Author Archives: Priyanka

Exercise Ex. 6.1

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Exercise Ex. 6.2

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Exercise Ex. 6.3

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x + y ≥ 4   … (1)

2x – y < 0  … (2)

The graph of the lines, x + y = 4 and 2x – y = 0 are drawn in the figure below.

Inequality (1) represents the region above the line x + y = 4. (including the line x + y = 4)

It is observed that (–1, 0) satisfies the inequality, 2x – y < 0.

[2(-1) – 0 = -2< 0]

Therefore, inequality (2) represents the half plane corresponding to the line, 2x – y = 0 containing the point (-1, 0). [excluding the line 2x – y < 0]

Hence, the solution of the given system of linear inequalities is represented by the common shaded region including the points on the line x + y = 4 and excluding the points on line 2x – y = 0 as follows:

Solution 5

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Solution 15

Exercise Misc. Ex.

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Exercise Ex. 1.1

Solution 1

 Hence: this collection is a set. (vii) The collection of all even integers is a well-defined collection because one can definitely identify an even integer that belongs to this collection.  Hence: this collection is a set.

(viii) The collection of questions in this chapter is a well-defined collection because one can definitely identify a question that belongs to this chapter.

 Hence: this collection is a set.

(ix) The collection ofmost dangerous animals of the world is not a well-defined collection because the criteria for determining the dangerousness of an animal can vary from person to person.

 Hence: this collection is not a set.Solution 2

Solution 3

(i) A = { x : x is an integer and -3 ≤ x < 7}.

The elements of this set are -3, -2, -1, 0, 1, 2, 3, 4, 5 and 6 only.

Therefore, the given set can written in roster form as

A = {-3, -2, -1, 0, 1, 2, 3, 4, 5, 6}

(ii) B = { x : x is a natural number less than 6}

The elements of this set are 1, 2, 3, 4 and 5 only.

Therefore, the given set can written in roster form as

B = {1, 2, 3, 4, 5}

(iii) C = { x : x is a two-digit number such that the sum of its digits is 8}

The elements of this set are 17, 26, 35, 44, 53, 62, 71 and 80 only.

Therefore, the given set can written in roster form as

C = {17, 26, 35, 44, 53, 62, 71, 80}

(iv) D = { x : x is a prime number which is divisor of 60}.

2	60
2	30
3	15
	5

∴ 60 = 2 × 2 × 3 × 5

The elements of this set are 2, 3 and 5 only.

Therefore, this set can written in roster form as D = {2, 3, 5}.

(v) E = The set of all letters in the word TRIGONOMETRY.

There are 12 letters in the word TRIGONOMETRY, out of which letters T, R, and O are repeated.

Therefore, this set can written in roster form as

E = {T, R, I, G, O, N, M, E, Y}

(vi) F = The set of all letters in the word BETTER.

There are 6 letters in the word BETTER, out of which letters E and T are repeated.

Therefore, this set can written in roster form as

F = {B, E, T, R}Solution 4

Solution 5

Solution 6

Exercise Ex. 1.2

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Exercise Ex. 1.3

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Exercise Ex. 1.4

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Exercise Ex. 1.5

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Exercise Ex. 1.6

Solution 1

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Solution 8

Exercise Misc. Ex.

Solution 1

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Solution 16

Exercise MCQ

Question 1

The length, breadth and height of a cuboid are 15 cm, 12 cm and 4.5 cm respectively. Its volume is

1. 243 cm³
2. 405 cm³
3. 810 cm³
4. 603 cm³

Solution 1

Question 2

A cuboid is 12 cm long, 9 cm broad and 8 cm high. Its total surface area is

1. 864 cm²
2. 552 cm²
3. 432 cm²
4. 276 cm²

Solution 2

Question 3

The length breadth and height of a cuboid are 15m, 6m, and 5 dm respectively. The lateral surface area of the cuboid is

1. 45 m²
2. 21 m²
3. 201 m²
4. 90 m²

Solution 3

Question 4

A beam 9 m long, 40 cm wide and 20 cm high is made up of iron which weighs 50 kg per cubic metre. The weight of the beam is

1. 27 kg
2. 48 kg
3. 36 kg
4. 56 kg

Solution 4

Question 5

The length of the longest rod that can be placed in a room of dimensions (10 m × 10 m × 5 m) is

1. 15 m
2. 16 m
3.  
4. 12 m

Solution 5

Question 6

What is the maximum length of a pencil that can be placed in a rectangular box of dimensions (8 cm × 6 cm × 5 cm)? 

1. 8 cm
2. 9.5 cm
3. 19 cm
4. 11.2 cm

Solution 6

Question 7

The number of planks of dimensions (4 m × 5 m × 2 m) that can be stored in a pit which is 40 m long, 12 m wide and 16 m deep is

1. 190
2. 192
3. 184
4. 180

Solution 7

Question 8

How many planks of dimensions (5 m × 25 cm × 10 cm) can be stored in a pit which is 20 m long, 6 m wide and 50 cm deep?

1. 480
2. 450
3. 320
4. 360

Solution 8

Question 9

How many bricks will be required to construct a wall 8 m long, 6 m high and 22.5 cm thick if each brick measures (25 cm × 11.25 cm × 6 cm)?

1. 4800
2. 5600
3. 6400
4. 5200

Solution 9

Question 10

How many persons can be accommodated in a dining hall of dimensions (20 m × 15 m × 4.5 m), assuming that each person requires 5 m³ of air?

1. 250
2. 270
3. 320
4. 300

Solution 10

Question 11

A river 1.5 m deep and 30 m wide is flowing at the rate of 3 km per hour. The volume of water that runs into the sea per minute is

1. 2000 m³
2. 2250 m³
3. 2500 m³
4. 2750 m³

Solution 11

Question 12

The lateral surface area of a cube is 256 m². The volume of the cube is

1. 64 m³
2. 216 m³
3. 256 m³
4. 512 m³

Solution 12

Question 13

The total surface area of a cube is 96m². The volume of the cube is

1. 8 cm³
2. 27cm³
3. 64cm³
4. 512 cm³

Solution 13

Question 14

The volume of a cube is 512 cm³. Its surface area is

1. 256 cm²
2. 384 cm²
3. 512 cm²
4. 64 cm²

Solution 14

Question 15

The length of the longest rod that can fit in a cubical vessel of side 10 cm is

1. 10 cm
2. 20 cm
3.  
4. 

Solution 15

Question 16

If the length of diagonal of a cube is  cm, then its surface area is

1. 192 cm²
2. 384 cm²
3. 512 cm²
4. 768 cm²

Solution 16

Question 17

If each edge of a cube is increased by 50%, then the percentage increase in its surface area is

1. 50%
2. 75%
3. 100%
4. 125%

Solution 17

Question 18

Three cubes of metal with edges 3 cm, 4 cm and 5 cm respectively are melted to form a single cube. The lateral surface area of the new cube formed is

1. 72 cm²
2. 144 cm²
3. 128 cm²
4. 256 cm²

Solution 18

Question 19

In a shower, 5 cm of rain falls, what is the volume of water that falls on 2 hectors of ground?

1. 500 m³
2. 750 m³
3. 800 m³
4. 1000 m³

Solution 19

Question 20

Two cubes have their volumes in the ratio 1:27. The ratio of their surface area is

1. 1:3
2. 1:8
3. 1:9
4. 1:18

Solution 20

Question 21

If each side of a cube is doubled, then its volume

1. is doubled
2. becomes 4 times
3. becomes 6 times
4. becomes 8 times

Solution 21

Question 22

The diameter of a base of a cylinder is 6 cm and its height is 14 cm. The volume of the cylinder is

a. 198 cm³

b. 396 cm³

c. 495 cm³

d. 297 cm³Solution 22

Question 23

The diameter of a cylinder is 28 cm and its height is 20 cm, then its curved surface area is

1. 880 cm²
2. 1760 cm²
3. 3520 cm²
4. 2640 cm²

Solution 23

Question 24

If the curved surface area of a cylinder is 1760 cm² and its base radius is 14 cm, then its height is

1. 10 cm
2. 15 cm
3. 20 cm
4. 40 cm

Solution 24

Question 25

The height of a cylinder is 14 cm and its curved surface area is 264 cm². The volume of the cylinder is

1. 308 cm²
2. 396 cm²
3. 1232 cm²
4. 1848 cm²

Solution 25

Question 26

The curved surface area of the cylindrical pillar is 264 m² and its volume is 924m³. The height of the pillar is

1. 4 m
2. 5 m
3. 6 m
4. 7 m

Solution 26

Question 27

The radii of two cylinders are in the ratio 2:3 and their heights are in the ratio 5:3. The ratio of their surface area is

1. 2:5
2. 8:7
3. 10:9
4. 16:9

Solution 27

Question 28

The radii of two cylinders are in the ratio 2:3 and their heights are in the ratio 5:3. The ratio of their volumes is

1. 27:20
2. 20:27
3. 4:9
4. 9:4

Solution 28

Question 29

The ratio between the radius of the base and height of a cylinder is 2:3. If its volume is 1617 cm³, then its total surface area is

1. 308 cm²
2. 462 cm²
3. 540 cm²
4. 770 cm²

Solution 29

Question 30

Two circular cylinders of equal volume have their heights in the ratio 1:2. The ratio of their radii is

Solution 30

Question 31

The ratio between the curved surface area and the total surface area of a right circular cylinder is 1:2. If the total surface area is 616 cm², then the volume of the cylinder is

1. 1078 cm³
2. 1232 cm³
3. 1848 cm³
4. 924 cm³

Solution 31

Question 32

In a cylinder, if the radius is halved and the height is doubled, then the volume will be

1. The same
2. Doubled
3. Halved
4. Four times

Solution 32

Question 33

The number of coins 1.5 cm in diameter and 0.2 cm thick to be melted to form a right circular cylinder of height 10 cm and diameter 4.5 cm is

1. 540
2. 450
3. 380
4. 472

Solution 33

Question 34

The radius of a wire is decreased to one-third. If volume remains the same, the length will become

1. 2 times
2. 3 times
3. 6 times
4. 9 times

Solution 34

Question 35

The diameter of a roller, 1m long is 84 cm. If it takes 500 complete revolutions to level a playground, the area of the playground is

1. 1440 m²
2. 1320 m²
3. 1260 m²
4. 1550 m²

Solution 35

Question 36

2.2 dm³ of lead is to be drawn into a cylindrical wire 0.50 cm in diameter. The length of the wire is

1. 110 m
2. 112 m
3. 98 m
4. 124 m

Solution 36

Question 37

The lateral surface area of a cylindrical is

Solution 37

Question 38

The height of a cone is 24 cm and the diameter of its base is 14 cm. The curved surface area of the cone is

1. 528 cm²
2. 550 cm²
3. 616 cm²
4. 704 cm²

Solution 38

Question 39

The volume of a right circular cone of height is 12 cm and base radius 6 cm, is

1. (12π) cm³
2. (36π) cm³
3. (72π) cm³
4. (144π) cm³

Solution 39

Question 40

How much cloth 2.5 m wide will be required to make a conical tent having base radius 7 m and height 24 m?

1. 120 m
2. 180 m
3. 220 m
4. 550 m

Solution 40

Question 41

The volume of a cone is 1570 cm³ and its height is 15 cm. What is the radius of the cone? (Use π = 3.14)

1. 10 cm
2. 9 cm
3. 12 cm
4. 8.5 cm

Solution 41

Question 42

The height of cone is 21 cm and its slant height is 28 cm. The volume of the cone is

1. 7356 cm³
2. 7546 cm³
3. 7506 cm³
4. 7564 cm³

Solution 42

Correct option: (b)

Question 43

The volume of a right circular cone of height 24 cm is 1232 cm³. Its curved surface area is

1. 1254 cm²
2. 704 cm²
3. 550 cm²
4. 462 cm²

Solution 43

Question 44

If the volumes of two cones be in the ratio 1:4 and the radii of their bases be in the ratio 4:5, then the ratio of their heights is

1. 1:5
2. 5:4
3. 25:16
4. 25:64

Solution 44

Question 45

If the height of a cone is doubled, then its volume is increased by

1. 100%
2. 200 %
3. 300 %
4. 400 %

Solution 45

Question 46

The curved surface area of the cone is twice that of the other while the slant height of the latter is twice that of the former. The ratio of their radii is

1. 2:1
2. 4:1
3. 8:1
4. 1:1

Solution 46

Question 47

The ratio of the volumes of a right circular cylinder and a right circular cone of the same base and same height will be

1. 1:3
2. 3:1
3. 4:3
4. 3:4

Solution 47

Question 48

A right circular cylinder and a right circular cone have the same radius and the same volume. The ratio of the height of the cylinder to that of the cone is

1. 3:5
2. 2:5
3. 3:1
4. 1:3

Solution 48

Question 49

The radii of the bases of a cylinder and a cone are in the ratio 3:4 and their heights are in the ratio 2:3. Then their volumes are in the ratio

1. 9:8
2. 8:9
3. 3:4
4. 4:3

Solution 49

Question 50

If the height and the radius of cone are doubled, the volume of the cone becomes

1. 3 times
2. 4 times
3. 6 times
4. 8 times

Solution 50

Question 51

A solid metallic cylinder of base radius 3 cm and height 5 cm is melted to make n solid cones of height 1 cm and base radius 1 mm. The value of n is

1. 450
2. 1350
3. 4500
4. 13500

Solution 51

Question 52

A conical tent is to accommodate 11 persons such that each person occupies 4 m² of space on the ground. They have 220m³ of air to breathe. The height of the cone is

1. 14m
2. 15 m
3. 16 m
4. 20 m

Solution 52

Question 53

The volume of a sphere of radius 2r is

Solution 53

Question 54

The volume of a sphere of a radius 10.5 cm is

1. 9702 cm³
2. 4851 cm³
3. 19404 cm³
4. 14553 cm³

Solution 54

Question 55

The surface area of a sphere of radius 21 cm is

1. 2772 cm²
2. 1386 cm²
3. 4158 cm²
4. 5544 cm²

Solution 55

Question 56

The surface area of a sphere is 1386 cm². Its volume is

1. 1617 cm³
2. 3234 cm³
3. 4851 cm³
4. 9702 cm³

Solution 56

Question 57

If the surface area of a sphere is (144 π) m², then its volume is

1. (288 π) m³
2. (188 π) m³
3. (300 π) m³
4. (316 π) m³

Solution 57

Question 58

The volume of a sphere is 38808 cm³. Its curved surface area is

1. 5544 cm²
2. 8316 cm²
3. 4158 cm²
4. 1386 cm²

Solution 58

Question 59

If the ratio of the volumes of two spheres is 1:8, then the ratio of their surface area is

1. 1:2
2. 1:4
3. 1:8
4. 1:16

Solution 59

Question 60

A solid metal ball of radius 8 cm is melted and cast into smaller balls, each of radius 2 cm, The number of such balls is

1. 8
2. 16
3. 32
4. 64

Solution 60

Question 61

A cone is 8.4 cm high and the radius of its base is 2.1 cm. It is melted and recast into a sphere. The radius of the sphere is

1. 4.2 cm
2. 2.1 cm
3. 2.4 cm
4. 1.6 cm

Solution 61

Question 62

A solid lead ball of radius 6 cm is melted and then drawn into a wire of diameter 0.2 cm. The length of wire is

1. 272 m
2. 288 m
3. 292 m
4. 296 m

Solution 62

Question 63

A metallic sphere of radius 10.5 cm is melted and then recast into small cones, each of radius 3.5 cm and height 3 cm. The number of such cones will be

1. 21
2. 63
3. 126
4. 130

Solution 63

Question 64

How many lead shots, each 0.3 cm in diameter, can be made from a cuboid of dimensions 9 cm × 11 cm × 12 cm?

