 10 cm subtends an angle of 90° at the centre. Find the area of the corresponding major segment of the circle. (Use π = 3.14).Solution 7

Question 8

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

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 1

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

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

A rectangular piece is 20 m long and 15 m wide. Form its four corners, quadrants of radii 3.5 m have been cut. Find the area of the remaining part.Solution 4

Question 5

In fig., PQRS is a square of side 4 cm. Find the area of the shaded region.

Solution 5

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?

Solution 6

Question 7

A cow is tied with a rope of length 14 m at the corner of a rectangular field of dimensions 20m × 16m, find the area of the field in which the cow can graze.Solution 7

Question 8

A calf is tied with a rope of length 6 m at the corner of a square grassy lawn of side 20 m. If the length of the rope is increased by 5.5 m, Find the increase in area of the grassy lawn in which the calf can graze.Solution 8

Question 9

Solution 9

Question 10

A rectangular park is 100 m by 50 m. It is surrounded by semi-circular flower beds all round. Find the cost of levelling the semi-circular flower beds at 60 paise per square metre (Ise = 3.14).Solution 10

Question 11

The inside perimeter of a running track (shown in Fig.) 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.

Solution 11

Question 12

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

Solution 12

Question 13

From a rectangular region ABCD with AB = 20 cm, a right angle AED with AE = 9 cm and DE = 12 cm, is cut off. On the other end, taking BC as diameter, a semicircle is added on outside the region. Find the area of the shaded region. (π = 22/7)

Solution 13

table attributes columnalign left end attributes row cell text Area of the rectangle ABCD = AB end text cross times text AD end text end cell row cell text AD is the hypotenuse of the right angled end text capital delta text AED. end text end cell row cell A D equals square root of A E squared plus E D squared end root equals square root of 9 squared plus 12 squared end root equals square root of 225 equals 15 text cm end text end cell row cell text Hence , area of the rectangle = end text 20 cross times 15 equals 300 text cm end text to the power of text 2 end text end exponent end cell row cell text Area of the right end text capital delta text AED = end text 1 half cross times A E cross times D E equals 1 half cross times 9 cross times 12 end cell row cell equals 54 text cm end text to the power of text 2 end text end exponent end cell row cell text Area of the semicircle = end text fraction numerator text 1 end text over denominator text 2 end text end fraction cross times pi cross times open parentheses 15 over 2 close parentheses squared end cell row cell equals 1 half cross times 3.14 cross times 7.5 squared equals 88.31 text cm end text to the power of text 2 end text end exponent end cell row cell text Area of the shaded region end text end cell row cell text = Area of the rectangle + Area of the semicircle end text end cell row cell negative text Area of the right triangle end text end cell row cell text = end text 300 plus 88.31 minus 54 equals 334.31 text cm end text to the power of text 2 end text end exponent end cell end table

Question 14

From each of the two opposite corners of a square of side 8.8 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. Find the area of the remaining (shaded) portion of the square. (Use π = 22/7).Solution 14

Question 15

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.

Solution 15

Question 16

ABCD is 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.

Solution 16

Question 17

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.

Solution 17

Question 18

In figure, find the area of the shaded region.

(Use π = 3.14)

Solution 18

Question 19

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.

Solution 19

Question 20

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

Solution 20

table attributes columnalign left end attributes row cell text Area of the square = end text O A squared equals 21 squared equals 441 text cm end text to the power of text 2 end text end exponent end cell row cell text Diagonal of the square OB = end text square root of text 2 end text end root O A equals 21 square root of 2 text cm end text end cell row cell text Diagonal of the square is equal to the radius end text end cell row cell text of the circle. end text end cell row cell text Hence , area of the quadrant = end text fraction numerator text 1 end text over denominator text 4 end text end fraction cross times pi r squared equals fraction numerator text 1 end text over denominator text 4 end text end fraction cross times 22 over 7 cross times left parenthesis 21 square root of 2 right parenthesis squared end cell row cell equals fraction numerator text 1 end text over denominator text 4 end text end fraction cross times 22 over 7 cross times 441 cross times 2 equals 693 end cell row cell text Hence , area of the shaded region end text end cell row cell text = Area of the quadrant end text minus text Area of the square = 693 end text minus 441 end cell row cell equals 252 text cm end text to the power of text 2 end text end exponent end cell end table

Question 21

Solution 21

Question 22

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

Solution 22

Question 23

Solution 23

Question 24

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

Solution 24

Question 25

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

Solution 25

*Answer is not matching with textbook.Question 26

Solution 26

Question 27

Find the area of a shaded region in the given figure, where a circular arc of radius 7 cm has been drawn with vertex A of an equilateral triangle ABC of side 14 cm as centre.

Solution 27

Question 28

A regular hexagon is inscribed in a circle. If the area of hexagon is , find the area of the circle. (Use π it = 3.14)Solution 28

Consider the following figure:

Question 29

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

Solution 29

(i)

According to the figure in the question, there are 6 triangles.

Area of one triangle is 9 cm².

Area of hexagon = 6 × 9 = 54 cm²

(ii)

Area of the equilateral triangle = 9 cm²

Area of the circle in which the hexagon is inscribed

= 65.26 cm²

NOTE: Answer not matching with back answer.Question 30

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

Solution 30

Question 31

Solution 31

Question 32

A child makes a poster on a chart paper drawing a square ABCD of side 14 cm. She draws four circles with centre A, B, C and D in which she suggests different ways to save energy. The circles are drawn in such a way that each circle touches externally two of the three remaining circles. In the shaded region she write a message ‘Save Energy’. Find the perimeter and area of the shaded region. (Use π = 22/7)

Solution 32

Question 33

The diameter of a coin is 1 cm. 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)

Solution 33

Question 34

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. (π = 22/7)Solution 34

table attributes columnalign left end attributes row cell text Area of the rectangular card = 14 end text cross times text 7 = 98 cm end text to the power of text 2 end text end exponent end cell row cell text Maximum diameter of a circle inscribed in end text end cell row cell text the given rectangle = 7 cm. end text end cell row cell text Area of the circles = end text pi cross times left parenthesis 7 divided by 2 right parenthesis squared end cell row cell text Area of two such circles = 2 end text cross times pi cross times left parenthesis 7 divided by 2 right parenthesis squared equals 77 text cm end text to the power of text 2 end text end exponent end cell row cell text Area of the remaining card = 98 cm end text to the power of text 2 end text end exponent minus 77 text cm end text to the power of text 2 end text end exponent equals 21 text cm end text to the power of text 2 end text end exponent end cell end table

Question 35

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.

Solution 35

Question 36

PSR, RTQ and PAQ are three semi-circles of diameters 10 cm, 3 cm and 7 cm respectively. Find the perimeter of the shaded region.

Solution 36

table attributes columnalign left end attributes row cell text Perimeter of the shaded region = end text end cell row cell text Length of end text stack P S R with overparenthesis on top plus text Length of end text stack P A Q with overparenthesis on top text end text plus text length of end text stack Q T R with overparenthesis on top end cell row cell equals pi cross times 5 plus pi cross times 3.5 plus pi cross times 1.5 end cell row cell equals pi cross times 10 end cell row cell equals 31.4 text cm end text end cell end table

Question 37

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.

Solution 37

Question 38

ABCD is a square of side 2a. Find the ratio between

(i) the circumferences

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

Solution 38

Question 39

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.

Solution 39

Question 40

Solution 40

Question 41

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)

Solution 41

Question 42

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.

Solution 42

Question 43

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.

Solution 43

Question 44

Solution 44

Question 45

Solution 45

Question 46

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.

Solution 46

Question 47

ABCD is a trapezium of area 24.5 cm². In it, AD ∥ BC, ∠DAB = 90°, AD = 10 cm and BC = 4 cm. If ABE is a quadrant of a circle, find the area of the shaded region.

(π = 22/7)

Solution 47

Question 48

ABCD is a trapezium with AB ∥ DC, AB = 18 cm, DC = 32 cm and the distance between AB and DC is 14 cm. Circles of equal radii 7 cm with centres A, B, C and D have been drawn. Then, find the area of the shaded region of the figure. (π = 22/7)

Solution 48

Since the data given in the question seems incomplete and inconsistent with the figure, we make the following assumptions to solve it:

1. ABCD a symmetric trapezium with AD = BC

2. AD = BC = 14 cm (the distance between AB and CD is not 14 cm)

Draw perpendiculars to CD from A and B to divide the trapezium into one rectangle and two congruent right angled triangles.

The base of the right angled triangle=(CD – AB) ÷ 2

=(32 – 18) ÷ 2=7 cm

cos∠D = base ÷ hypotenuse = 7 ÷ 14 =1/2

m∠D = 60°

Hence, m∠A = 120°

table attributes columnalign left end attributes row cell s i n angle D equals fraction numerator p e r p e n d i c u l a r over denominator h y p o t e n u s e end fraction equals p over 14 equals fraction numerator square root of 3 over denominator 2 end fraction end cell row cell rightwards double arrow p equals 7 square root of 3 text cm end text end cell row cell text Area of the trapezium = end text fraction numerator text 1 end text over denominator text 2 end text end fraction p left parenthesis A B plus C D right parenthesis equals 1 half cross times 7 square root of 3 left parenthesis 18 plus 32 right parenthesis end cell row cell equals 175 square root of 3 text cm end text to the power of text 2 end text end exponent end cell row cell text The trapezium has four parts of circles , end text end cell row cell text two with angles 60 end text degree text at centre and two with angles 120 end text degree text at centre. end text end cell row cell text So in total it has 2 end text cross times fraction numerator text 1 end text over denominator text 6 end text end fraction t h plus 2 cross times 1 third r d equals text one full circle. end text end cell row cell text Area of the circle = end text pi R squared equals 22 over 7 cross times 7 squared equals 154 text cm end text to the power of text 2 end text end exponent end cell row cell text Area of the shaded part end text end cell row cell text = Area of the trapezium end text minus text Area of the circle end text end cell row cell text = end text 175 square root of 3 text cm end text to the power of text 2 end text end exponent minus 154 text cm end text to the power of text 2 end text end exponent end cell row cell equals 149.1 text cm end text to the power of text 2 end text end exponent end cell end table

*Answer is not matching with textbook answer.Question 49

Solution 49

Question 50

Solution 50

Question 51

Sides of a triangular field are 15 m, 16 m and 17 m. With the three corners of the field a cow, a buffalo and a horse are tied separately with ropes of length 7 m each to graze in the field. Find the area of the field which cannot be grazed by three animals.Solution 51

Question 52

In the given Fig., the side of square is 28 cm, and radius of each circle is half of the length of the side of the square where O and O’ are centres of the circles. Find the area of shaded region.

Solution 52

According to the question,

Side of a square is 28 cm.

Radius of a circle is 14 cm.

Required area = Area of the square + Area of the two circles – Area of two quadrants …(i)

Area of the square = 28² = 784 cm²

Area of the two circles = 2πr²

= 1232 cm²

Area of two quadrants =

= 308 cm²

Required area = 784 + 1232 – 308 = 1708 cm²

NOTE: Answer not matching with back answer.Question 53

In a hospital used water is collected in a cylindrical tank of diameter 2 m and height 5 m. After recycling, this water is used to irrigate a park of hospital whose length is 25 m and breadth is 20 m. If tank is filled completely then what will be the height of standing water used for irrigating the park?Solution 53

According to the question,

For a cylindrical tank

d = 2 m, r = 1 m, h = 5 m

Volume of the tank = πr²h

After recycling, this water is used irrigate a park of a hospital with length 25 m and breadth 20 m.

If the tank is filled completely, then

Volume of cuboidal park = Volume of tank

h = 0.0314 m = 3.14 cm = p cmQuestion 54

In the figure, a square OABC is inscribed in a quadrant OPBQ. If OA = 15 cm, find the area of shaded region (use π = 3.14).

Solution 54

Join OB.

Here, is a right triangle.

By Pythagoras theorem,

Therefore, radius of the circle (r)

Area of the square

Area of the quadrant of a circle

Area of the shaded region = Area of quadrant – Area of square

= 128.25 cm²Question 55

In the figure, ABCD is a square with side and inscribed in a circle. Find the area of the shaded region. (Use π = 3.14).

Solution 55

Join AC.

Here, is a right triangle.

By Pythagoras theorem,

Therefore, diameter of the circle = 4 cm

So, the radius of the circle (r) = 2 cm

Area of the square

Area of the circle

Area of the shaded region = Area of the circle – Area of square

= 4.56 cm²

Question 1

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

begin mathsize 12px style left parenthesis straight a right parenthesis space straight pi over 2
left parenthesis straight b right parenthesis space 2 straight pi
left parenthesis straight c right parenthesis space 2
left parenthesis straight d right parenthesis space 4 end style

Solution 1

Correct Option :- (D)

begin mathsize 12px style Let space radivs space of space circle space be space straight r.
circumferance space of space circle space equals space 2 πr
area space of space circle space equals space πr squared
Given comma area space equals space circumference
space space space space space space space space space space space space space space space space space space space space space rightwards double arrow πr squared equals 2 πr
space space space space space space space space space space space space space space space space space space space space space rightwards double arrow space box enclose straight r equals 2 end enclose space
diameter space equals space 2 straight r
space space space space space space space space space space space space space space space space space equals space 4 space space space space space space space space space space space space space space end style

Question 2

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

(a) 154

(b) 44

(d) 7 Solution 2

According to the question,

Circumference of a circle =

= 44 cm Question 3

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

(a) 3520 cm²

(b) 6400 cm²

(d) 8800 cm²Solution 3

Correct option (c)

begin mathsize 12px style L e n g t h space o f space w i r e space equals space C i r c u m t e r n c e space o f space C i r c l e
left parenthesis l right parenthesis space equals space 2 πr
straight l equals 2 straight pi cross times left parenthesis 56 right parenthesis
equals space 2 cross times 22 over 7 cross times 56
equals 352 space cm
If space bent space in space form space of space square space then space lenth space of space wire space equals space Perimeter space of space squre space if space side space of space squre space is space straight a space
then space straight l space equals space 49
rightwards double arrow straight a space equals space 352 over 4
rightwards double arrow straight a space equals space 88 space cm
area space of space square space equals space straight a squared
equals space 7744 cm squared

end style

Question 4

begin mathsize 12px style If space straight a space wire space is space bent space into space the space shape space of space straight a space square comma space then space the space area space of space the space squre space 81 cm squared. space when space wire space is space bent space into space straight a space semi minus space circular space shape comma space then space area space of space the space semi space minus space circle space will space be
left parenthesis straight a right parenthesis space 22 cm squared
left parenthesis straight b right parenthesis space 44 cm squared
left parenthesis straight c right parenthesis space 77 space cm squared
left parenthesis straight d right parenthesis space 154 space cm squared end style