1. 7200
2. 8400
3. 72000
4. 84000

Solution 64

Question 65

The diameter of a sphere is 6 cm. It is melted and drawn into a wire of diameter 2 mm. The length of the wire is

1. 12 m
2. 18 m
3. 36 m
4. 66 m

Solution 65

Question 66

A sphere of diameter 12.6 cm is melted and cast into a right circular cone of height 25.2 cm. The radius of the base of the cone is

1. 6.3 cm
2. 2.1 cm
3. 6 cm
4. 4 cm

Solution 66

Question 67

A spherical ball of radius 3 cm is melted and recast into three spherical balls. The radii of two of these balls are 1.5 cm and 2 cm. The radius of the third ball is

1. 1 cm
2. 1.5 cm
3. 2.5 cm
4. 0.5 cm

Solution 67

Question 68

The radius of a hemispherical balloon increases from 6 cm to 12 cm as air is being pumped into it. The ratio of the surface areas of the balloons in two cases is

1. 1:4
2. 1:3
3. 2:3
4. 1:2

Solution 68

Question 69

The volumes of the two spheres are in the ratio 64:27 and the sum of their radii is 7 cm. The difference of their total surface areas is

1. 38 cm²
2. 58 cm²
3. 78 cm²
4. 88 cm²

Solution 69

Question 70

A hemispherical bowl of radius 9 cm contains a liquid. This liquid is to be filled into cylindrical small bottles of diameter 3 cm and height 4 cm. How many bottles will be needed to empty the bowl?

1. 27
2. 35
3. 54
4. 63

Solution 70

Question 71

A cone and a hemisphere have equal bases and equal volumes. The ratio of their heights is

1. 1:2
2. 2:1
3. 4:1
4. 

Solution 71

Question 72

A cone, a hemisphere and a cylinder stand on equal bases and have the same height. The ratio of their volumes is

1. 1:2:3
2. 2:1:3
3. 2:3:1
4. 3:2:1

Solution 72

Question 73

If the volumes and the surface area of sphere are numerically the same, then its radius is

1. 1 units
2. 2 units
3. 3 units
4. 4 units

Solution 73

Exercise Ex. 15C

Question 1

Find the curved surface area of a cone with base radius 5.25 cm and slant height 10 cm.Solution 1

Radius of a cone, r = 5.25 cm

Slant height of a cone, l = 10 cm

Question 2

Find the total surface area of a cone, if its slant height is 21 m and diameter of its base is 24 m.Solution 2

Radius of a cone, r = 12 m

Slant height of a cone, l = 21 cm

Question 3

A joker’s cap is in the form of a right circular cone of base radius 7 cm and height 24 cm. Find the area of the sheet required to make 10 such caps.Solution 3

Radius of a conical cap, r = 7 cm

Height of a conical cap, h = 24 cm

Thus, 5500 cm² sheet will be required to make 10 caps.Question 4

The curved surface area of a cone is 308 cm² and its slant height is 14 cm. Find the radius of the base and total surface area of the cone.Solution 4

Let r be the radius of a cone.

Slant height of a cone, l = 14 cm

Curved surface area of a cone = 308 cm²

Question 5

The slant height and base diameter of a conical tomb are 25 m and 14 m respectively. Find the cost of whitewashing its curved surface at the rate of Rs.12 per m².Solution 5

Radius of a cone, r = 7 m

Slant height of a cone, l = 25 m

Cost of whitewashing = Rs. 12 per m²

⇒ Cost of whitewashing 550 m² area = Rs. (12 × 550) = Rs. 6600 Question 6

A conical tent is 10 m high and radius of its base is 24 m. Find the slant height of the tent. If the cost of 1 m² canvas is Rs.70, find the cost of canvas required to make the tent.Solution 6

Radius of a conical tent, r = 24 m

Height of a conical tent, h = 10 m

Question 7

A bus stop is barricaded from the remaining part of the road by using 50 hollow cones made of recycled cardboard. Each one has a base diameter of 40 cm and height 1 m. If the outer side of each of the cones is to be painted and the cost of painting is Rs.25 per m², what will be the cost of painting all these cones? (Use π = 3.14 and 1.02.)Solution 7

Question 8

Find the volume, curved surface area and the total surface area of a cone having base radius 35 cm and height 12 cm.Solution 8

Question 9

_{Find the volume, curved surface area and the total surface area of a cone whose height and slant heights are 6 cm and 10 cm respectively. (Take =3.14)}Solution 9

Question 10

A conical pit of diameter 3.5 m is 12 m deep. What is its capacity in kilolitres?

HINT 1 m³ = 1 kilolitre.Solution 10

Question 11

A heap of wheat is in the form of a cone of diameter 9 m and height 3.5 m. Find its volume. How much canvas cloth is required to just cover the heap? (Use π = 3.14.)Solution 11

Radius of a conical heap, r = 4.5 m

Height of a conical tent, h = 3.5 m

Question 12

A man uses a piece of canvas having an area of 551 m², to make a conical tent of base radius 7 m. Assuming that all the stitching margins and wastage incurred while cutting, amount to approximately 1 m², find the volume of the tent that can be made with it.Solution 12

Radius of a conical tent, r = 7 m

Area of canvas used in making conical tent = (551 – 1) m² = 550 m²

⇒ Curved surface area of a conical tent = 550 m²

Question 13

_{How many meters of cloth , 2.5 m wide , will be required to make conical tent whose base radius is 7 m and height 24 metres?}Solution 13

Question 14

_{Two cones have their height in the ratio 1:3 and the radii of their bases in the ratio3: 1. Show that their volumes are in the ratio 3:1.}Solution 14

Question 15

_{A cylinder and a cone have equal radii of their bases and equal height s. If their curved surface areas are in the ratio 8:5, show that the radius and height of each has the ratio 3:4.}Solution 15

Question 16

_{A right circular cone is 3.6 cm height and the radius of its base is 1.6 cm. It is melted and recast into a right circular cone having base radius 1.2 cm. Find its height.}Solution 16

Question 17

_{A circus tent is cylindrical to a height of 3 meters and conical above it. If its diameter is 105 m and the slant height of the conical portion is 53 m, calculate the length of the canvas 5 m wide to make the required tent.}Solution 17

Question 18

_{An iron pillarconsistsof a cylindricalportion2.8 m highand 20cm indiameterand a cone42 cm high is surmounting it . Find the weight of the pillar, given that 1 cm³ of iron weights 7.5 g.}Solution 18

Question 19

_{From a solid right circular cylinder with height 10 cm and radius of the base 6 cm, a right circular cone of the same height and the base is removed .find the volume of the remaining solid. (Take =3.14)}Solution 19

Question 20

_{Water flows at the rate of 10 meters per minute through a cylindrical pipe 5 mm in diameter. How long would it take to fill a conical vessel whose diameter at the surface 40 cm and depth 24 cm?}Solution 20

Question 21

A cloth having an area of 165 m² is shaped into the form of a conical tent of radius 5 m. (i) How many students can sit in the tent if a student, on an average, occupies m² on the ground? (ii) Find the volume of the cone.Solution 21

Curved surface area of the tent = Area of the cloth = 165 m²

Exercise Ex. 15A

Question 1(iv)

_{Find the volume, the lateral surface area and the total surface area cuboid whose dimensions are:}

Length =24 m, breadth =25 cmand height =6mSolution 1(iv)

_{Length = 24 m, breadth = 25 cm =0.25 m, height = 6m.}

_{Volume of cuboid= l x b x h}

_{= (24 x 0.25 x 6) m³.}

_{= 36 m³.}

_{Lateral surface area= 2(l + b) x h}

_{= [2(24 +0.25) x 6] m²}

_{= (2 x 24.25 x 6) m²}

_{= 291 m².}

_{Total surface area =2(lb+ bh + lh)}

_{=2(24 x 0.25+0.25x 6 +24 x 6) m²}

_{= 2(6+1.5+144) m²}

_{= (2 x151.5) m²=303 m².}Question 1(iii)

_{Find the volume, the lateral surface area and the total surface area cuboid whose dimensions are:}

_{Length =15m, breadth =6 m and height =5 dm}Solution 1(iii)

_{Length = 15 m, breadth = 6m and height = 5 dm = 0.5 m}

_{Volume of a cuboid = l x b x h}

_{= (15 x 6 x 0.5) m³=45 m³.}

_{Lateral surface area = 2(l + b) x h}

_{= [2(15 + 6) x 0.5] m²}

_{= (2 x 21×0.5) m²=21 m²}

_{Total surface area =2(lb+ bh + lh)}

_{= 2(15 x 6 +6 x 0.5+ 15 x 0.5) m²}

_{= 2(90+3+7.5) m²}

_{= (2 x 100.5) m²}

_{=201 m²}Question 1(ii)

_{Find the volume, the lateral surface area and the total surface area cuboid whose dimensions are:}

_{Length =26 m, breadth =14m and height =6.5m}Solution 1(ii)

Length 26 m, breadth =14 m and height =6.5 m

Volume of a cuboid= l x b x h

= (26 x 14 x 6.5) m³

= 2366 m³

Lateral surface area of a cuboid =2 (l + b) x h

= [2(26+14) x 6.5] m²

= (2 x 40 x 6.5) m²

= 520 m²

_{Total surface area= 2(lb+ bh + lh)}

_{= 2(26 x 14+14 x6.5 +26 x6.5)}

_{= 2 (364+91+169) m²}

_{= (2 x 624) m2= 1248 m².}Question 1(i)

_{Find the volume, the lateral surface area and the total surface area cuboid whose dimensions are:}

_{Length=12 cm,breadth=8 cm and height =4.5 cm}Solution 1(i)

length =12cm, breadth = 8 cm and height = 4.5 cm

Volume of cuboid = l x b x h

= (12 x 8 x 4.5) cm³= 432 cm³

Lateral surface area of a cuboid = 2(l + b) x h

= [2(12 + 8) x 4.5] cm²

= (2 x 20 x 4.5) cm² = 180 cm²

Total surface area cuboid = 2(lb +b h+ l h)

= 2(12 x 8 + 8 x 4.5 + 12 x 4.5) cm²

= 2(96 +36 +54) cm²

= (2 x186) cm²

= 372 cm²Question 2

A matchbox measures 4 cm × 2.5 cm × 1.5 cm. What is the volume of a packet containing 12 such matchboxes?Solution 2

For a matchbox,

Length = 4 cm

Breadth = 2.5 cm

Height = 1.5 cm

Volume of one matchbox = Volume of cuboid

= Length × Breadth × Height

= (4 × 2.5 × 1.5) cm³

= 15 cm³

Hence, volume of 12 such matchboxes = 12 × 15 = 180 cm³Question 3

A cuboidal water tank is 6 m long, 5 m wide and 4.5 m deep. How many litres of water can it hold? (Given, 1 m³ = 1000 litres.)Solution 3

For a cuboidal water tank,

Length = 6 m

Breadth = 5 m

Height = 4.5 m

Now,

Volume of a cuboidal water tank = Length × Breadth × Height

= (6 × 5 × 4.5) m³

= 135 m³

= 135 × 1000 litres

= 135000 litres

Thus, a tank can hold 135000 litres of water. Question 4

The capacity of a cuboidal tank is 50000 litres of water. Find the breadth of the tank if its length and depth are respectively 10 m and 2.5 m. (Given, 1000 litres = 1 m³.)Solution 4

For a cuboidal water tank,

Length = 10 m

Breadth = 2.5 m

Volume = 50000 litres = 50 m³

Now,

Volume of a cuboidal tank = Length × Breadth × Height

⇒ 50 = 10 × 2.5 × Height

⇒ Height = 2 m = Depth

Thus, the depth of a tank is 2 m. Question 5

A godown measures 40 m × 25 m × 15 m. Find the maximum number of wooden crates, each measuring 1.5 m × 1.25 m × 0.5 m, that can be stored in godown.Solution 5

For a godown,

Length = 40 m

Breadth = 25 m

Height = 15 m

Volume of a godown = Length × Breadth × Height

= (40 × 25 × 15) m³

For each wooden crate,

Length = 1.5 m

Breadth = 1.25 m

Height = 0.5 m

Volume of each wooden crate = Length × Breadth × Height

= (1.5 × 1.25 × 0.5) m³

Question 6

_{How many planks of dimensions (5mx25cmX10cm) can be stored in a pit which is 20 m long , 6 m wide and 80 cm deep ?}Solution 6

Question 7

_{How many bricks will be required to construct a wall 8 m long , 6 m high and 22.5 cm thick if each brick measures (25cm x11.25cm x 6cm)?}Solution 7

Question 8

_{Find the capacity of a closed rectangular cistern whose length is 8 m, breadth 6 m and depth 2.5 m. Also, find the area of the iron sheet required to make the cistern.}Solution 8