Solution 4

correct option – (c)

begin mathsize 12px style a r e a space o f space s q u a r e space equals space 81 space c m squared
s i d e space o f space s q u a r e space equals space square root of 81
equals space 9 space c m
L e n g t h space o f space w i r e space equals space p e r i m e t e r space o f space s q u a r e
equals space 4 cross times 9
equals 36 space c m
w i r e space i s space b e n t space i n t o space s e m i minus c i r c l e space t h e n space
L e n g h t space o f space w i r e space equals space h a l f space o f space c i r c u m f e r n c e space plus 2 r
rightwards double arrow 36 equals πr plus 2 straight r
rightwards double arrow box enclose straight r equals fraction numerator 36 over denominator straight pi plus 2 end fraction end enclose
area space of space semi minus circle space equals space πr squared over 2
equals space straight pi over 2 cross times open parentheses fraction numerator 36 over denominator straight pi plus 2 end fraction close parentheses squared
equals straight pi over 2 open parentheses fraction numerator 36 over denominator straight pi plus 2 end fraction close parentheses squared
equals straight pi over 2 open parentheses fraction numerator 36 cross times 7 over denominator 36 end fraction close parentheses squared
equals fraction numerator 22 over denominator 7 cross times 2 end fraction cross times 7 cross times 7
equals 77 cm squared end style

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

(d) 24 mSolution 5

correct option – (b)

begin mathsize 12px style L e t space o u t e r space r a d i u s space i s space b space a n d space i n n e r space r a d i u s space i s space a
G i v e n space comma space 2 pi b minus 2 pi a equals 132
rightwards double arrow 2 pi open parentheses b minus a close parentheses equals 132
rightwards double arrow open parentheses b minus a close parentheses space equals space fraction numerator 132 over denominator 2 cross times 22 end fraction cross times 7
w i d t h space o f space p a t h space rightwards double arrow space open parentheses b minus a close parentheses equals 21 end style

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

(d) 7000Solution 6

Correct Option: d

begin mathsize 12px style Let space wheel space take space straight n space revolutions
space After space 1. space revolution space distance space covered space by space wheel space equals space 2 πr
rightwards double arrow After space straight n space revolution space distance space covered space by space wheel space equals space 2 πrn
rightwards double arrow space Given comma space 2 πrn space equals space 11000 space straight m
rightwards double arrow 2 straight pi space cross times space open parentheses 0.25 close parentheses straight n space equals space 11000
rightwards double arrow straight n space equals space fraction numerator 11000 over denominator 2 cross times begin display style 22 over 7 end style cross times begin display style 1 fourth end style end fraction
straight n equals space 7000 end style

Question 7

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

(a) 55m

(b) 110 m

(d) 230 mSolution 7

Correct Option: (c)

Perimeter space of space circular space path space equals space 2 space πr
Let space outer space radius space be space straight b space and space inner space radius space be space straight a space
rightwards double arrow space given space fraction numerator 2 πb over denominator 2 πa end fraction equals 23 over 22 rightwards double arrow straight b over straight a equals 23 over 22 rightwards double arrow box enclose straight b equals 23 over 22 straight a end enclose
also space width space comma space straight b minus straight a space equals space 5
from space circle enclose 1 comma circle enclose 2
23 over 22 straight a minus straight a space equals space 5
rightwards double arrow straight a over 22 equals 5
rightwards double arrow straight a space equals space 110 space straight m
diameter space equals space 2 straight a
equals space 220 space straight m

Question 8

begin mathsize 12px style The space circumference space of space straight a space circle space is space 100 space cm. space The space side space of space straight a space square space inscribed space in space the space circle space is
left parenthesis straight a right parenthesis space 50 square root of 2 space cm
left parenthesis straight b right parenthesis space 100 over straight pi space cm
left parenthesis straight C right parenthesis space 50 square root of 2 cm
left parenthesis straight d right parenthesis space fraction numerator 100 square root of 2 over denominator straight pi end fraction cm end style

Solution 8

Correct option – (c)

begin mathsize 12px style 2 πr space equals space 100
rightwards double arrow straight r space equals space 50 over straight pi cm
Image space Pending space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space
OA squared plus OB squared equals space AB squared
OA space equals space OB space equals space straight r
rightwards double arrow 2 straight r squared equals AB squared
rightwards double arrow AB space equals space square root of 2 straight r end root
equals space fraction numerator 50 square root of 2 over denominator straight pi end fraction cm
end style

Question 9

begin mathsize 12px style The space area space of space the space incircle space of space an space equilateral space triangle space of space side space 42 space cm space is
left parenthesis straight a right parenthesis space 22 square root of 3 cm squared
left parenthesis straight b right parenthesis space 231 space cm squared
left parenthesis straight c right parenthesis space 462 space cm squared
left parenthesis straight d right parenthesis space 924 space cm squared end style

Solution 9

Correct option (c)

begin mathsize 12px style Imege space pending space
AB space equals space 42 space cm
straight E space is space mid space point space of space AB
Hence space EB space equals space 21 space cm
Also space less than ABC space equals space 60 to the power of 0
But space less than space ABO space equals space 1 half less than ABC
equals space 30 to the power of 0
In space increment OEB
tan space 30 to the power of 0 equals space OE over EB
rightwards double arrow OE space equals space EB space tan space 30 to the power of 0
equals space fraction numerator 21 over denominator square root of 3 end fraction
radius space space space straight r space equals space OE
straight r space equals space fraction numerator 21 over denominator square root of 3 end fraction
area space of space circle space equals space πr squared
equals space straight pi open parentheses fraction numerator 21 over denominator square root of 3 end fraction close parentheses squared
equals space straight pi cross times fraction numerator 21 cross times 21 over denominator 3 end fraction
equals space 22 over 7 cross times fraction numerator 21 cross times 21 over denominator 3 end fraction
equals space 462 space cm squared end style

Question 10

Solution 10

Correct Option ( d )

begin mathsize 12px style πr squared equals 154
straight r squared space equals space fraction numerator 154 over denominator straight pi space end fraction space space equals space 154 over 22 cross times 7
straight r squared space equals space 49
straight r space equals space 7 space cm
OE space equals space straight r
Image space pending space
AB equals BC equals space AC
OE perpendicular AB space and space AE space equals EB
and space angle space EBC space equals space 60 to the power of 0
But space angle space EBO space equals space 1 half space angle space EBC space equals space 30 to the power of 0
In space increment space OEB space comma space tan space 30 to the power of 0 equals space OE over EB
rightwards double arrow EB space equals space fraction numerator OE over denominator tan space 30 to the power of 0 end fraction
equals 7 square root of 3
rightwards double arrow AB space equals space 2 EB
equals space 14 space square root of 3
perimeter space equals space AB space plus BC space plus CA
equals space 3 space AB
equals space 42 square root of 3
equals space 72.7 space cm
end style

Question 11

begin mathsize 12px style The space area space of space the space largest space triangle space that space can space be space inscribed space in space straight a space semi minus space circle space of space radius space straight r space comma space is
left parenthesis straight a right parenthesis space straight r squared
left parenthesis straight b right parenthesis space 2 space straight r squared
left parenthesis straight c right parenthesis space straight r cubed
left parenthesis straight d right parenthesis space 2 space straight r cubed end style

Solution 11

Correct option (a)

begin mathsize 12px style Image space pending space space
For space the space largest space triangle. space with space largest space area. space we space need space that space base space of space triangle space must space be space large.
hence space AB space is space Base space of space triangle
Let space straight C space be space 3 rd space vertex space of space triangle.
area space equals 1 half cross times AB cross times space straight h
equals space 1 half cross times 2 straight r space cross times straight h
equals space straight r space straight h
largest space possible space value space of space straight h space is space straight r space
Hence space area space equals space straight r space cross times straight r
equals space straight r squared end style

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 Solution 12

Let r be the radius of the circle.