_{Length of Cistern = 8 m}

_{Breadth of Cistern = 6 m}

_{And Height (depth) of Cistern =2.5 m}

 _{Capacity of the Cistern = Volume of cistern}

 _{Volume of Cistern = (l x b x h)}

_{= (8 x 6 x2.5) m³}

_{=120 m³}

_{Area of the iron sheet required = Total surface area of the cistem.}

 _{Total surface area = 2(lb +bh +lh)}

_{= 2(8 x 6 + 6×2.5+ 2.5×8) m²}

_{= 2(48 + 15 + 20) m²}

_{= (2 x 83) m²=166 m²}Question 9

The dimensions of a room are (9 m × 8 m × 6.5 m). It has one door of dimensions (2 m × 1.5 m) and two windows, each of dimensions (1.5 m × 1 m). Find the cost of whitewashing the walls at Rs.25 per square metre.Solution 9

Area of four walls of the room = 2(length + breadth) × Height

= [2(9 + 8) × 6.5] m²

= (34 × 6.5) m²

= 221 m²

Area of one door = Length × Breadth = (2 × 1.5) m² = 3 m²

Area of two windows = 2 × (Length × Breadth)

= [2 × (1.5 × 1)] m²

= (2 × 1.5) m²

= 3 m²

Area to be whitewashed

= Area of four walls of the room – Area of one door – Area of two windows

= (221 – 3 – 3) m²

= 215 m²

Cost of whitewashing = Rs. 25 per square metre

⇒ Cost of whitewashing 215 m² = Rs. (25 × 215) = Rs. 5375Question 10

_{A wall 15 m long , 30 cm wide and 4 m high is made of bricks, each measuring (22cm x12.5cm x7.5cm) if  of the total volume of the wall consists of mortar , how many bricks are there in the wall ?}Solution 10

_LQuestion 11

How many cubic centimetres of iron are there in an open box whose external dimensions are 36 cm, 25 cm, 16.5 cm, the iron being 1.5 cm thick throughout? If 1 cm³ of iron weighs 15 g, find the weight of the empty box in kilograms.Solution 11

External length of the box = 36 cm

External breadth of the box = 25 cm

External height of the box = 16.5 cm

∴ External volume of the box = (36 × 25 × 16.5) cm³ = 14850 cm³

Internal length of the box = [36 – (1.5 × 2)] cm = 33 cm

Internal breadth of the box = [25 – (1.5 × 2)] cm = 22 cm

Internal height of the box = (16.5 – 1.5) cm = 15 cm

∴ Internal volume of the box = (33 × 22 × 15) cm³ = 10890 cm³

Thus, volume of iron used in the box

= External volume of the box – Internal volume of the box

= (14850 – 10890) cm³

= 3960 cm³

Question 12

A box made of sheet metal costs Rs.6480 at Rs.120 per square metre. If the box is 5 m long and 3 m wide, find its height.Solution 12

Question 13

_{The volume of a cuboid is 1536m³. Its length is 16m, and its breadth and height are in the ratio 3:2. Find the breadth and height of the cuboid.}Solution 13

Question 14

_{How many persons can be accommodated in a dining hall of dimensions (20m x16mx4.5m), assuring that each person’s requires 5 cubic metres of air?}Solution 14

Question 15

_{A classroom is 10m long, 6.4 m wide and 5m high. If each student be given 1.6 m² of the floor area, how many students can be accommodated in the room? How many cubic metres of air would each student get?}Solution 15

Question 16

_{The surface of the area of a cuboid is 758 cm². Its length and breadth are 14 cm and 11cm respectively. Find its height.}Solution 16

Question 17In  shower, 5 cm of rain falls. Find the volume of water that falls on 2 hectares of ground.Solution 17

Question 18

_{Find the volume, the lateral surface area, the total surface area and the diagonal of cube, each of whose edges measures 9m. [Take ]}Solution 18

Question 19

_{The total surface area of a cube is 1176 cm². Find its volume.}Solution 19

Question 20

_{The lateral surface area of a cube is 900 cm². Find its volume.}Solution 20

Question 21

_{The volume of a cube is 512 cm³. Find its surface area.}Solution 21

Question 22

_{Three cubes of metal with edges 3cm, 4 cm and 5 cm respectively are melted to form a single cube. Find the lateral surface area of the new cube formed.}
Solution 22

Question 23

_{Find the length of the longest pole that can be put in a room of dimensions (10mx 10m x5m).}Solution 23

Question 24

The sum of length, breadth and depth of a cuboid is 19 cm and length of its diagonal is 11 cm. Find the surface area of the cuboid.Solution 24

Question 25

Each edge of a cube is increased by 50%. Find the percentage increase in the surface area of the cube.Solution 25

Let the edge of the cube = ‘a’ cm

Then, surface area of cube = 6a² cm²

Question 26

If V is the volume of a cuboid of dimensions a, b, c and S is its surface area then prove that Solution 26

Question 27

Water in a canal, 30 dm wide and 12 dm deep, is flowing with a velocity of 20 km per hour. How much area will it irrigate, if 9 cm of standing water is desired?Solution 27

Question 28

A solid metallic cuboid of dimensions (9 m × 8 m × 2 m) is melted and recast into solid cubes of edge 2 m. Find the number of cubes so formed.Solution 28

Volume of a cuboid = (9 × 8 × 2) m³ = 144 m³

Volume of each cube of edge 2 m = (2 m)³ = 8 m³

Exercise Ex. 15B

Question 1

_{The diameter of a cylinder is 28 cm and its height is 40 cm. find the curved surface area, total surface area and the volume of the cylinder.}Solution 1

Question 2

A patient in a hospital is given soup daily in a cylindrical bowl of diameter 7 cm. If the bowl is filled with soup to a height of 4 cm, how much soup the hospital has to prepare daily to serve 250 patients?Solution 2

Radius (r) of cylindrical bowl =

Height (h) up to which the bowl is filled with soup = 4 cm

Volume of soup in 1 bowl = pr²h  = 154 cm³

Hence, volume of soup in 250 bowls = (250 × 154) cm³ = 38500 cm³ = 38.5 litres

Thus, the hospital will have to prepare 38.5 litres of soup daily to serve 250 patients.  Question 3

The pillars of a temple are cylindrically shaped. Each pillar has a circular base of radius 20 cm and height 10 m. How much concrete mixture would be required to build 14 such pillars?Solution 3

Radius (r) of pillar = 20 cm =  m

Height (h) of pillar = 10 m

Question 4

A soft drink is available in two packs: (i) a tin can with a rectangular base of length 5 cm, breadth 4 cm and height 15 cm, and (ii) a plastic cylinder with circular base of diameter 7 cm and height 10 cm. Which container has greater capacity and by how much?Solution 4

For a tin can of rectangular base,

Length = 5 cm

Breadth = 4 cm

Height = 15 cm

∴ Volume of a tin can = Length × Breadth × Height

= (5 × 4 × 15) cm³

= 300 cm³

For a cylinder with circular base,

Diameter = 7 ⇒ Radius = r = cm

Height = h = 10 cm

⇒ Volume of plastic cylinder is greater than volume of a tin can.

Difference in volume = (385 – 300) = 85 cm³

Thus, a plastic cylinder has more capacity that a tin can by 85 cm³.Question 5

There are 20 cylindrical pillars in a building, each having a diameter of 50 cm and height 4 m. Find the cost of cleaning them at Rs.14 per m².Solution 5

Radius (r) of 1 pillar = 

Height (h) of 1 pillar = 4 m

Question 6

The curved surface area of a right circular cylinder is 4.4 m². If the radius of its base is 0.7 m, find its (i) height and (ii) volume.Solution 6

Curved surface area of a cylinder = 4.4 m²

Radius (r) of a cylinder = 0.7 m

Question 7

The lateral surface area of a cylinder is 94.2 cm² and its height is 5 cm. Find (i) the radius of its base and (ii) its volume. (Take π = 3.14.)Solution 7

Lateral surface area of a cylinder = 94.2 cm²

Height (h) of a cylinder = 5 cm

Question 8

The capacity of a closed cylindrical vessel of height 1 m is 15.4 litres. Find the area of the metal sheet needed to make it.Solution 8

Volume of a cylinder = 15.4 litres = 15400 cm³

Height (h) of a cylinder = 1 m = 100 cm

Question 9

The inner diameter of a cylindrical wooden pipe is 24 cm and its outer diameter is 28 cm. The length of the pipe is 35 cm. Find the mass of the pipe, if 1 cm³ of wood has a mass of 0.6 g.Solution 9

Internal diameter of a cylinder = 24 cm

⇒ Internal radius of a cylinder, r = 12 cm

External diameter of a cylinder = 28 cm

⇒ External radius of a cylinder, R = 14 cm

Length of the pipe, i.e height, h = 35 cm

Question 10

In a water heating system, there is a cylindrical pipe of length 28 m and diameter 5 cm. Find the total radiating surface in the system.Solution 10

Diameter of a cylindrical pipe = 5 cm

⇒ Radius (r) of a cylindrical pipe = 2.5 cm

Height (h) of a cylindrical pipe = 28 m = 2800 cm

Question 11

_{Find the weight of a solid cylinder of radius10.5 cm and height 60 cm if the material of the cylinder weights 5 g per cm²}Solution 11

Question 12

_{The curved surface area of a cylinder is 1210 cm² and its diameter is 20 cm. find its height and volume.}Solution 12

Question 13

_{The curved surface area of a cylinder is 4400 cm}² _{and the circumferences of its base are 110 cm. Find the height and the volume of the cylinder.}Solution 13

Question 14

_{The radius of the base and the height of a cylinder are in the ratio 2:3. If its volume is 1617 cm³, find the total surface area of the cylinder}Solution 14

Question 15

_{The total surface area of the cylinder is 462 cm}²_{. And its curved surface area is one third of its total surface area. Find the volume of the cylinder.}Solution 15

Question 16

_{The total surface area of the solid cylinder is 231 cm² and its curved surface area is  of the total surface area. Find the volume of the cylinder.}Solution 16

Question 17

_{The ratio between the curved surface area and the total surface area of a right circular cylinder is 1:2. Find the volume of the cylinder if its total surface area is 616 cm².}Solution 17

Question 18

_{A cylindrical bucket , 28 cm in diameter and 72 cm high , is full of water .The water is emptied into a rectangular tank, 66 cm long and 28 cm wide. Find the height of the water level in the tank}Solution 18

Question 19

_{The barrel of a fountain pen, cylindrical in shape, is 7 cm long and 5 mm in diameter. A full barrel of ink in the pen will be used up on writing 330 words on an average. How many words would use up a bottle of ink containing one fifth of a liter?}Solution 19

Question 20

_{1 cm³ of gold is drawn into a wire 0.1 mm is diameter. Find the length of a wire.}Solution 20

Question 21

_{Ifs 1 cm³ of cast iron weighs 21 g, find the weight of a cast iron pipe of length 1 m with a bore of 3 cm in which the thickness of the metal is 1 cm.}Solution 21

Question 22

_{A cylindrical tube, open at both ends, is made of metal. The internal diameter of the tube is 10.4 cm and its length is 25 cm. The thickness of the metal is 8 mm everywhere. Calculate the volume of the metal.}Solution 22

Question 23

It is required to make a closed cylindrical tank of height 1 m and base diameter 140 cm from a metal sheet. How many square metres of the sheet are required for the same?Solution 23

Diameter of a cylinder = 140 cm

⇒ Radius, r = 70 cm

Height (h) of a cylinder = 1 m = 100 cm

Question 24

A juiceseller has a large cylindrical vessel of base radius 15 cm filled up to a height of 32 cm with orange juice. The juice is filled in small cylindrical glasses of radius 3 cm up to a height of 8 cm, and sold for Rs.15 each, How much money does he receive by selling the juice completely?Solution 24

Radius (r) of cylindrical vessel = 15 cm

Height (h) of cylindrical vessel = 32 m

Radius of small cylindrical glass = 3 cm

Height of a small cylindrical glass = 8 cm

Question 25

A well with inside diameter 10 m is dug 8.4 m deep. Earth taken out of it is spread all around it to a width of 7.5 m to form an embankment. Find the height of the embankment.Solution 25

Radius of the well = 5 m

Depth of the well = 8.4 m

Width of the embankment = 7.5 m

External radius of the embankment, R = (5 + 7.5) m = 12.5 m

Internal radius of the embankment, r = 5 m

Area of the embankment = π (R² – r²)

Volume of the embankment = Volume of the earth dug out = 660 m²

Question 26

How many litres of water flows out of a pipe having an area of cross section of 5 cm² in 1 minute, if the speed of water in the pipe is 30 cm/sec?Solution 26

Speed of water = 30 cm/sec

∴ Volume of water that flows out of the pipe in one second

= Area of cross-section × Length of water flown in one second

= (5 × 30) cm³

= 150 cm³

Hence, volume of water that flows out of the pipe in 1 minute

= (150 × 60) cm³

= 9000 cm³

= 9 litresQuestion 27

A cylindrical water tank of diameter 1.4 m and height 2.1 m is being fed by a pipe of diameter 3.5 cm through which water flows at the rate of 2 m per second. In how much time will the tank be filled?Solution 27

Suppose the tank is filled in x minutes. Then,

Volume of the water that flows out through the pipe in x minutes

= Volume of the tank

Hence, the tank will be filled in 28 minutes.Question 28

A cylindrical container with diameter of base 56 cm contains sufficient water to submerge a rectangular solid of iron with dimensions (32 cm × 22 cm × 14 cm). Find the rise in the level of water when the solid is completely submerged.Solution 28

Let the rise in the level of water = h cm

Then,

Volume of the cylinder of height h and base radius 28 cm

= Volume of rectangular iron solid

Thus, the rise in the level of water is 4 cm.Question 29

Find the cost of sinking a tube-well 280 m deep, having a diameter 3 m at the rate of Rs.15 per cubic metre. Find also the cost of cementing its inner curved surface at Rs.10 per square metre.Solution 29

Radius, r = 1.5 m

Height, h = 280 m

Question 30

Find the length of 13.2 kg of copper wire of diameter 4 mm, when 1 cubic centimetre of copper weights 8.4 g.Solution 30

Let the length of the wire = ‘h’ metres

Then,

Volume of the wire × 8.4 g = (13.2 × 1000) g

Thus, the length of the wire is 125 m.Question 31

It costs Rs.3300 to paint the inner curved surface of a cylindrical vessel 10 m deep at the rate of Rs.30 per m². Find the

(i) inner curved surface area of the vessel,

(ii) inner radius of the base, and

(iii) capacity of the vessel.Solution 31

Question 32

The difference between inside and outside surfaces of a cylindrical tube 14 cm long, is 88 cm². If the volume of the tube is 176 cm³, find the inner and outer radii of the tube.Solution 32

Let R cm and r cm be the outer and inner radii of the cylindrical tube.

We have, length of tube = h = 14 cm

Now,

Outside surface area – Inner surface area = 88 cm²

⇒ 2πRh – 2πrh = 88

⇒ 2π(R – r)h = 88

It is given that the volume of the tube = 176 cm³

⇒ External volume – Internal volume = 176 cm³

⇒ πR²h – πr²h = 176

⇒ π (R² – r²)h = 176

Adding (i) and (ii), we get

2R = 5

⇒ R = 2.5 cm

⇒ 2.5 – r = 1

⇒ r = 1.5 cm

Thus, the inner and outer radii of the tube are 1.5 cm and 2.5 cm respectively.Question 33

A rectangular sheet of paper 30 cm × 18 cm can be transformed into the curved surface of a right circular cylinder in two ways namely, either by rolling the paper along its length or by rolling it along its breadth. Find the ratio of the volumes of the two cylinders, thus formed.Solution 33

When the sheet is folded along its length, it forms a cylinder of height, h₁ = 18 cm and perimeter of base equal to 30 cm.

Let r₁ be the radius of the base and V₁ be is volume.

Then,

Again, when the sheet is folded along its breadth, it forms a cylinder of height, h₂ = 30 cm and perimeter of base equal to 18 cm.

Let r₂ be the radius of the base and V₂ be is volume.

Then,

Exercise Ex. 15D

Question 1(iii)

_{Find the volume and the surface area of a sphere whose radius is:}

_{5 m}Solution 1(iii)

Question 1(ii)

_{Find the volume and the surface area of a sphere whose radius is}

_{4.2 cm}Solution 1(ii)

Question 1(i)

_{Find the volume and the surface area of a sphere whose radius is}

_{3.5 cm}Solution 1(i)

Question 2

_{The volume of a sphere is 38808 cm³. Find the radius and hence its surface area.}Solution 2

Question 3

_{Find the surface area of a sphere whose volume is 606.375 m³}Solution 3

Question 4

Find the volume of a sphere whose surface area is 154 cm².Solution 4

Surface area of sphere = 154 cm²

⇒ 4πr² = 154

Question 5

_{The surface area of a sphere is (576) cm². Find its volume.}Solution 5

Question 6

_{How many leads shots, each 3 mm in diameter, can be made from cuboid with dimensions  (12cm x 11cm x 9cm)?}Solution 6

Question 7

_{How many lead balls, each of radius 1 cm, can be made from a sphere of radius 8 cm?}Solution 7

Question 8

_{A solid sphere of radius 3 cm is melted and then cast into smaller spherical balls, each of diameters 0.6 cm. find the number of small balls thus obtained.}Solution 8

Question 9

_{A metallic sphere of radius 10.5 cm is melted an then recast into smaller cones , each of radius 3.5 cm and height 3 cm. How many cones are obtained?}Solution 9

Question 10

_{How many spheres 12 cm in diameter can be made from a metallic cylinder of diameter 8 cm and}

_{height 90 cm ?}Solution 10

Question 11

_{The diameter of sphere is 6 cm. It is melted and drawn into wire of diameter 2 mm. Find the length of the wire.}Solution 11

Question 12

_{The diameter of the copper sphere is 18cm. It is melted and drawn into a long wire of uniform cross section. If the length of the wire is 108 m, find its diameter.}Solution 12

Question 13

_{A sphere of a diameter 15.6 cm is melted and cast into a right circular cone of height 31.2 cm. find the diameter of the base of the cone.}Solution 13

Question 14

_{A spherical cannonball 28 cm in diameter is melted and recast into a right circular cone mould,  whose base is 35 cm in diameter. Find the height of the cone.}Solution 14

Question 15

_{A spherical ball of radius 3 cm is melted and recast into three spherical balls. The radii of two of these balls are 1.5 cm and 2cm. Find the radius of the third ball.}Solution 15

Question 16

The radii of two spheres are in the ratio 1:2. Find the ratio of their surface areas.Solution 16

Question 17

The surface areas of two spheres are in the ratio 1:4. Find the ratio of their volumes.Solution 17

Question 18

A cylindrical tub of a radius 12 cm contains water to a depth of 20 cm. A spherical iron ball is dropped into the tub and thus the level of water is raised by 6.75cm.what is the radius of the ball?Solution 18

Question 19

A cylindrical bucket with base radius 15 cm is filled with water to up height of 20 cm. a heavy iron spherical ball of radius 9 cm is dropped into the bucket to submerge completely in the water . Find the increase in the level of waterSolution 19

Question 20

_{The outer diameter of a spherical shell is 12 cm and its inner diameter is 8 cm. Find the volume of metal contained in the shell. Also, find its outer surface area.}Solution 20

Question 21

A hollow spherical shell is made of a metal of density 4.5 g per cm³. If it’s internal and external radii are 8 cm and 9cm respectively, find the weight of the shell.Solution 21

Question 22

A hemisphere of lead of radius 9 cm is cast into a right circular cone of height 72 cm . Find the radius of the base of the cone.Solution 22

Question 23

A hemisphere bowl of internal radius 9 cm contains a liquid. This liquid is to be filled into cylindrical shaped bottles of diameter 3 cm and height 4 cm. How many bottles are required to empty the bowl?Solution 23

Question 24

A hemispherical bowl is made of steel 0.5 cm thick. The inside radius of the bowl is 4 cm. Find the volume of steel used in making the bowl.Solution 24

Question 25

A hemispherical bowl is made of steel, 0.25 cm thick. The inner radius of the bowl is 5 cm. Find the outer curved surface area of the bowl.Solution 25

Inner radius = 5 cm

⇒ Outer radius = 5 + 0.25 = 5.25 cm

Question 26

A hemispherical bowl made of brass has inner diameter 10.5 cm. Find the cost of tin-plating it on the inside at the rate of Rs.32 per 100 cm².Solution 26

Inner diameter of the hemispherical bowl = 10.5 cm

Question 27

The diameter of the moon is approximately one fourth of the diameter of the earth. What fraction of the volume of the earth is the volume of the moon?Solution 27

Let the diameter of earth = d

⇒ Radius of the earth = 

Then, diameter of moon = .

⇒ Radius of moon = 

Volume of moon 

Volume of earth 

Thus, the volume of moon is of volume of earth.Question 28

Volume and surface area of a solid hemisphere are numerically equal. What is the diameter of the hemisphere?Solution 28

Volume of a solid hemisphere = Surface area of a solid hemisphere

Exercise MCQ

Question 1

The range of the data

12,25,15,18,17,20,22,6,16,11,8,19,10,30,20,32 is

1. 10
2. 15
3. 18
4. 26

Solution 1

Correct option: (d)

Range = maximum value – minimum value

 = 32 – 6

 = 26Question 2

The class mark of the class 100-120 is

1. 100
2. 110
3. 115
4. 120

Solution 2

Question 3

In the class intervals 10-20, 20-30, the number 20 is included in

1. 10-20
2. 20-30
3. In each of 10-20 and 20-30
4. In none of 10-20 and 20-30

Solution 3

Question 4

The class marks of a frequency distribution are 15, 20, 25, 30………. The class corresponding to the mark 20 is

1. 12.5-17.5
2. 17.5-22.5
3. 18.5-21.5
4. 19.5-20.5

Solution 4

Question 5

In a frequency distribution, the mid-value of a class is 10 and width of each class is 6. The lower limit of the class is

1. 6
2. 7
3. 8
4. 12

Solution 5

Question 6

The mid – value of a class interval is 42 and the class size is 10. The lower and upper limits are

1. 37-47
2. 37.5-47.5
3. 36.5-47.5
4. 36.5-46.5

Solution 6

Question 7

Let m be in the midpoint and u be the upper class limit of a class in a continuous frequency distribution. The lower class limit of the class is

1. 2m – u
2. 2m + u
3. m – u
4. m + u

Solution 7

Question 8

The width of each of the five continuous classes in a frequency distribution is 5 and the lower class limit of the class is

1. 45
2. 25
3. 35
4. 40

Solution 8

Question 9

Let L be the lower class boundary of a class in a frequency distribution and m be the midpoint of the class. Which one of the following is the upper class boundary of the class?

Solution 9

Exercise Ex. 16

Question 1

Define statistics as a subject.Solution 1

Statistics is a branch of science which deals with the collection, presentation, analysis and interpretation of numerical data.Question 2

Define some fundamental characteristics of statistics.Solution 2

Fundamental characteristics of statistics :

(i) It deals only with the numerical data.

(ii) Qualitative characteristic such as illiteracy, intelligence, poverty etc cannot be measured numerically

(iii) Statistical inferences are not exact.Question 3

What are the primary data and secondary data? Which of the two is more reliable and why?Solution 3

Primary data: Primary data is the data collected by the investigator himself with a definite plan in his mind. These data are very accurate and reliable as these being collected by the investigator himself.

Secondary Data: Secondary data is the data collected by a person other than the investigator.

Secondary Data is not very reliable as these are collected by others with purpose other than the investigator and may not be fully relevant to the investigation.
Question 4

Explain the meaning of each of the following terms.

(i)Variate(ii) Class interval(iii)Class size

(iv)Class mark (v)Class limit(vi)True class limits

(vii)Frequency of a class(viii) Cumulative frequency of a classSolution 4

(i)Variate : Any character which can assume many different values is called a variate.

(ii)Class Interval :Each group or class in which data is condensed is calleda class interval.

(iii)Class-Size :The difference between the true upper limitand the true lower limit of a class is called class size.

(iv)Classmark : The average of upper and lower limit of a class interval is called its class mark.

i.e Class mark=

(v) Class limit: Class limits are the two figures by which a class is bounded . The figure on the left side of a class is called lower lower limit and on the right side is called itsupper limit.

(vi)True class limits: In the case of exclusive form of frequency distribution, the upper class limits and lower classlimits are the true upper limits and thetrue lower limits. But in the case of inclusive form of frequency distribution , the true lower limit of a class is obtained by subtracting 0.5 from the lower limit of the class. And the true upper limit of the class is obtained by adding 0.5 to the upper limit.

(vii)Frequency of a class : The number of observations falling in aclass determines its frequency.

(viii)Cumulative frequency of a class: The sum of all frequenciesup to and including that class is called , the cumulative frequency of that class.Question 5

The blood groups of 30 students of a class are recorded as under:

A, B, O, O, AB, O, A, O, A, B, O, B, A, O, O, A, AB, O, A, A, O, O, AB, B, A, O, B, A, B, O.

(i) Represent this data in the form of a frequency distribution table.

(ii) Find out which is the most common and which is the rarest blood group among these students.Solution 5

(i) Frequency Distribution Table:

(ii) The most common blood group is ‘O’ and the rarest blood group is ‘AB’.Question 6

Three coins are tossed 30 times. Each time the number of heads occurring was noted down as follows:

0, 1, 2, 2, 1, 2, 3, 1, 3, 0, 1, 3, 1, 1, 2, 2, 0, 1, 2, 1, 0, 3, 0, 2, 1, 1, 3, 2, 0, 2.

Prepare a frequency distribution table.Solution 6

Frequency Distribution Table:

Question 7

Following data gives the number of children in 40 families :

1, 2, 6, 5, 1, 5, 1, 3, 2, 6, 2, 3, 4, 2, 0, 4, 4, 3, 2, 2, 0, 0, 1,2, 2, 4, 3, 2, 1, 0, 5, 1, 2, 4, 3, 4, 1, 6, 2, 2.

Represent in the form of a frequency distribution, taking classes 0-2, 2-4, etc.Solution 7nimum observation is 0 and maximum observation is 6. The classes of equal size covering the given data are : (0-2), (2-4), (4-6) and (6-8).

Thus , the frequency distribution may be given as under:

Question 8

Thirty children were asked about the number of hours they watched TV programmes in the previous week. The results were found as under:

8, 4, 8, 5, 1, 6, 2, 5, 3, 12, 3, 10, 4, 12, 2, 8, 15, 1, 6, 17, 5, 8, 2, 3, 9, 6, 7, 8, 14, 12.

(i) Make a grouped frequency distribution table for this data, taking class width 5 and one of the class interval as 5 – 10.

(ii) How many children watched television for 15 or more hours a week?Solution 8

(i) Grouped Frequency Distribution Table:

(ii) 2 children watch television for 15 or more hours a week. Question 9

The marks obtained by 40 students of a class in an examination are given below .

3, 20, 13, 1, 21, 13, 3, 23, 16, 13, 18, 12, 5, 12, 5, 24, 9, 2, 7, 18, 20, 3, 10, 12, 7, 18, 2, 5, 7, 10, 16, 8, 16, 17, 8, 23, 24, 6, 23, 15.

Present the data in the form of a frequency distribution using equal class size, one such class being 10-15(15 not included).Solution 9

Minimum observation is 1 and minimum observation is 24. The classes of equal size converging the given data are : (0-5), (5-10), (10-15), (15-20), (20-25)

Thus, the frequency distribution may be given as under :Question 10

Construct a frequency table for the following ages (in years) of 30 students using equal class intervals, one of them being 9-12, where 12 is not included.

18, 12, 7, 6, 11, 15, 21, 9, 8, 13, 15, 17, 22, 19, 14, 21, 23, 8, 12, 17, 15, 6, 18, 23, 22, 16, 9, 21, 11, 16.Solution 10

Grouped Frequency Distribution Table:

Question 11

Construct a frequency table with equal class intervals from the following data on the monthly wages (in rupees) of 28 labourers working in a factory, taking one of the class intervals as 210-230 (230 not included).