2pr = 88

Perimeter of a triangle = 30 cm

Semi-perimeter = 15 cm

Hence,

Area of a triangle = r × s …(r = incircle radius, s =semi perimeter)

= 14 × 15

= 210 cm² Question 13

begin mathsize 12px style The space area space of space straight a space circle space is space 220 space cm squared. space The space area space of space straight a space square space inscribed space in space it space is
left parenthesis straight a right parenthesis space 49 space cm squared space space
left parenthesis straight b right parenthesis space 70 space cm squared space space space
left parenthesis straight c right parenthesis space 140 space cm squared
left parenthesis straight d right parenthesis space 150 space cm squared space end style

Solution 13

Correct option – (c)

begin mathsize 12px style πr squared equals space 220
rightwards double arrow straight r squared space equals fraction numerator space 220 over denominator 22 end fraction cross times 7
straight r squared equals space 70
straight r space equals space square root of 70 cm
Image space Pending space
OA space equals space OB equals straight r
ABCD space is space straight a space squre.
OA squared plus OB squared equals space AB squared
rightwards double arrow AB to the power of 2 space equals space 2 straight r squared end exponent
rightwards double arrow AB space equals space square root of 2 straight r end root
equals square root of 140
area space equals space left parenthesis AB right parenthesis squared
equals space 140 space cm squared

end style

Question 14

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

(a) halved

(b) doubled

(d) quadrupledSolution 14

begin mathsize 12px style Correct space option colon space open parentheses straight d close parentheses
Initially comma space 2 πr subscript 1 space equals space 4 straight pi
straight r subscript 1 space equals space 2
finally comma space 2 πr subscript 2 space equals space 8 straight pi
straight r subscript 2 space equals space 4
Initially space area space equals space πr subscript 1 superscript 2 space equals space 4 straight pi
final space area space equals space πr subscript 2 superscript 2
space space space space space space space space space space space space space space space space space space space space equals space 16 space straight pi
fraction numerator final space area over denominator Initial space area end fraction space equals space fraction numerator 16 straight pi over denominator 4 straight pi end fraction space equals space 4
area space is space quadrupled. end style

Question 15

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

(a) 10%

(b) 19%

(d) 36%Solution 15

begin mathsize 12px style Coerrect space option colon space open parentheses straight b close parentheses
Initial space radius space equals space straight r
final space radius space equals space diminshed space by space 10 percent sign
space space space space space space space space space space space space space space space space space space space space space space space space equals space 90 percent sign space of space initial
space space space space space space space space space space space space space space space space space space space space space space space space equals space 0.9 space straight r
Initial space area space equals space πr squared
final space area space equals space 0.81 space straight r squared straight pi
area space is space diminished space by space equals space fraction numerator Initial space area space space minus space final space area space over denominator Initial space area end fraction space cross times space 100
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space.19 space cross times space 100
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space 19 percent sign
end style

Question 16

begin mathsize 12px style If space the space area space of space straight a space square space is space same space as space the space area space of space straight a space circle comma space then space the space ratio space of space their space perimetres comma space in space terms space of space straight pi comma space is
open parentheses straight a close parentheses space straight pi space colon space square root of 3
open parentheses straight b close parentheses space 2 space colon space square root of straight pi
open parentheses straight c close parentheses space 3 space colon space straight pi
open parentheses straight d close parentheses space straight pi space colon space square root of 2 end style

Solution 16

begin mathsize 12px style Correct space option colon space open parentheses straight b close parentheses
area space of space square space equals space straight a squared
area space of space circle space equals space πr squared
given space straight a squared space equals space πr squared
straight a space equals space square root of straight pi straight r space space space space space space space space..... open parentheses 1 close parentheses
fraction numerator perimeter space of space square over denominator perimeter space of space circle space end fraction space equals space fraction numerator 4 straight a over denominator 2 πr end fraction
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space fraction numerator 4 square root of straight pi straight r over denominator 2 πr end fraction space space space space space space space space from space open parentheses 1 close parentheses
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space fraction numerator 2 over denominator square root of straight pi end fraction end style

Question 17

Solution 17

begin mathsize 12px style Correct space option colon space open parentheses straight b close parentheses
For space the space largest space triangle. space with space largest space area. space we space need space that space base space of space triangle space must space be space large.
hence space AB space is space Base space of space triangle
Let space straight C space be space 3 rd space vertex space of space triangle.
area space equals 1 half cross times AB cross times space straight h
equals space 1 half cross times 2 straight r space cross times straight h
equals space straight r space straight h
largest space possible space value space of space straight h space is space straight r space
Hence space area space equals space straight r space cross times straight r
equals space straight r squared end style

Question 18

begin mathsize 12px style The space ratio space of space the space areas space of space straight a space circle space and space an space equilateral space triangle space whose space diameter space and space straight a space sides space are space respectively space equal comma space is
left parenthesis straight a right parenthesis space straight pi space colon space square root of 2
left parenthesis straight b right parenthesis space straight pi space colon space square root of 3
open parentheses straight c close parentheses space square root of 3 space colon space straight pi
open parentheses straight d close parentheses space square root of 2 space colon space straight pi end style

Solution 18

begin mathsize 12px style Correct space option colon space open parentheses straight b close parentheses
Let space radius space of space circle space equals space straight r
Let space side space of space triangle space equals space straight a
given comma space 2 straight r space equals space straight a
fraction numerator area space of space circle over denominator area space of space equilateral space triangle end fraction space equals space fraction numerator πr squared over denominator begin display style fraction numerator square root of 3 over denominator 4 end fraction end style straight a squared end fraction
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space fraction numerator πr squared over denominator begin display style fraction numerator square root of 3 over denominator 4 end fraction end style space cross times space 4 straight r squared end fraction
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space bevelled fraction numerator straight pi over denominator square root of 3 end fraction
end style

Question 19

begin mathsize 12px style If space space the space sum space of space the space areas space of space two space circles space with space radii space straight r subscript 1 space and space straight r subscript 2 space is space equal space to space the space area space of space straight a space circle space of space radius space straight r comma space then space straight r subscript 1 superscript 2 space plus space straight r subscript 2 superscript 2
left parenthesis straight a right parenthesis space greater than straight r squared
left parenthesis straight b right parenthesis space equals space straight r squared
open parentheses straight c close parentheses space less than straight r squared
open parentheses straight d close parentheses space None space of space these end style

Solution 19

begin mathsize 12px style Correct space option colon space open parentheses straight b close parentheses
area space of space circle space 1 space equals space πr subscript 1 superscript 2
area space of space circle space 2 space equals space πr subscript 2 superscript 2
area space of space circle space 3 space equals space πr squared
Given comma space space πr subscript 1 superscript 2 space plus space space πr subscript 2 superscript 2 space equals space πr squared
straight r subscript 1 superscript 2 space plus space straight r subscript 2 superscript 2 space equals space straight r squared end style

Question 20

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

(a) 10 cm

(b) 12 cm

(d) 16 cmSolution 20

begin mathsize 12px style Correct space option colon space open parentheses straight c close parentheses
perimeter space of space semicircle space equals space πr space plus space 2 straight r
2 straight r space plus space πr space equals space 36
straight r space equals space fraction numerator 36 over denominator straight pi space plus space 2 end fraction
diameter space equals space 2 straight r
space space space space space space space space space space space space space space space space space space space equals space fraction numerator 72 over denominator straight pi space plus space 2 end fraction
space space space space space space space space space space space space space space space space space space space equals space 14 space cm end style