220, 268, 258, 242, 210, 268, 272, 242, 311, 290, 300, 320, 319, 304, 302, 318, 306, 292, 254, 278, 210, 240, 280, 316, 306, 215, 256, 236.Solution 11

Minimum observation is 210 and maximum observation =320

So the range is (320-210)=110

The classes of equal size covering the given data are :

(210-230), (230-250), (250-270) , (270-290), (290-310), (310-330)

Thus the frequency distribution may be given as under :

Question 12

The weights (in grams ) of 40 oranges picked at random from a basket are as follow :

40, 50, 60, 65, 45, 55, 30, 90, 75, 85, 70,85, 75, 80, 100, 110, 70, 55, 30, 35, 45, 70, 80, 85, 95, 70, 60, 70, 75, 40, 100, 65, 60, 40, 100, 75, 110, 30, 45, 84.

Construct a frequency table as well as a cumulative frequency table.Solution 12

Minimum observation is 30 and maximum observation is 110

So, range is 100-30=80

The classes of equal size covering the given data are :

(30-40) ,(40-50) , (50-60) ,(60-70) , (70-80), (80-90),(90-100),(100-110), (110-120)

Thus , the frequency and cumulative frequency table may be given as under :

Question 13

The heights (in cm) of 30 students of a class are given below:

161, 155, 159, 153, 150, 158, 154, 158, 160, 148, 149, 162, 163, 159, 148, 153, 157, 151, 154, 157, 153, 156, 152, 156, 160, 152, 147, 155, 155, 157.

Prepare a frequency table as well as a cumulative frequency table with 160-165 (165 not included) as one of the class intervals.Solution 13

Grouped Frequency Distribution Table and Cumulative Frequency Table:

Question 14

Following are the ages (in years ) of 360 patients , getting medical treatment in a hospital:

Age (in years)	10-20	20-30	30-40	40-50	50-60	60-70
Number of patients	90	50	60	80	50	30

Construct the cumulative frequency table for the above data.Solution 14

Age (in years)(age)	No of patients (Frequency)	Cumulative Frequency
10-2020-3030-4040-5050-6060-70	905060805030	90140200280330360
Total	360

Question 15

Present the following as an ordinary grouped frequency table :

Marks(below)	10	20	30	40	50	60
Number of students	5	12	32	40	45	48

Solution 15

Marks (below)	No of students(Cumulative Frequency.)	Class Intervals	Frequency
102030405060	51232404548	0-1010-2020-3030-4040-5050-60	512 – 5 = 732 – 12 = 2040 – 32 = 845 – 40 = 548 – 45 = 3
		Total	48

Question 16

Given below is a cumulative frequency table ;

Marks	Number of students
Below 10	17
Below 20	22
Below 30	29
Below 40	37
Below 50	50
Below 60	60

Extract a frequency table from the above .Solution 16

Marks (below)	No of students(Cumulative Frequency)	Class Intervals	Frequency
102030405060	172229375060	0-1010-2020-3030-4040-5050-60	1722 – 17 = 529 – 22 = 737 – 29 = 850 – 37 = 1360 – 50 = 10
		Total	60

Question 17

Make a frequency table from the following ;

Marks obtained	Number of students
More than 60	0
More than50	16
More than40	40
More than30	75
More than20	87
More than10	92
More than0	100

Solution 17

Marks (below)	No of student s(C.F.)	Class Intervals	Frequency
More than 60More than 50More than 40More than 30More than 20More than 10More than 0	01640758792100	More than 6050-6040-5030-4020-3010-200-10	016-0=1640-16=2475-40=3587-75=1292-87=5100-92=8
		Total	100

Question 18

The marks obtained by 17 students in a mathematics test (out of 100) are given below:

90, 79, 76, 82, 65, 96, 100, 91, 82, 100, 49, 46, 64, 48, 72, 66, 68.

Find the range of the above data.Solution 18

Arranging data in ascending order, we have

46, 48, 49, 64, 65, 66, 68, 72, 76, 79, 82, 82, 90, 91, 96, 100, 100

Minimum marks = 46

Maximum Marks = 100

∴ Range of the above data = Maximum Marks – Minimum Marks

 = 100 – 46

= 54Question 19

(i) Find the class mark of the class 90 – 120.

(ii) In a frequency distribution, the mid-value of the class is 10 and width of the class is 6. Find the lower limit of the class.

(iii) The width of each of five continuous classes in a frequency distribution is 5 and lower class limit of the lowest class is 10. What is the upper class limit of the highest class?

(iv) The class marks of a frequency distribution are 15, 20, 25, … Find the class corresponding to the class mark 20.

(v) In the class intervals 10-20, 20-30, find the class in which 20 is included.Solution 19

Question 20

Find the values of a, b, c, d, e, f, g from the following frequency distribution of the heights of 50 students in a class:

Height (in cm)	Frequency	Cumulativefrequency
160-165	15	a
165-170	b	35
170-175	12	c
175-180	d	50
180-185	e	55
185-190	5	f
	g

Solution 20

Height(in cm)	Frequency	Cumulativefrequency
160-165	15	a = 15
165-170	b = 35 – 15 = 20	35
170-175	12	c = 35 + 12 = 47
175-180	d = 50 – 47 = 3	50
180-185	e = 55 – 50 = 5	55
185-190	5	f = 55 + 5 = 60
	g = 60

Exercise Ex. 17A

Question 1

The following table shows the number of students participating in various games in a school.

Game	Cricket	Football	Basketball	Tennis
Number of students	27	36	18	12

Draw a bar graph to represent the above data.

Hint: Along the y-axis, take 1 small square=3 units.Solution 1

Take the various types of games along the x-axis and the number of students along the y-axis.

 Along the y-axis, take 1 small square=3 units.

All the bars should be of same width and same space should be left between the consecutive bars.

Now we shall draw the bar chart, as shown below:

Question 2

On a certain day, the tempreture in a city was recorded as under.

Times	5 a.m	8 a.m	11a.m	3p.m	6p.m
Tempreture (in 0C)	20	24	26	22	18

Illustrate the data by a bar graph.Solution 2

Take the timings along the x-axis and the temperatures along the y-axis.

Along the y-axis, take 1 small square=5 units.

All the bars should be of same width and same space should be left between the consecutive bars.

Now we shall draw the bar chart, as shown below:

Question 3

The approximate velocities of some vehicles are given below:

Name of vehicle	Bicycle	Scooter	Car	Bus	Train
Velocity (in km/hr)	27	45	90	72	63

Draw bar graph to represent the above data.Solution 3

Question 4

The following table shows the favorite sports of 250 students of a school. Represent the data by a bar graph.

Sports	Cricket	Football	Tennis	Badminton	Swimming
No. of Students	75	35	50	25	65

Solution 4

Take the various types of sports along the x-axis and the number of students along the y-axis.

Along the y-axis, take 1 small square=10 units.

All the bars should be of same width and same space should be left between the consecutive bars.

Now we shall draw the bar chart, as shown below:

Question 5

Given below is a table which shows the year wise strength of a school. Represent this data by a bar graph.

Year	2012-13	2013-14	2014-15	2015-16	2016-17
No of students	800	975	1100	1400	1625

Solution 5

Take the academic year along the x-axis and the number of students along the y-axis.

Along the y-axis, take 1 big division =200 units.

All the bars should be of same width and same space should be left between the consecutive bars.

Now we shall draw the bar chart, as shown below:

Question 6

The following table shows the number of scooters sold by a dealer during six consecutive years. Draw a bar graph to represent this data.

Year	2011	2012	2013	2014	2015	2016
Number of scooters sold (in thousand)	16	20	32	36	40	48

Solution 6

Question 7

The air distances of four cities from Delhi (in km) are given below :

City	Kolkata	Mumbai	Chennai	Hyderabad
Distance from Delhi(in km)	1340	1100	1700	1220

Draw a bar graph to represent the above data.Solution 7

Take city along the x-axis and distance from Delhi (in Km) along the y-axis.

Along the y-axis, take 1 big division =200 units.

All the bars should be of same width and same space should be left between the consecutive bars.

Now we shall draw the bar chart, as shown below:

Question 8

The birth rate per thousand in five countries over a period of time shown below:

Country	China	India	Germany	UK	Sweden
Birth ratePer thousand	42	35	14	28	21

Represent the above data by a bar graph.Solution 8

Take the countries along the x-axis and the birth rate (per thousand) along the y-axis.

Along the y-axis, take 1 big division = 5 units.

All the bars should be of same width and same space should be left between the consecutive bars.

Now we shall draw the bar chart, as shown below:

Question 9

The following table shows the life expectancy (average age to which people live) in various countries in a particular year. Represent the data by a bar graph.

Country	Japan	India	Britain	Ethiopia	Cambodia	UK
Life expectancy(in years)	84	68	80	64	62	73

Solution 9

Question 10

Given below are the seats won by different political parties in the polling outcome of a state assembly elections:

Political party	A	B	C	D	E	F
Seats won	65	52	34	28	10	31

Draw a bar graph to represent the polling results.Solution 10

Question 11

Various modes of transport used by 1850 students of a school are given below.

School bus	Private bus	Bicycle	Rickshaw	By foot
640	360	490	210	150

Draw a bar graph to represent the above data.Solution 11

Take themode of transport along the x-axis and the number of students along the y-axis.

Along the y-axis, take 1 big division = 100 units.

All the bars should be of same width and same space should be left between the consecutive bars.

Now we shall draw the bar chart, as shown below:

Question 12

Look at the bar graph given below.

Read it carefully and answer the following questions.

(i) What information does the bar graph give?

(ii) In which subject does the student very good?

(iii) In which subject is he poor?

(iv) What is the average of the marks?Solution 12

(i) The bar graph shows the marks obtained by a student in various subject in an examination.

(ii) The student is very good in mathematics.

(iii) He is poor in Hindi

        (iv)  Average marks =

Exercise Ex. 17B

Question 1

The daily wages of 50 workers in a factory are given below :

Daily wages in rupees	340-380	380-420	420-460	460-500	500-540	540-580
Number ofworkers	16	9	12	2	7	4

Construct a histogram to represent the above frequency distribution.Solution 1

Given frequency distribution is as below :

Daily wages (in Rs)	340-380	380-420	420-460	460-500	500-540	540-580
No. of workers	16	9	12	2	7	4

In the class intervals, if the upper limit of one class is the lower limit of the next class, it is known as the exclusive method of classification.

Clearly, the given frequency distribution is in the exclusive form.

To draw the required histogram , take class intervals , i.e. daily wages (in Rs. ) along x-axis and frequencies i.e.no.of workers alongy-axisand draw rectangles . So , we get the requiredhistogram .

Since the scale on X-axis starts at 340, a kink(break) is indicated near the origin to show that the graph is drawn to scale beginning at 340.

Question 2

The following table shows the average daily earnings of 40 general stores in a market, during a certain week.

Daily earning (in rupees)	700-750	750-800	800-850	850-900	900-950	950-1000
Number ofStores	6	9	2	7	11	5

Draw a histogram to represent the above data.Solution 2

Given frequency distribution is as below :

Daily earnings (in Rs)	700-750	750-800	800-850	850-900	900-950	950-1000
No of stores	6	9	2	7	11	5

In the class intervals, if the upper limit of one class is the lower limit of the next class, it is known as the exclusive method of classification.

Clearly, the given frequency distribution is in the exclusive form.

We take class intervals, i.e. daily earnings (in Rs .) along x-axis and frequencies i.e. number of stores along y-axis. So , we get the required histogram .

Since the scale on X-axis starts at 700, a kink(break) is indicated near the origin to show that the graph is drawn to scale beginning at 700. 

Question 3

the heights of 75 students in a school are given below :

Height(in cm)	130-136	136-142	142-148	148-154	154-160	160-166
Number of students	9	12	18	23	10	3

Draw a histogram to represent the above data.Solution 3

Height(in cm)	130-136	136-142	142-148	148-154	154-160	160-166
No. of students	9	12	18	23	10	3

In the class intervals, if the upper limit of one class is the lower limit of the next class, it is known as the exclusive method of classification.

Clearly, the given frequency distribution is in the exclusive form.

We take class intervals, i.e. height (in cm ) along x-axis and frequencies i.e. number of student s along y-axis . So we get the required histogram.

Since the scale on X-axis starts at 130, a kink(break) is indicated near the origin to show that the graph is drawn to scale beginning at 130.

Question 4

The following table gives the lifetimes of 400 neon lamps:

Lifetime(in hr )	300-400	400-500	500-600	600-700	700-800	800-900	900-1000
Number of lamps	14	56	60	86	74	62	48

(i) Represent the given information with the help of a histogram.

(ii) How many lamps have a lifetime of more than 700 hours?Solution 4

(i) Histogram is as follows:

(ii) Number of lamps having lifetime more than 700 hours = 74 + 62 + 48 = 184Question 5

Draw a histogram for frequency distribution of the following data.

Class -Interval	8-13	13-18	18-23	23-28	28-33	33-38	38-43
Frequency	320	780	160	540	260	100	80

Solution 5

Give frequency distribution is as below :

Class interval	8-13	13-18	18-23	23-28	28-33	33-38	38-43
Frequency	320	780	160	540	260	100	80

In the class intervals, if the upper limit of one class is the lower limit of the next class, it is known as the exclusive method of classification.

Clearly, the given frequency distribution is in the exclusive form.

We take class intervals along x-axis and frequency along y-axis . So , we get the required histogram.

Since the scale on X-axis starts at 8, a kink(break) is indicated near the origin to show that the graph is drawn to scale beginning at 8.

Question 6

Construct a histogram for the following frequency distribution.

Class interval	5-12	13-20	21-28	29-36	37-44	45-52
Frequency	6	15	24	18	4	9

Solution 6

Histogram is the graphical representation of a frequency distribution in the form of rectangles, such that there is no gap between any two successive rectangles.

Clearly the given frequency distribution is in inclusive form, that is there is a gap between the upper limit of a class and the lower limit of the next class.

Therefore, we need to convert the given frequency distribution into exclusive form, as shown below:

Class interval	4.5-12.5	12.5-20.5	20.5-28.5	28.5-36.5	36.5-44.5	44.5-52.5
Frequency	6	15	24	18	4	9

To draw the required histogram , take class intervals, along x-axis and frequencies along y-axis and draw rectangles . So, we get the required histogram .

Since the scale on X-axis starts at 4.5, a kink(break) is indicated near the origin to show that the graph is drawn to scale beginning at 4.5.

Question 7

The following table shows the number of illiterate persons in the age group (10-58 years) in a town:

Age group(in years)	10-16	17-23	24-30	31-37	38-44	45-51	52-58
Number of illiterate persons	175	325	100	150	250	400	525

Draw a histogram to represent the above data.Solution 7

Given frequency distribution is as below :

Age group (in years )	10-16	17-23	24-30	31-37	38-44	45-51	52-58
No. of illiterate persons	175	325	100	150	250	400	525

Histogram is the graphical representation of a frequency distribution in the form of rectangles, such that there is no gap between any two successive rectangles.

Clearly the given frequency distribution is in inclusive form, that is there is a gap between the upper limit of a class and the lower limit of the next class.

Therefore, we need to convert the frequency distribution in exclusive form, as shown below:

Age group(in years)	9.5-16.5	16.5-23.5	23.5-30.5	30.5-37.5	37.5-44.4	44.5-51.5	51.5-58.5
No of illiterate persons	175	325	100	150	250	400	525

To draw the required histogram , take class intervals, that is age group, along x-axis and frequencies, that is number of illiterate persons along y-axis and draw rectangles . So , we get the required histogram.

Since the scale on X-axis starts at 9.5, a kink(break) is indicated near the origin to show that the graph is drawn to scale beginning at 9.5.

Question 8

Draw a histogram to represent the following data.

Class -Interval	10-14	14-20	20-32	32-52	52-80
Frequency	5	6	9	25	21

Solution 8

given frequency distribution is as below :

Class interval	10-14	14-20	20-32	32-52	52-80
Frequency	5	6	9	25	21

In the above table , class intervals are of unequal size, so we calculate the adjusted frequency by using the following formula :

Thus , the adjusted frequency table is

Class intervals	frequency	Adjusted Frequency
10-14 14-20 20-32 32-52 52-80	5 6 9 25 21

Now take class intervals along x-axis and adjusted frequency along y-axis and constant rectangles having their bases as class size and heights as the corresponding adjusted frequencies.

Thus, we obtain the histogram as shown below:

Question 9

100 surnames were randomly picked up from a local telephone directory and frequency distribution of the number of letters in the English alphabet in the surnames was found as follows:

Number of letters	1-4	4-6	6-8	8-12	12-20
Number of surnames	6	30	44	16	4

(i) Draw a histogram to depict the given information.

(ii) Write the class interval in which the maximum number of surnames lie.Solution 9

(i) Minimum class size = 6 – 4 = 2

(ii) Maximum number of surnames lies in the class interval 6 – 8.Question 10

Draw a histogram to represent the following information:

Class interval	5-10	10-15	15-25	25-45	45-75
Frequency	6	12	10	8	18

Solution 10

Minimum class size = 10 – 5 = 5

Question 11

Draw a histogram to represent the following information:

Marks	0-10	10-30	30-45	45-50	50-60
Number of students	8	32	18	10	6

Solution 11

Minimum class size = 50 – 45 = 5

Question 12

In a study of diabetic patients in a village , the following observations were noted.

Age in years	10-20	20-30	30-40	40-50	50-60	60-70
Number of patients	2	5	12	19	9	4

Represent the above data by a frequency polygon.Solution 12

The given frequency distribution is as below:

Age in years	10-20	20-30	30-40	40-50	50-60	60-70
No of patients	2	5	12	19	9	4

In order to draw, frequency polygon, we require class marks.

The class mark of a class interval is:

The frequency distribution table with class marks is given below:

Class- intervals	Class marks	Frequency
0-1010-2020-3030-4040-5050-6060-7070-80	515253545556575	0251219940

In the above table, we have taken imaginary class intervals 0-10 at beginning and 70-80 at the end, each with frequency zero . Now take class marks along x-axis and the corresponding frequencies along y-axis.

Plot points (5,0), (15,2), (25, 5), (35, 12), (45, 19), (55, 9), (65, 4) and (75, 0) and draw line segments.

Question 13

Draw a frequency polygon for the following frequency distribution

Class-interval	1-10	11-20	21-30	31-40	41-50	51-60
Frequency	8	3	6	12	2	7

Solution 13

The given frequency distribution table is as below:

Class intervals	1-10	11-20	21-30	31-40	41-50	51-60
Frequency	8	3	6	12	2	7

This table has inclusive class intervals and so these are to be converted into exclusive class intervals (i.e true class limits).

These are (0.5-10.5), (10.5-20.5), (20.5-30.5), (30.5-40.5),
(40.5-50.5), and (50.5-60.5)

In order to draw a frequency polygon, we need to determine the class marks. Class marks of a class interval =

Take imaginary class interval ( -9.5-0.5) at the beginning and (60.5-70.5) at the end , each with frequency zero. So we have the following table

Class intervals	True class intervals	Class marks	Frequency
(-9)-01-1011-2021-3031-4041-5051-6061-70	(-9.5)-0.50.5-10.510.5-20.520.5-30.530.5-40.540.5-50.550.5-60.560.5-70.5	-4.55.515.525.535.545.555.565.5	083612270

Now, take class marks along x-axis and their corresponding frequencies along y-axis.

Mark the points and join them.

Thus, we obtain a complete frequency polygon as shown below:

Question 14

The ages (in years) of 360 patients treated in a hospital on a particular dayare given below.

Age in years	10-20	20-30	30-40	40-50	50-60	60-70
Number of patients	90	40	60	20	120	30

Draw a histogram and a frequency polygon on the same graph to represent the above data.Solution 14

The given frequency distribution is as under

Age in years	10-20	20-30	30-40	40-50	50-60	60-70
Numbers of patients	90	40	60	20	120	30

Take class intervals i.e age in years along x-axis and number of patients of width equal to the size of the class intervals and height equal to the corresponding frequencies.

Thus we get the required histogram.

In order to draw frequency polygon,we take imaginary intervals 0-10 at the beginning and 70-80 at the end each with frequency zero and join the mid-points of top of the rectangles.

Thus, we obtain a complete frequency polygon, shown below:

Question 15

Draw a histogram and frequency polygon from the following data.

Class interval	20-25	25-30	30-35	35-40	40-45	45-50
Frequency	30	24	52	28	46	10

Solution 15

The given frequency distribution is as below :

Class intervals	20-25	25-30	30-35	35-40	40-45	45-50
Frequency	30	24	52	28	46	10

Take class intervals along x-axis and frequencies along y-axis and draw rectangle s of width equal to the size of the class intervals and heights equal to the corresponding frequencies.

Thus we get required histogram.

Now take imaginary class intervals 15-20 at the beginning and 50-55 at the end , each with frequency zero and join the mid points of top of the rectangles to get the required frequency polygon.

Question 16

Draw a histogram for the following data:

Class interval	600-640	640-680	680-720	720-760	760-800	800-840
Frequency	18	45	153	288	171	63

Usingthis histogram, draw the frequencypolygon on the same graph.Solution 16

The given frequency distribution table is given below :

Class interval	600-640	640-680	680-720	720-760	760-800	800-840
Frequency	18	45	153	288	171	63

Take class intervals along x-axis and frequencies along y-axis and draw rectangles of width equal to to size of class intervals and height equal to their corresponding frequencies. 

Thus we get the requiredhistogram.

Take imaginary class intervals 560-600 at the beginning and 840-880 at the end, each with frequency zero.

Now join the mid points of the top of the rectangles to get the required frequency polygon.

Exercise MCQ

Question 1

If the mean of x, x+2, x+4, x+6 and x+8 is 11, the value of x is

Solution 1

Question 2

If the mean of x, x + 3, x + 5, x + 10 is 9, the mean of the last three observations is

Solution 2

Question 3

Solution 3

Question 4

If each observation of the data is decreased by 8 then their mean

(a) remains the same

(b) is decreased by 8

(c) is increased by 5

(d) becomes 8 times the original meanSolution 4

Correct option: (b)

If each observation of the data is decreased by 8 then their mean is also decreased by 8. Question 5

The mean weight of six boys in a group is 48 kg. The individual weights of five them are 51 kg, 45 kg, 48 kg and 44 kg. The weight of 6^th boy is

1. 52 kg
2. 52.8 kg
3. 53 kg
4. 47 kg

Solution 5

Question 6

The mean of the marks scored by 50 students was found to be 39. Later on it was discovered that a score of 43 was misread as 23. The correct mean is

1. 38.6
2. 39.4
3. 39.8
4. 39.2

Solution 6

Question 7

The mean of 100 items was found to be 64. Later on it was discovered that two items were misread as 26 and 9 instead of 36 and 90 respectively. The correct mean is

1. 64.86
2. 65.31
3. 64.91
4. 64.61

Solution 7

Question 8

The mean of 100 observations is 50. If one of the observations 50 is replaced by 150, the resulting mean will be

1. 50.5
2. 51
3. 51.5
4. 52

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11

Solution 11

Question 12

The mean of the following data is 8

X	3	5	7	9	11	13
Y	6	8	15	P	8	4

The value of p is

1. 23
2. 24
3. 25
4. 21

Solution 12

Question 13

The runs scored by 11 members of a cricket team are

15, 34, 56, 27, 43, 29, 31, 13, 50, 20, 0

The median score is

1. 27
2. 29
3. 31
4. 20

Solution 13

Question 14

The weight of 10 students (in kgs) are

55,40,35,52,60,38,36,45,31,44

The median weight is

1. 40 kg
2. 41 kg
3. 42 kg
4. 44 kg

Solution 14

Question 15

The median of the numbers 4,4,5,7,6,7,7,12,3 is

1. 4
2. 5
3. 6
4. 7

Solution 15

Question 16

The median of the numbers 84,78,54,56,68,22,34,45,39,54 is

1. 45
2. 49.5
3. 54
4. 56

Solution 16

Question 17

Mode of the data 15,17,15,19,14,18,15,14,16,15,14,20,19,14,15 is

1. 14
2. 15
3. 16
4. 17

Solution 17

Question 18

The median of the data arranged in ascending order

8, 9, 12, 18, (x +2), (x + 4), 30, 31, 34, 39 is 24. The value of x is.

1. 22
2. 21
3. 20
4. 24

Solution 18

Exercise Ex. 18A

Question 1(iv)

Find the arithmetic mean of:

All the factors of 20Solution 1(iv)

Factors of 20 are: 1,2,4,5,10,20

Question 1(iii)

Find the arithmetic mean of:

The first seven multiple of 5Solution 1(iii)

First seven multiples of 5 are: 5,10,15, 20, 25, 30, 35

Therefore, Mean =20Question 1(ii)

Find the arithmetic mean of:

The first ten odd numbersSolution 1(ii)

First ten odd numbers are:

1,3,5,7,9,11,13,15, 17, and 19

Question 1(i)

Find the arithmetic mean of:

The first eight natural numbersSolution 1(i)

first eight natural numbers are:

1,2,3,4,5,6,7and 8

Question 1(v)

Find the mean of:

all prime numbers between 50 and 80.Solution 1(v)

Prime numbers between 50 and 80 are as follows:

53, 59, 61, 67, 71, 73, 79

Total prime numbers between 50 and 80 = 7

Question 2

The number of children in 10 families of a locality are

2,4,3,4,2,0,3,5,1,6.

Find the mean number of children per family.Solution 2

Question 3

The following are number of books issued in a school library during a week:

105, 216, 322, 167, 273, 405, and 346.

Find the average number of books issued per day.Solution 3

Sol.3 

Question 4

The daily minimum temperature recorded (in degree F) at a place during a week was as under:

Monday	Tuesday	Wednesday	Thursday	Friday	Saturday
35.5	30.8	27.3	32.1	23.8	29.9

Find the mean temperature.Solution 4

Question 5

If the mean of five observations x, x + 2, x + 4, x + 6, x + 8 is 13, find the value of x and hence find the mean of the last three observations.Solution 5

Total numbers of observations = 5

Thus, last three observations are (9 + 4), (9 + 6) and (9 + 8),

i.e. 13, 15 and 17

Question 6

The mean weight of 6 boys in a group is 48 kg. The individual weights of five of them are 51 kg, 45 kg, 49 kg, 46 kg and 44 kg. Find the weight of the sixth boy.Solution 6

Mean weight of the boys =48 kg

Sum of the weight of6 boys =(48×6)kg =288kg

Sum of the weights of 5 boys=(51+45+49+46+44)kg=235kg

Weight of the sixth boy=(sum of the weights of 6 boys ) – (sum of the weights of 5 boys)

=(288-235)=53kg.