Question 21

begin mathsize 12px style The space perimeter space of space the space sector space OAB space shown space in space figure comma space is
open parentheses straight a close parentheses space 64 over 3 space cm
open parentheses straight b close parentheses space 26 space cm
open parentheses straight c close parentheses space 64 over 5 space cm
open parentheses straight d close parentheses space 19 space cm end style

Solution 21

begin mathsize 12px style Correct space option colon space open parentheses straight a close parentheses
OA space equals space 7 space cm
circle space with space angle space 2 straight pi space has space circumference
equals space 2 πr
straight pi over 3 space radian space equals space 60 degree
for space angle space 1 space radian space equals space straight r
for space straight pi over 3 space radian space equals space straight pi over 3 straight r
straight pi over 3 space cross times space 7
22 over 7 space cm
perimeter space equals space OA space plus space OB space plus space circumference
space space space space space space space space space space space space space space space space space space space space space equals space 7 space plus space 7 space plus space 22 over 3
space space space space space space space space space space space space space space space space space space space space space equals space 14 space plus space 22 over 3
space space space space space space space space space space space space space space space space space space space space space equals space 64 over 3 space cm
end style

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

(d) 56 cm²Solution 22

begin mathsize 12px style Correct space option space colon space open parentheses straight b close parentheses
any space sector space of space circle space with space radius space straight r.
circumference space open parentheses AB close parentheses space equals space open parentheses fraction numerator straight theta over denominator 2 straight pi end fraction close parentheses space 2 πr
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space θr space space space space space space space space space space...... open parentheses 1 close parentheses
area space equals space open parentheses fraction numerator straight theta over denominator 2 straight pi end fraction close parentheses πr squared
space space space space space space space space space space space equals space straight theta over 2 space straight r squared space space space space space...... open parentheses 2 close parentheses
perimeter space of space OAB space equals space OA space plus space AB space plus space OB
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space straight r space plus space straight r space plus space AB
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space 2 straight r space plus space AB
2 straight a space equals space 13 space space plus space AB
AB space equals space 16
From space open parentheses 1 close parentheses space θr space equals space 16
straight theta space equals space fraction numerator 16 over denominator 6.5 end fraction space radian
from space open parentheses 2 close parentheses
area space equals space straight theta over 2 space straight r squared
space space space space space space space space space space space equals space fraction numerator 16 over denominator 6.5 space cross times space 2 end fraction space cross times space 6.5 space cross times 6.5
space space space space space space space space space space space equals space 8 space cross times space 6.5
space space space space space space space space space space space equals space 52 space cm squared

end style

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

(d) 10 cmSolution 23

begin mathsize 12px style Correct space option colon space open parentheses straight c close parentheses
arc space equals space 5 straight pi comma space area space equals space 20 straight pi
We space know comma space length space of space arc space equals space θr
and space area space of space sector space equals space straight theta over 2 space straight r squared
Hence space θr space equals space 5 space straight pi space and space straight theta over 2 straight r squared space equals space 20 straight pi
fraction numerator begin display style straight theta over 2 end style straight r squared over denominator θr end fraction space equals space fraction numerator 20 straight pi over denominator 5 straight pi end fraction
straight r over 2 space equals space 4
straight r space equals space 8 space cm end style

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

(d) 25 π cm²Solution 24

Correct option: (d)

Diameter of circle = side of square

2r = 10

r = 5 cm

Area of circle = πr² = 25 π cm²

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

(d) 150 cm²Solution 25

begin mathsize 12px style Correct space option colon space open parentheses straight a close parentheses
2 πr space minus space straight r space equals space 37
straight r space equals space fraction numerator 37 over denominator 2 straight pi space minus space 1 end fraction equals space fraction numerator 37 over denominator 2 space cross times space begin display style 22 over 7 end style space minus space 1 end fraction space equals space fraction numerator 37 over denominator begin display style 37 over 7 end style end fraction
straight r space equals space 7 space cm
area space equals space πr squared
space space space space space space space space space space space equals space 49 space straight pi space cm squared
space space space space space space space space space space space equals space 154 space cm squared end style

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

(d) π (h + r) hSolution 26

Correct option: (b)

Inner radius = r

outer radius = r + h

area of shaded region = area of outer circle – area of inner circle

= π (r + h)² – πr²

= π {(r + h)² – r²}

= π (r + h – r) (r + h + r)

= π (2r + h) h

Question 27

begin mathsize 12px style If space AB space is space straight a space chord space of space length space 5 square root of 3 space cm space of space straight a space circle space with space centre space straight O space and space radius space 5 space cm comma space then space area space of space sector space OAB space is
open parentheses straight a close parentheses space fraction numerator 3 straight pi over denominator 8 end fraction space cm squared
open parentheses straight b close parentheses space fraction numerator 8 straight pi over denominator 3 end fraction space cm squared
open parentheses straight c close parentheses space 25 straight pi space cm squared
open parentheses straight d close parentheses space fraction numerator 25 straight pi over denominator 3 end fraction space cm squared end style

Solution 27

begin mathsize 12px style Correct space option space colon space open parentheses straight d close parentheses
OO apostrophe space is space perpendicular space bisector space of space AB.
Hence space AO apostrophe space equals space 1 half AB
space space space space space space space space space space space space space space space space space space space space space space equals space fraction numerator 5 square root of 3 over denominator 2 end fraction
OA space equals space 5 space cm
In space triangle space OAO apostrophe comma space sin space straight theta space equals space fraction numerator AO apostrophe over denominator AO end fraction space equals space fraction numerator 5 square root of 3 over denominator 5 space cross times space 2 end fraction space equals space fraction numerator square root of 3 over denominator 2 end fraction
straight theta space equals space 60 degree
angle AOB space equals space 2 space angle AOO apostrophe
angle AOB space equals space 120 degree
space space space space space space space space space space space space space space equals space fraction numerator 2 straight pi over denominator 3 end fraction space radian
area space of space sector space with space angle space straight theta space equals space straight theta over 2 straight r squared
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space open parentheses fraction numerator 2 straight pi over denominator 3 end fraction close parentheses space cross times space 1 half space cross times space open parentheses 5 close parentheses squared
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space straight pi over 3 cross times space 25
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space fraction numerator 25 straight pi over denominator 3 end fraction end style

Question 28

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

(a) 2π sq. units

(b) 4π sq. units

(d) 8π sq. unitsSolution 28

Correct option: (b)

area = circumference

πr² = 2πr

r = 2 units

area = πr²

= 4π sq. unitsQuestion 29

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

(a) 96%

(b) 40%

(d) 48%Solution 29

begin mathsize 12px style Correct space option colon space open parentheses straight a close parentheses
Initial space diameter space equals space straight d
Final space diameter space equals space straight d space plus space 40 percent sign space of space straight d
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space straight d space plus space 40 over 100 straight d space equals space 1.4 straight d
Initial space area space equals space straight pi open parentheses straight d over 2 close parentheses squared space equals space πd squared over 4
Final space area space equals space straight pi open parentheses fraction numerator 1.4 space straight d over denominator 2 end fraction close parentheses squared space equals space 1.96 πd squared over 4
percent sign space increase space in space area space equals space fraction numerator Final space minus space initial over denominator initial end fraction space cross times space 100
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space 96 percent sign space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space end style