Question 7

The mean of the marks scored by 50 students was found to be 39. Later on it was discovered that a score of 43 was misread as 23. Find the correct mean.Solution 7

Calculated mean marks of 50 students =39

 calculated sum of these marks=(39x 50)=1950

Corrected sum of these marks

=[1950-(wrong number)+(correct number)]

=(1950-23+43) =1970

correct mean =Question 8

The mean of 24 numbers is 35. If 3 is added to each number, what will bethe new mean?Solution 8

Let the given numbers be x₁,X_2……X₂₄

Question 9

The mean of 20 numbers is 43. If 6 is subtracted from each of the numbers , what will be the new mean ?Solution 9

Let the given numbers be x₁, x₂…..x₂₀

Then , the mean of these numbers =

Question 10

The mean of 15 numbers is 27. If each numbers is multiplied by 4, what will be the mean of the new numbers ?Solution 10

Let the given numbers be x₁, x₂…….x₁₅

Then the mean of these numbers=27

Question 11

The mean of 12 numbers is 40. If each number is divided by 8, what will be the mean of the new numbers?Solution 11

Question 12

The mean of 20 number is 18. If 3 is added to each of the first ten numbers , find the mean of the new set of 20 numbers.Solution 12

Question 13

The mean of six numbers is 23 . If one of the numbers is excluded , the mean of the remaining numbers is 20. Find the excluded number.Solution 13

Mean of 6 numbers = 23

Sum of 6 numbers =(23×6 )=138

Again , mean of 5 numbers =20

Sum of 5 numbers=(20x 5 ) =100

 The excluded number= (sum of 6 numbers )-(sum of 5 numbers)

=(138-100) =38

 The excluded number=38.Question 14

The average height of 30 boys was calculated to be 150 cm. It was detected later that one value of 165 cm was wrongly copied as 135 cm for the computation of the mean. Find the correct mean.Solution 14

Mean height of 30 boys = 150 cm

⇒ Total height of 30 boys = 150 × 30 = 4500 cm

Correct sum = 4500 – incorrect value + correct value

= 4500 – 135 + 165

= 4530

Question 15

The mean weight of a class of 34 students is 46.5 kg. If the of the teacher is included, the mean rises by 500 g. Find the weight of the teacherSolution 15

_{Mean weight of 34 students = 46.5 kg}

_{Total weight of 34 students =(34×46.5)kg =1581 kg}

_{Mean weight of 34 students and the teacher =(46.5+0.5)kg=47kg} 

_{Total weight of 34 students and the teacher}

_{=(47×35)kg =1645kg}

 _{Weight of the teacher =(1645-1581)kg= 64kg}Question 16

The mean weight of a class of 36 students is 41 kg. If one of the students leaves the class then the mean is decreased by 200 g. find the weight of the student who left.Solution 16

_{Mean weight of 36 students = 41 kg}

_{Total weight of 36 students = 41x 36 kg = 1476kg}

_{One student leaves the class mean is decreased by 200 g.}

 _{New mean =(41-0.2)kg = 40.8 kg}      

_{Total weight of 35 students = 40.8×35 kg = 1428 kg.}

_{the weight of the student who left =(1476-1428)kg =48 kg.}Question 17

The average weight of a class of 39 students is 40 kg . When a new student is admitted to the class , the average decreases by 200 g . find the weight of the new student.Solution 17

_{Mean weight of 39 students =40 kg}

_{Total weight of 39 students = 40x 39) = 1560 kg}

_{One student joins the class mean is decreased by 200 g.}

 _{New mean =(40-0.2)kg = 39.8 kg}               

_{Total weight of 40 students =(39.8×40)kg=1592 kg.}

_{the weight of new student}

_{= Total weight of 40 students – Total weight of 39 students}

_{= 1592-1560 = 32 kg}Question 18

The average weight of 10 oarsmen in a boat is increased by 1.5 kg when one of the crew who weighs 58 kg is replaced by a new man . find the weight of a new man.Solution 18

The increase in the average of 10 oarsmen = 1.5 kg

_{Total weight increased =(1.5×10) kg=15 kg}

Since the man weighing 58 kg has been replaced,

 _{Weight of the new man =(58+15)kg =73kg.}Question 19

The mean of 8 numbers is 35 . if a number is excluded then the mean is reduced by 3 . find the excluded number.Solution 19

_{Mean of 8 numbers=35}

_{Total sum of 8 numbers = 35×8 = 280}

 _{Since One number is excluded, New mean = 35 – 3 = 32}

_{Total sum of 7 numbers = 32×7 = 224}

_{the excluded number = Sum of 8 numbers – Sum of 7 numbers}

_{= 280 – 224 = 56}Question 20

The mean of 150 items was found to be 60. Later on , it was discovered that the values of two items were misread as 52 and 8 instead of 152 and 88 respectively. Find the correct mean.Solution 20

_{Mean of 150 items = 60}

_{Total Sum of 150 items = 150×60 = 9000}

_{Correct sum of items =[(sum of 150 items)-(sum of wrong items)+(sum of right items)]}

_{= [9000 – (52 + 8) + (152 + 88)]}

_{= [9000-(52+8)+(152+88)]}

_{= 9180}

 _{Correct mean =}Question 21

The mean of 31 results 60. If the mean of the first 16 results is 58 and that of the last 16 numbers is 62, find the 16^th result.Solution 21

_{Mean of 31 results=60}

_{Total sum of 31 results = 31×60 = 1860}

_{Mean of the first 16 results =16×58=928}

_{Total sum of the first 16 results=16×58=928}

_{Mean of the last 16 results=62}

_{Total sum of the last 16 results=16×62=992}

_{The 16th result = 928 + 992 – 1860}

                          _{= 1920 – 1860 = 60}

_{The 16th result  = 60.}Question 22

The mean of 11 numbers is 42. If the mean of the first 6 numbers is 37 and that of the last 6 numbers is 46 . find the 6^th number.Solution 22

_{Mean of 11 numbers = 42}

_{Total sum of 11 numbers = 42×11 = 462}

_{Mean of the first 6 numbers = 37}

_{Total sum of first 6 numbers = 37×6 = 222}

_{Mean of the last 6 numbers = 46}  

_{Total sum of last 6 numbers = 6×46 = 276}

_{The 6th number= 276 + 222 – 462}

                          _{= 498 – 462 = 36}

_{The 6th number = 36}Question 23

The mean weight of 25 students of a class is 52 kg . If the mean weight of the first 13 students of the class is 48 kg that of the last 13 students is 55 kg . find the weight of the 13^th student.Solution 23

_{Mean weight of 25 students = 52kg}

_{Total weight of 25 students = 52×25 kg=1300 kg}

_{Mean of the first 13 students = 48 kg}

_{Total weight of the first 13 students = 48×13 kg = 624kg}

_{Mean of the last 13 students = 55 kg}

_{Total weight of the last 13 students = 55×13 kg = 715 kg}

_{The weight of 13th student}

_{= Total weight of the first 13 students +  Total weight of the last 13 students – Total weight of 25 students}

_{= 624+715-1300 kg}

_{= 39 kg.}

Therefore, the weight of 13th student is 39 kg.Question 24

The mean score of 25 observations is 80 and the mean score of another 55 observations is 60. Determine the mean score of the whole set of observations .Solution 24

_{Mean score of 25 observations = 80}

_{Total score of 25 observations = 80×25 = 2000}

_{Mean score of 55 observations = 60}

_{Total score of 55 observations = 60×55 =3300}

Total no. of observations = 25+55 =80 observations

_{Total score = 2000+3300 = 5300}

_{Mean score =}Question 25

Arun scored 36 marks in English , 44 marks in hindi, 75 marks in mathematics and x marks in science . If he has secured an average of 50 marks , find the value of x.Solution 25

_{Average marks of 4 subjects = 50}

_{Total marks of 4 subjects = 50×4 = 200}

_{36 + 44 + 75 + x = 200}

 _{155  + x = 200}

 _{x = 200 – 155 = 45}

_{The value of x = 45}Question 26

A ship sails out to an island at the rate of 15 km/h and the sails back to the starting point at 10 km /h . find the average sailing speed for the whole journey .Solution 26

_{Let the distance of mark from the staring point be x km.}

_{Then , time taken by the ship reaching the marks=}

_{Time taken by the ship reaching the starting point from the marks =}

_{Total time taken =}

_{Total distance covered =x+x=2x km.}

Question 27

There are 50 students in a class, of which 40 are boys . The average weight of the class is 44 kg and that of the girls is 40 kg . find the average weight of the boys.Solution 27

_{Total number of students = 50}

_{Total number of girls = 50-40 = 10}

_{Average weight of the class = 44 kg}

_{Total weight of 50 students= 44x 50 kg = 2200kg}

_{Average weight of 10 girls = 40 kg}

_{Total weight of 10 girls = 40×10 kg = 400 kg}

_{Total weight of 40 boys = 2200-400 kg =1800 kg}

_{the average weight of the boys =} Question 28

The aggregate monthly expenditure of a family was Rs.18720 during the first 3 months, Rs.20340 during the next 4 months and Rs.21708 during the last 5 months of a year. If the total saving during the year be Rs.35340. Find the average monthly income of the family.Solution 28

Total earnings of the year

= Rs. (3 × 18720 + 4 × 20340 + 5 ×21708 + 35340)

= Rs. (56160 + 81360 + 108540 + 35340)

= Rs. 281400

Number of months = 12

Question 29

The average weekly payment to 75 workers in a factory is Rs.5680. The mean weekly payment to 25 of them is Rs.5400 and that of 30 others is Rs.5700. Find the mean weekly payment of the remaining workers.Solution 29

Average weekly payment of 75 workers = Rs. 5680

⇒ Total weekly payment of 75 workers = Rs. (75 × 5680) = Rs. 426000

Mean weekly payment of 25 workers = Rs. 5400

⇒ Total weekly payment of 25 workers = Rs. (25 × 5400) = Rs. 135000

Mean weekly payment of 30 workers = Rs. 5700

⇒ Total weekly payment of 30 workers = Rs. (30 × 5700) = Rs. 171000

Number of remaining workers = 75 – 25 – 30 = 20

Therefore, Total weekly payment of remaining 20 workers

= Rs. (426000 – 135000 – 171000)

= Rs. 120000

Question 30

The mean marks (out of 100) of boys and girls in an examination are 70 and 73 respectively. If the mean marks of all the students in that examination is 71, find the ratio of the number of boys to the number of girls.Solution 30

Let the ratio of number of boys to the number of girls be x : 1.

Then,

Sum of marks of boys = 70x

Sum of marks of girls = 73 × 1 = 73

And, sum of marks of boys and girls = 71 × (x + 1)

⇒ 70x + 73 = 71(x + 1)

⇒ 70x + 73 = 71x + 71

⇒ x = 2

Hence, the ratio of number of boys to the number of girls is 2 : 1.Question 31

The average monthly salary of 20 workers in an office is Rs.45900. If the manager’s salary is added, the average salary becomes Rs.49200 per month. What’s manager’s monthly salary?Solution 31

Mean monthly salary of 20 workers = Rs. 45900

⇒ Total monthly salary of 20 workers = Rs. (20 × 45900) = Rs. 918000

Mean monthly salary of 20 workers + manager = Rs. 49200

⇒ Total monthly salary of 20 workers + manager = Rs. (21 × 49200) = Rs. 1033200

Therefore, manager’s monthly salary = Rs. (1033200 – 918000) = Rs. 115200

Exercise Ex. 18C

Question 1(i)

_{Find the median of:}

_{2,10, 9, 9, 5, 2, 3, 7, 11}Solution 1(i)

_{Arranging the data in accending order, we have}

_{2,2,3, 5, 7, 9, 9, 10, 11}

_{Here n = 9, which is odd}

Question 1(ii)

_{Find the median of:}

_{15, 6, 16, 8, 22, 21, 9, 18, 25}Solution 1(ii)

_{Arranging the data in ascending order , we have}

_{6, 8, 9, 15, 16, 18, 21, 22, 25}

_{Here n = 9, which is odd}

Question 1(iii)

_{Find the median of}

_{20, 13, 18, 25, 6, 15, 21, 9, 16, 8, 22}Solution 1(iii)

_{Arranging data in ascending order:}

_{6, 8, 9, 13, 15, 16, 18, 20, 21, 22, 25}

_{Here n = 11 odd}

Question 1(iv)

_{Find the median of:}

_{7, 4, 2, 5, 1, 4, 0, 10, 3, 8, 5, 9, 2}Solution 1(iv)

_{Arranging the data in ascending order , we have}

_{0, 1, 2, 2, 3, 4, 4, 5, 5, 7, 8, 9, 10}

_{Here n = 13, which is odd}

Question 2(iii)

_{Find the median of:}

_{10, 75, 3, 15, 9, 47, 12, 48, 4, 81, 17, 27}Solution 2(iii)

_{Arranging the data in ascending order , we have}

_{3, 4, 9, 10, 12, 15, 17, 27, 47, 48, 75, 81 Here n = 12, which is even}

Question 2(ii)

_{Find the median of:}

_{72, 63, 29, 51, 35, 60, 55, 91, 85, 82}Solution 2(ii)

Arranging the data in ascending order , we have

_{29, 35, 51, 55, 60, 63, 72, 82, 85, 91}

_{Here n = 10, which is even}

Question 2(i)

_{Find the median of:}

_{17, 19, 32, 10, 22, 21, 9, 35}Solution 2(i)

_{Arranging the data in ascending order , we have}

_{9, 10, 17, 19, 21, 22, 32, 35}

_{Here n = 8, which is even}

Question 3

_{The marks of 15 students in an examination are :}

25,19,17,24,23,29,31,40,19,20,22,26,17,35,21

Find the median score.
Solution 3

_{Arranging the data in ascending order , we have}

_{17, 17, 19, 19, 20, 21, 22, 23, 24, 25, 26, 29, 31, 35, 40}

_{Here n = 15, which is odd}

Thus, the median score is 23.

Question 4

The heights (in cm) of 9 students of a class are

148, 144, 152, 155, 160, 147, 150, 149, 145.

Find the median heightSolution 4

Total number of students = n = 9 (odd)

Arranging heights (in cm) in ascending order, we have

144, 145, 147, 148, 149, 150, 152, 155, 160

Question 5

_{The weights (in kg ) of 8 children are:}

_{13.4, 10.6, 12.7, 17.2, 14.3, 15, 16.5, 9.8}

_{Find the median weight.}Solution 5

_{Arranging the weights of 8 children in ascending order, we have}

_{9.8, 10.6, 12.7, 13.4, 14.3, 15, 16.5, 17.2}

_{Here , n= 8 , which is even}

Question 6

_{The ages (in years ) of 10 teachers in a school are:}

_{32, 44, 53, 47, 37, 54, 34, 36, 40, 50}

_{Fid the median age.}Solution 6

_{Arranging the ages of teachers in ascending order , we have}

_{32, 34, 36, 37, 40, 44, 47, 50, 53, 54}

_{Here, n =10, which is even}

Question 7

_{If 10, 13, 15, 18, x+1, x+3, 30, 32, 35, 41 are ten observations in an ascending order with median 24, find the value of x.}Solution 7

  _{The  ten observations in ascending order:}

_{10, 13, 15, 18, x+1, x+3, 30, 32, 35, 41}

_{Here, n =10, which is even}

Question 8

The following observations are arranged in ascending order:

26, 29, 42, 53, x, x + 2, 70, 75, 82, 93.

If the median is 65, find the value of x.Solution 8

Total number of observations = n = 10 (even)

Median = 65

Question 9

The numbers 50, 42, 35, (2x + 10), (2x – 8), 12, 11, 8 have been written in a descending order. If their median is 25, find the value of x.Solution 9

Total number of observations = n = 8 (even)

Median = 25

Question 10

Find the median of the data

46, 41, 77, 58, 35, 64, 87, 92, 33, 55, 90.