Question 30

In figure, the shaded area is

(a) 50 (π – 2) cm²

(b) 25 (π – 2) cm²

(d) 5 (π – 2) cm²

Solution 30

** img pending

begin mathsize 12px style Correct space option colon space open parentheses straight b close parentheses
area space of space OAB space sector space equals space straight theta over 2 space straight r squared
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space straight pi over 4 straight r squared
area space of space triangle OAB space equals space 1 half space cross times space OA space cross times space OB
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space straight r squared over 2
area space of space shaded space region space equals space area space of space sector space minus space area space of space OAB
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space straight pi over 4 straight r squared space space minus space straight r squared over 2
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space open parentheses fraction numerator straight pi space minus space 2 over denominator 4 end fraction close parentheses straight r squared
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space space open parentheses fraction numerator straight pi space minus space 2 over denominator 4 end fraction close parentheses space open parentheses 10 close parentheses squared
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space 25 open parentheses straight pi space minus space 2 close parentheses space cm squared space space space space space space space space space space space space space space space space space space space space end style

Question 31

begin mathsize 12px style In space figure comma space the space area space of space the space segment space PAQ space is
open parentheses straight a close parentheses space straight a squared over 4 open parentheses straight pi space plus space 2 close parentheses
open parentheses straight b close parentheses space straight a squared over 4 open parentheses straight pi space minus space 2 close parentheses
open parentheses straight c close parentheses space straight a squared over 4 open parentheses straight pi space minus space 1 close parentheses
open parentheses straight d close parentheses space straight a squared over 4 open parentheses straight pi space plus space 1 close parentheses end style

Solution 31

begin mathsize 12px style Correct space option space colon space open parentheses straight b close parentheses
same space as space straight Q space 30 comma
Here space radius space is space straight a space
Hence space straight a squared over 4 open parentheses straight pi space minus space 2 close parentheses end style

Question 32

begin mathsize 12px style In space figure comma space the space area space of space segment space ACB space is
open parentheses straight a close parentheses space open parentheses straight pi over 3 space minus space fraction numerator square root of 3 over denominator 2 end fraction close parentheses straight r squared
open parentheses straight b close parentheses space open parentheses straight pi over 3 space plus space fraction numerator square root of 3 over denominator 2 end fraction close parentheses straight r squared
open parentheses straight c close parentheses space open parentheses straight pi over 3 space minus space fraction numerator 2 over denominator square root of 3 end fraction close parentheses straight r squared
open parentheses straight d close parentheses space None space of space these end style

Solution 32

begin mathsize 12px style Correct space option colon space open parentheses straight d close parentheses
straight theta space equals space 120 degree
space space space space equals space fraction numerator 2 straight pi over denominator 3 end fraction space radian
area space of space sector space OACBO space equals space straight theta over 2 space straight r squared
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space straight pi over 3 space straight r squared
space area space of space triangle OAB
In space triangle space OO apostrophe straight A
sin space 30 degree space equals space fraction numerator OO apostrophe over denominator OA end fraction
OO apostrophe space equals space straight r over 2
and space tan space 30 degree space equals space fraction numerator OO apostrophe over denominator AO apostrophe end fraction
AO apostrophe space equals space fraction numerator OO apostrophe over denominator tan space 30 degree end fraction space equals space fraction numerator square root of 3 straight r over denominator 2 end fraction
AB space equals space 2 space AO apostrophe
space space space space space space equals space square root of 3 straight r space
area space of space space triangle OAB space equals space 1 half space cross times space AB space cross times space OO apostrophe
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space 1 half space cross times space square root of 3 straight r space cross times space straight r over 2 space equals space fraction numerator square root of 3 straight r squared over denominator 4 end fraction
area space of space segment space ACB space equals space straight pi over 3 straight r squared space minus space fraction numerator square root of 3 over denominator 4 end fraction straight r squared
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space open parentheses straight pi over 3 space minus space fraction numerator square root of 3 over denominator 4 end fraction close parentheses straight r squared end style

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 is

(a) 12 cm

(b) 16 cm

(d) 10 cmSolution 33

begin mathsize 12px style Correct space option colon space open parentheses straight c close parentheses
length space of space arc space equals space θr
area space of space sector space equals space straight theta over 2 straight r squared
θr space equals space 5 space and space straight theta over 2 straight r squared space equals space 20 straight pi
fraction numerator θr over denominator begin display style straight theta over 2 end style straight r squared end fraction equals space fraction numerator 5 straight pi over denominator 20 straight pi end fraction
2 over straight r equals space 1 fourth

straight r space equals space 8 space cm end style

Question 34

In Figure, the ratio of the areas of two sectors S₁and S₂ is

(a) 5 : 2

(b) 3 : 5

(d) 4 : 5Solution 34

begin mathsize 12px style Correct space option colon space open parentheses straight d close parentheses
straight theta space of space sector space straight S subscript 1 space equals space 120 degree
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space fraction numerator 2 straight pi over denominator 3 end fraction space radian
area space equals space open parentheses straight theta over 2 close parentheses straight r squared
space space space space space space space space space space space equals space straight pi over 3 straight r squared
straight theta space of space sector space straight S subscript 2 space equals space 150 degree
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space fraction numerator 5 straight pi over denominator 6 end fraction space radian
area space equals space open parentheses fraction numerator 5 straight pi over denominator 12 end fraction close parentheses straight r squared
fraction numerator area space straight S subscript 1 over denominator area space straight S subscript 2 end fraction space equals space fraction numerator open parentheses begin display style straight pi over 3 end style close parentheses straight r squared over denominator open parentheses begin display style fraction numerator 5 straight pi over denominator 12 end fraction end style close parentheses straight r squared end fraction space equals space 12 over 15 space equals space 4 over 5 end style

Question 35

begin mathsize 12px style If space the space area space of space sector space of space straight a space circle space is space 5 over 18 space of space the space area space of space the space circle comma space then space the space sector space angle space is space equal space to
open parentheses straight a close parentheses space 60 degree
open parentheses straight b close parentheses space 90 degree
open parentheses straight c close parentheses space 100 degree
open parentheses straight d close parentheses space 120 degree end style

Solution 35

begin mathsize 12px style Correct space option colon space open parentheses straight c close parentheses
area space of space sector space equals space straight theta over 2 straight r squared
area space of space circle space equals space πr squared
Given space fraction numerator begin display style straight theta over 2 end style straight r squared over denominator πr squared end fraction equals space 5 over 18
fraction numerator straight theta over denominator 2 straight pi end fraction equals space 5 over 18
straight theta space equals space fraction numerator 5 straight pi over denominator 9 end fraction
space space space space space equals space 100 degree end style

Question 36

begin mathsize 12px style If space the space area space of space sector space of space straight a space circle space is space 7 over 20 space of space the space area space of space the space circle comma space then space the space sector space angle space is space equal space to
open parentheses straight a close parentheses space 110 degree
open parentheses straight b close parentheses space 130 degree
open parentheses straight c close parentheses space 100 degree
open parentheses straight d close parentheses space 126 degree end style

Solution 36

begin mathsize 12px style Correct space option space colon space open parentheses straight d close parentheses
fraction numerator begin display style straight theta over 2 end style straight r squared over denominator πr squared end fraction equals 7 over 20
fraction numerator straight theta over denominator 2 straight pi end fraction equals 7 over 20 rightwards double arrow straight theta space equals space fraction numerator 7 straight pi over denominator 10 end fraction
straight theta space equals space 126 degree end style

Question 37

begin mathsize 12px style In space figure comma space if space ABC space is space an space equilateral space triangle comma space then space shaded space area space is space equal space to
open parentheses straight a close parentheses space open parentheses straight pi over 3 space minus space fraction numerator square root of 3 over denominator 4 end fraction close parentheses straight r squared
open parentheses straight b close parentheses space open parentheses straight pi over 3 space minus space fraction numerator square root of 3 over denominator 2 end fraction close parentheses straight r squared
open parentheses straight c close parentheses space space open parentheses straight pi over 3 space plus space fraction numerator square root of 3 over denominator 4 end fraction close parentheses straight r squared
open parentheses straight d close parentheses space open parentheses straight pi over 3 space plus space square root of 3 close parentheses straight r squared end style

Solution 37

begin mathsize 12px style Correct space option colon space open parentheses straight a close parentheses
If space angle BAC space equals space straight theta
then space angle BOC space equals space 2 straight theta
But space angle BAC space equals space 60 degree
as space ABC thin space is space equilateral space
Hence space straight theta space equals space 60 degree
In space triangle OBC
sin space straight theta space equals space BE over OB
BE space equals space OB space sin space 60 degree
space space space space space space equals space fraction numerator square root of 3 straight r over denominator 2 end fraction
and space BC space equals space 2 BE
space space space space space space space space space space space space space space space space equals space square root of 3 straight r
Also space cos space straight theta space equals space OE over OB
OE space equals space OB space cos space 60 degree
space space space space space space space equals space straight r over 2
area space of space triangle OBC space equals space 1 half space cross times space BC space cross times space OE
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space 1 half space cross times space square root of 3 space straight r space cross times space straight r over 2 space equals space fraction numerator square root of 3 space straight r squared over denominator 4 end fraction
area space of space sector space OBC space equals space straight theta over 2 space straight r squared
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space θr squared
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space straight pi over 3 space straight r squared
area space of space shaded space region space equals space straight pi over 3 space straight r squared space minus space fraction numerator square root of 3 over denominator 4 end fraction space straight r squared
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space open parentheses straight pi over 3 space minus space fraction numerator square root of 3 over denominator 4 end fraction close parentheses space straight r squared space space space space space end style

Question 38

In figure, the area of the shaded region is

(a) 3π cm²

(b) 6π cm²

(d) 7π cm²

Solution 38

begin mathsize 12px style Correct space option colon space left parenthesis straight a right parenthesis
PA space equals space 3 space cm
angle straight A space plus space angle straight B space plus space angle straight C space plus space angle straight D space equals space 360 degree
angle straight A space plus space 90 degree space plus space 90 degree space plus space 60 degree space equals space 360 degree
angle straight A space equals space 120 degree
space space space space space space space equals space fraction numerator 2 straight pi over denominator 3 end fraction space space radian
area space of space sector space with space angle space fraction numerator 2 straight pi over denominator 3 end fraction space equals space straight theta over 2 straight r squared
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space open parentheses fraction numerator 2 straight pi over denominator 3 end fraction close parentheses space 1 half space straight r squared
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space straight pi over 3 open parentheses PA close parentheses squared
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space straight pi over 3 open parentheses 3 close parentheses squared
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space space space 3 straight pi space cm squared space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space end style

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

(d) 11 : 14Solution 39

begin mathsize 12px style Correct space option space colon space open parentheses straight b close parentheses
perimeter space of space circle space equals space perimeter space of space square
2 πr space equals space 4 straight a
straight a space equals space πr over 2
fraction numerator area space of space circle over denominator area space of space square end fraction space equals space πr squared over straight a squared space equals space fraction numerator πr squared over denominator begin display style fraction numerator straight pi squared straight r squared over denominator 4 end fraction end style end fraction equals space 4 over straight pi
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space fraction numerator 4 over denominator begin display style 22 over 7 end style end fraction space equals space 14 space colon space 11
end style

Question 40

begin mathsize 12px style The space radius space of space straight a space circle space is space 20 space cm. space It space is space divided space into space four space parts space of space equal space area space by space drawing space three space concentric space circles space inside space it space. space Then comma space the space radius space of space the space largest space of space three space concentric space space circles space drawn space is
open parentheses straight a close parentheses space 10 square root of 5 space cm
open parentheses straight b close parentheses space 10 square root of 3 space cm
open parentheses straight c close parentheses space 10 space cm
open parentheses straight d close parentheses space 10 square root of 2 space cm end style

Solution 40

begin mathsize 12px style Correct space option colon space open parentheses straight b close parentheses
OA space equals space 20 space cm
area space of space 4 space parts space are space equal
Area space of space straight C 4 space equals space πr squared
space space space space space space space space space space space space space space space space space space space space space space space equals space straight pi open parentheses OA close parentheses squared
space space space space space space space space space space space space space space space space space space space space space space space equals space 400 space straight pi
area space of space each space part space equals space 1 fourth space cross times space 400 space straight pi
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space 100 space straight pi
straight C 1 space circle space area space equals space straight pi space open parentheses OD close parentheses squared
100 space straight pi space equals space straight pi space open parentheses OD close parentheses squared
open parentheses OD close parentheses squared space equals space 100
OD space equals space 10 space cm
area space of space straight C 2 space equals space 2 space cross times space area space of space circle space straight C 1
straight pi open parentheses OC close parentheses squared space equals space 2 space cross times space 100 straight pi
open parentheses OC close parentheses squared space equals space 2 space cross times space 100
OC space equals space 10 square root of 2
area space of space straight C 3 space equals space 3 space cross times space area space of space circle space straight C 1
straight pi open parentheses OB close parentheses squared space equals space 3 space cross times space 100 space straight pi
OB space equals space 10 square root of 3 space cm
end style

Question 41

begin mathsize 12px style The space area space of space straight a space sector space whose space perimeter space is space four space space times space its space radius space straight r space units comma space is
open parentheses straight a close parentheses space straight r squared over 4 space sq. space units
open parentheses straight b close parentheses space 2 straight r to the power of 2 space end exponent sq. space units
open parentheses straight c close parentheses space straight r squared space sq. space units
open parentheses straight d close parentheses space straight r squared over 2 space sq. space units end style

Solution 41

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

(d) 2240 cm²Solution 42

begin mathsize 12px style Correct space option colon space open parentheses straight C close parentheses
radius space equals space 28 space cm
area space of space sector space straight O space ABD space equals space straight theta over 2 space straight r squared
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space straight pi over 4 space straight r squared
area space of space circle space equals space πr squared
area space of space major space segment space equals space πr squared space minus space straight pi over 4 straight r squared
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space fraction numerator 3 straight pi over denominator 4 end fraction space straight r squared
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space 3 over 4 cross times 22 over 7 cross times space 28 space cross times space 28
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space 1848 space cm squared space space
end style

Question 43

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

begin mathsize 12px style open parentheses straight a close parentheses space 17 square root of 3 space units
open parentheses straight b close parentheses space 36 space units
open parentheses straight c close parentheses space 72 space units
open parentheses straight d close parentheses space 48 square root of 3 space units end style

Solution 43

begin mathsize 12px style Correct space opton colon space open parentheses straight c close parentheses
ABC space is space an space equilateral space triangle
area space of space circle space equals space πr squared
πr squared space equals space 48 space straight pi
straight r squared space equals space 48
straight r space equals space 4 square root of 3
OE space equals space straight r space equals space 4 square root of 3
angle BCA space equals space 60 degree
and space angle ECO space equals space 1 half angle BCA
angle ECO space equals space 30 degree
In space triangle OEC
tan space 30 degree space equals space OE over EC
EC space equals space square root of 3 space OE space equals space 12
BC space equals space 2 space EC
space space space space space space equals space 24
perimeter space equals space AB space plus space BC space plus space CA
space space space space space space space space space space space space space space space space space space space space space equals space 3 space BC
space space space space space space space space space space space space space space space space space space space space space equals space 72 space end style

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

(d) 10 cm²Solution 44

begin mathsize 12px style Correct space option colon space open parentheses straight b close parentheses
hour space hand space is space 6 space cm space long.
when space minute space hand space complete space 60 space minute space or space 2 straight pi space radian comma space hour space hand space lower space straight pi over 6 space radian
minute space hand space 2 straight pi space radian space rightwards arrow hour space hand space straight pi over 6 rad
minute space hand space 1 space radian space rightwards arrow space hour space hand space 1 over 12 rad
minute space hand space goes space 11 space colon space 20 space to space 11 space colon space 55 space equals space 35 space min
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space fraction numerator 2 straight pi over denominator 60 end fraction space cross times space 35
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space fraction numerator 35 straight pi over denominator 30 end fraction
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space fraction numerator 7 straight pi over denominator 6 end fraction space rad
minute space hand space fraction numerator 7 straight pi over denominator 6 end fraction space radian space rightwards arrow space hour space hand space 1 over 12 space cross times space fraction numerator 7 straight pi over denominator 6 end fraction space rad
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space rightwards arrow space fraction numerator 7 straight pi over denominator 72 end fraction space rad
area space equals straight Q over 2 straight r squared
space space space space space space space space space space space space space equals space open parentheses fraction numerator 7 straight pi over denominator 72 end fraction close parentheses space cross times space 1 half space cross times space open parentheses 6 close parentheses squared
space space space space space space space space space space space space space equals space fraction numerator 7 straight pi over denominator 144 end fraction space cross times space 36
space space space space space space space space space space space space space equals space fraction numerator 7 straight pi over denominator 4 end fraction space equals space 7 over 4 space cross times space 22 over 7 space equals space 5.5 space cm squared

space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space end style

Question 45

begin mathsize 12px style ABCD space is space straight a space square space of space side space 4 space cm. space If space straight E space is space straight a space point space in space the space interior space of space the space square space such space that space triangle CED space is space equilateral comma space then space area space of space triangle ACE space is
open parentheses straight a close parentheses space 2 open parentheses square root of 3 space minus space 1 close parentheses space cm squared
open parentheses straight b close parentheses space 4 open parentheses square root of 3 space minus space 1 close parentheses space cm squared
open parentheses straight c close parentheses space 6 open parentheses square root of 3 space minus space 1 close parentheses space cm squared
open parentheses straight d close parentheses space 8 open parentheses square root of 3 space minus space 1 close parentheses space cm squared end style

Solution 45

begin mathsize 12px style Correct space option colon space open parentheses straight b close parentheses
AB space equals space 4 space cm
triangle CED space is space an space equilateral space triangle.
Hence space CD space equals space ED space equals space EC space equals space 4 space cm
In space triangle OED
sin space 30 degree space equals space OE over ED
OE space equals space ED over 2 space equals space 2 space cm
In space triangle EO apostrophe straight C
sin space 60 degree space equals space fraction numerator straight O apostrophe straight E over denominator EC end fraction
straight O apostrophe straight E space equals space EC space cross times space fraction numerator square root of 3 over denominator 2 end fraction
space space space space space space space space equals space 2 square root of 3
area space of space triangle EDC space equals space 1 half cross times DC cross times EO apostrophe
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space 1 half cross times 4 cross times 2 square root of 3
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space 4 square root of 3
area space of space triangle AED space equals space 1 half cross times space AD space cross times space OE
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space 1 half cross times space 4 space cross times space 2
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space 4
area space of space triangle ADC space equals space 1 half cross times space AD space cross times space DC
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space 1 half cross times space 4 space cross times space 4
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space 8
area space of space triangle AEC space equals space area space open parentheses triangle AED close parentheses space plus space area space open parentheses triangle DEC close parentheses space minus space area space open parentheses triangle ADC close parentheses
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space space 4 space plus space 4 square root of 3 space minus space 8
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space 4 square root of 3 space minus space 4
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space 4 open parentheses square root of 3 space minus space 1 close parentheses
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space end style

Question 46

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

(a) 34

(b) 26

(d) 14Solution 46

Correct option: (b)

radius of Circle = 5 cm

area = π (5)²

= 25 π

rdius of circle 2 = 12 cm

area = π (12)²

= 144 π

area of larger circle = 144 π + 25π

= 169 π

πr² = 169 π

r² = 169

r = 13

diameter = 2r

= 26Question 47

If Π is taken as 22/7, the distance (in metres) covered by a wheel of diameter 35 cm, in one revolution, is

(a) 2.2

(b) 1.1

(d) 96.25 Solution 47

begin mathsize 12px style Correct space option colon space open parentheses straight b close parentheses
radius space of space wheel space equals space diameter over 2 equals space 35 over 2 cm
distance space covered space by space wheel space in space one space revolution space equals space 2 πr
equals space 2 space cross times space 22 over 7 cross times 35 over 2
equals space 110 space cm
equals space 1.1 space straight m end style

Question 48

ABCD is a rectangle whose three vertices are B(4, 0), C(4, 3) and D(0, 3). The length of one of its diagonals is

(a) 5

(b) 4

(d) 25Solution 48

begin mathsize 12px style Correct space option colon space open parentheses straight a close parentheses
Other space co minus ordinate space of space rectangle space is space open parentheses 0 comma space 0 close parentheses
Diagonals space of space rectangle space are space equal.
Hence space OC space equals space OB
OC space equals space square root of open parentheses 4 space minus space 0 close parentheses squared space plus space open parentheses 3 space minus space 0 close parentheses squared end root
space space space space space space space equals space square root of 16 space plus space 9 end root
space space space space space space space equals space square root of 25
space space space space space space space equals space 5 space space end style

Question 49

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

a. r² sq. units

c. 2r² sq. units

d. Solution 49

Question 50

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 = r₁ + r₂

c. r₁ + r₂ < r

d. Solution 50

Question 51

If the sum of the circumference of two circles with radii r₁ and r₂ is equal to the circumference of a circle of radius r, then

a. r = r₁ + r₂

b. r₁ + r₂ > r

c. r₁ + r₂ < 2

d. None of theseSolution 51

Question 52

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. Nothing definite can be saidSolution 52

Question 53

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

a. 22 : 7

b. 14 : 11

c. 7 : 22

d. 11 : 14Solution 53

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RD SHARMA SOLUTION CHAPTER- 15 Areas Related to Circles | CLASS 10TH MATHEMATICS-EDUGROWN

Chapter 15 – Areas Related to Circles Exercise Ex. 15.1

Chapter 15 – Areas Related to Circles Exercise Ex. 15.2

Chapter 15 – Areas Related to Circles Exercise Ex. 15.3

Chapter 15 – Areas Related to Circles Exercise Ex. 15.4

Chapter 15 – Areas Related to Circles Exercise 15.69

Chapter 15 – Areas Related to Circles Exercise 15.70

Chapter 15 – Areas Related to Circles Exercise 15.71

Chapter 15 – Areas Related to Circles Exercise 15.72

Chapter 15 – Areas Related to Circles Exercise 15.73

Chapter 13 – Areas Related to Circles Exercise 13.74

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