In the above data, if 41 and 55 are replaced by 61 and 75 respectively, what will be the new median?Solution 10

Total number of observations = n = 11 (odd)

Arranging data in ascending order, we have

33, 35, 41, 46, 55, 58, 64, 77, 87, 90, 92

Now, 41 and 55 are replaced by 61 and 75 respectively. 

Arranging new data in ascending order, we have

33, 35, 46, 58, 61, 64, 75, 77, 87, 90, 92

Exercise Ex. 18D

Question 1

_{Find the mode of the following items.}

_{0, 6, 5, 1, 6, 4, 3, 0, 2, 6, 5, 6}Solution 1

_{Arrange the given data in ascending order we have}

_{0, 0, 1, 2, 3, 4, 5, 5, 6, 6, 6, 6}

Let us prepare the following table:

Observations(x)	0	1	2	3	4	5	6
Frequency	2	1	1	1	1	2	4

_{As 6 ocurs the maximum number of times i.e. 4, mode = 6}Question 2

_{Determine the mode of the following values of a variable.}

_{23, 15, 25, 40, 27, 25, 22, 25, 20}Solution 2

_{Arranging the given data in ascending order , we have:}

_{15, 20, 22, 23, 25, 25, 25, 27, 40}

_{The frequency table of the data is :}

Observations(x)	15	20	22	23	25	27	40
Frequency	1	1	1	1	3	1	1

_{As 25 ocurs the maximum number of times i.e. 3, mode = 25}Question 3

_{Calculate the mode of the following sizes of shoes by a shop on a particular day}

_{5, 9, 8, 6, 9, 4, 3, 9, 1, 6, 3, 9, 7, 1, 2, 5, 9}Solution 3

_{Arranging the given data in ascending order , we have:}

_{1, 1, 2, 3, 3, 4, 5, 5, 6, 6, 7, 8, 9, 9, 9, 9, 9,}

_{The frequency table of the data is :}

Observations(x)	1	2	3	4	5	6	7	8	9
Frequency	2	1	2	1	2	2	1	1	5

_{As 9, occurs the maximum number of times i.e. 5, mode = 9}Question 4

_{A cricket player scored the following runs in 12 one-day matches:}

_{50, 30, 9, 32, 60, 50, 28, 50, 19, 50, 27, 35.}

_{Find his modal score.}Solution 4

_{Arranging the given data in ascending order , we have:}

_{9, 19, 27, 28, 30, 32, 35, 50, 50, 50, 50, 60}

_{The frequency table of the data is :}

Observations(x)	9	19	27	28	30	32	35	50	60
Frequency	1	1	1	1	1	1	1	4	1

_{As 50, ocurs the maximum number of times i.e. 4, mode = 50}

Thus, the modal score of the cricket player is 50.
Question 5

If the mean of the data 3, 21, 25, 17, (x + 3), 19, (x – 4) is 18, find the value of x. Using this value of x, find the mode of the data.Solution 5

Total number of observations = n = 7

Mean = 18

Thus, data is as follows:

3, 21, 25, 17, 24, 19, 17

The most occurring value is 17.

Hence, the mode of the data is 17.Question 6

The numbers 52, 53, 54, 54, (2x + 1), 55, 55, 56, 57 have been arranged in an ascending order and their median is 55. Find the value of x and hence find the mode of the given data.Solution 6

Number of values = n = 9 (odd)

Numbers in ascending order:

52, 53, 54, 54, (2x + 1), 55, 55, 56, 57 

Thus, we have

52, 53, 54, 54, 55, 55, 55, 56, 57

The most occurring number is 55.

Hence, the mode of the data is 55. Question 7

For what value of x is the mode of the data 24, 15, 40, 23, 27, 26, 22, 25, 20, x + 3 found 25? Using this value of x, find the median.Solution 7

Mode of the data = 25

So, we should have the value 25 occurring maximum number of times in the given data.

That means, x + 3 = 25

⇒ x = 22

Thus, we have 24, 15, 40, 23, 27, 26, 22, 25, 20, 25.

Arranging data in ascending order, we have

15, 20, 22, 23, 24, 25, 25, 26, 27, 40

Number of observations = 10 (even)

Question 8

The numbers 42, 43, 44, 44, (2x + 3), 45, 45, 46, 47 have been arranged in an ascending order and their median is 45. Find the value of x. Hence, find the mode of the above data.Solution 8

Total number of observations = n = 9 (odd)

Median = 45

Thus, we have

42, 43, 44, 44, 45, 45, 45, 46, 47

The most occurring value is 45.

Hence, the mode of the data is 45.

Exercise Ex. 18B

Question 1

Obtain the mean of the following distribution:

Variable (xi)	4	6	8	10	12
Frequency (fi)	4	8	14	11	3

Solution 1

Question 2

_{The following table shows the weights of 12 workers in a factory :}

Weight (in Kg)	60	63	66	69	72
No of workers	4	3	2	2	1

_{Find the mean weight of the workers.}Solution 2

For calculating the mean , we prepare the following frequency table :

Weight (in kg)(Xi)	No of workers(fi)	fiXi
6063666972	43221	24018913213872
		771

Question 3

The measurements (in mm) of the diameters of the heads of 50 screws are given below:

Diameter (in mm) (xi)	34	37	40	43	46
Number of screws (fi)	5	10	17	12	6

Calculate the mean diameter of the heads of the screws.Solution 3

Question 4

_{The following data give the number of boys of a particular age in a class of 40 students.}

Age (in years)	15	16	17	18	19	20
Frequency (f)	3	8	9	11	6	3

_{Calculate the mean age of the students}Solution 4

For calculating the mean , we prepare the following frequency table :

Age (in years)(Xi)	Frequency(fi)	fiXi
151617181920	3891163	4512815319811460
		698

Question 5

_{Find the mean of the following frequency distribution :}

Variable (xi)	10	30	50	70	89
Frequency(fi)	7	8	10	15	10

Solution 5

For calculating  the mean , we prepare  the following frequency table :

Variable(Xi)	Frequency(fi)	fiXi
1030507089	78101510	702405001050890

Question 6

Find the mean of daily wages of 40 workers in a factory as per data given below:

Daily wages (in Rs.) (xi)	250	300	350	400	450
Number of workers (fi)	8	11	6	10	5

Solution 6

Question 7

If the mean of the following data is 20.2, find the value of p.

Variable (xi)	10	15	20	25	30
Frequency (fi)	6	8	p	10	6

Solution 7

Question 8

_{If the mean of the following frequency distribution is 8, find the value of p.}

X	3	5	7	9	11	13
F	6	8	15	p	8	4

Solution 8

We prepare  the following frequency table :

(Xi)	(fi)	fiXi
35791113	6815P84	18401059P8852

 303 + 9p = 8(41+p)

 303 + 9p= 328 + 8p

 9p – 8p = 328 -303

 P=25

 the value of P=25Question 9

_{Find the missing frequency p for the following frequency distribution whose mean is 28.25.}

X	15	20	25	30	35	40
F	8	7	p	14	15	6

Solution 9

We prepare the following frequency distribution table:

(Xi)	(fi)	fiXi
152025303540	87P14156	12014025p420525240

 1445 + 25p = (28.25)(50+p)

 1445 + 25p = 1412.50 + 28.25p

 -28.25p + 25p = -1445 + 1412.50

 -3.25p = -32.5

 the value of p=10Question 10

_{Find the value of p for the following frequency distribution whose mean is 16.6.}

X	8	12	15	p	20	25	30
F	12	16	20	24	16	8	4

Solution 10

We prepare the following frequency distribution table:

(Xi)	(fi)	fiXi
81215P202530	121620241684	9619230024p320200120

 1228 + 24p = 1660

 24p = 1660-1228

 24p = 432

 the value of p =18Question 11

Find the missing frequencies in the following frequency distribution whose mean is 34.

x	10	20	30	40	50	60	Total
f	4	f1	8	f2	3	4	35

Solution 11

Question 12

_{Find the missing frequencies in the following frequency distribution, whose mean is 50.}

x	10	30	50	70	90	Total
f	17	f1	32	f2	19	120

Solution 12

Let f₁  and f₂ be the missing frequencies. 

We prepare the following frequency distribution table.

(Xi)	(fi)	fixi
1030507090	17f132f219	17030f1160070f21710
Total	120	3480 + 30f1 + 70f2

Here, 

Thus,  …….(1)

Also,

Substituting the value of f₁ in equation 1, we have,

f₂=52 – 28 = 24

Thus, the missing frequencies are f₁ =28 and f₂=24 respectively.Question 13

Find the value of p, when the mean of the following distribution is 20.

x	15	17	19	20 + p	23
f	2	3	4	5p	6

Solution 13

Question 14

The mean of the following distribution is 50.

x	10	30	50	70	90
f	17	5a + 3	32	7a – 11	19

Find the value of a and hence the frequencies of 30 and 70.Solution 14

Exercise MCQ

Question 1

In a sample survey of 645 people, it was found that 516 people have a high school certificate. If a person is chosen at random, what is the probability that he/she has a high school certificate?

(a) 

(b) 

(c) 

(d) Solution 1

Correct option: (d)

Total number of people = 645

Number of people having high school certificate = 516

Question 2

In a medical examination of students of a class, the following blood groups are recorded:

Blood group	A	B	AB	O
Number of students	11	15	8	6

From this class, a student is chosen at random. What is the probability that the chosen student has blood group AB?

(a) 

(b) 

(c) 

(d) Solution 2

Correct option: (c)

Total number of students = 11 + 15 + 8 + 6 = 40

Number of students having blood group AB = 8 

Question 3

80 bulbs are selected at random from a lot and their lifetime in hours is recorded as under.

Lifetime (in hours)	300	500	700	900	1100
Frequency	10	12	23	25	10

One bulb is selected at random from the lot. What is the probability that its life is 1150 hours?

(a) 

(b) 

(c) 1

(d) 0Solution 3

Correct option: (d)

Total number of bulbs = 80

Number of bulbs having life of 1150 hours = 0

∴ Required probability = 0 Question 4

In a survey of 364 children aged 19 – 36 months, it was found that 91 liked to eat potato chips. If a child is selected at random, the probability that he/she does not like to eat potato chips is

(a) 

(b) 

(c) 

(d) Solution 4

Correct option: (c)

Total number of children = 364

Number of children who like to eat potato chips = 91

⇒ Number of children who do not like to eat potato chips = 364 – 91 = 273

Question 5

Two coins are tossed 1000 times and the outcomes are recorded as given below:

Number of heads	2	1	0
Frequency	200	550	250

Now, if two coins are tossed at random, what is the probability of getting at most one head?

(a) 

(b) 

(c) 

(d) Solution 5

Correct option: (b)

Total number of outcomes = 1000

Question 6

80 bulbs are selected at random from a lot and their lifetime in hours is recorded as under.

Lifetime (in hours)	300	500	700	900	1100
Frequency	10	12	23	25	10

One bulb is selected at random from the lot. What is the probability that selected bulb has a life more than 500 hours?

(a) 

(b) 

(c) 

(d) Solution 6

Correct option: (b)

Total number of bulbs = 80

Question 7

To know the opinion of the students about the subject Sanskrit, a survey of 200 students was conducted. The data is recorded as under.

Opinion	like	dislike
Number of students	135	65

What is the probability that a student chosen at random does not like it?

(a) 

(b) 

(c) 

(d) Solution 7

Correct option: (c)

Total number of students = 200

Number of students who does not like Sanskrit = 65

Question 8

A coin is tossed 60 times and the tail appears 35 times. What is the probability of getting a head?

Solution 8

Question 9

It is given that the probability of winning a game is 0.7. What is the probability of losing the game?

(a) 0.8

(b) 0.3

(c) 0.35

(d) 0.15Solution 9

Question 10

In a cricket match, a batsman hits a boundary 6 times out of 30 balls he plays. What is the probability that in given throw, the ball does not hit the boundary?

Solution 10

Question 11

A bag contains 16 cards bearing number 1, 2, 3,…, 16 respectively. One card is chosen at random. What is the probability that the chosen card bears a number which is divisible by 3?

Solution 11

Question 12

A bag contains 5 red, 8 black and 7 white balls. One ball is chosen at random. What is the probability that the chosen ball is black

Solution 12

Question 13

In 65 throws of a die, the outcomes were noted as under:

Outcomes	1	2	3	4	5	6
Number of times	8	10	12	16	9	10

A die is thrown at random. What is probability of getting a prime number

Solution 13

Question 14

In 50 throws of a die, the outcomes were notes as under:

Outcome	1	2	3	4	5	6
Number of times	8	9	6	7	12	8

A die is thrown at random. What is the probability of getting an even number?

Solution 14

Question 15

The table given below shows the month of birth of 36 students of a class :

Month of birth	Jan	Feb	March	April	May	June	July	Aug	Sept	Oct	Nov	Dec
No.of students	4	3	5	0	1	6	1	3	4	3	4	2

A student is chosen at random from the class. What is the probability that the chosen student was born in October?

Solution 15

Question 16

Two coins are tossed simultaneously 600 times to get

2 heads: 234 times, 1 head: 206 times, 0 head: 160 times.

If two coins are tossed at random, what is the probability of getting at least one head?

(a) 

(b) 

(c) 

(d) Solution 16

Correct option: (c)

Total number of outcomes = 600

Exercise Ex. 19

Question 1

_{A coin is tossed 500 times we get}

_{Head: 285 times, Tail: 215 times.}

_{When a coin is tossed at random, what is the probability of getting (i) A head? (ii) A tail?}Solution 1

Question 2

_{Two coin are tossed 400 times and we get}

_{two heads:112 times; one head :160 times; 0 head :128 times.}

_{When two coins are tossed at random, what is the probability of getting}

_{(i) 2 heads? (ii)1 head? (iii)0 head?}Solution 2

Question 3

Three coins are tossed 200 times and we get

three heads: 39 times; two heads: 58 times;

one head: 67 times; 0 head : 36 times.

When three coins are tossed at random, what is the probability of getting

(i) 3 heads? (ii) 1 head? (iii) 0 head? (iv) 2 heads?Solution 3

Question 4

A die is thrown 300 times and the outcomes are noted as given below:

Outcome	1	2	3	4	5	6
frequency	60	72	54	42	39	33

When a die is thrown at random, what is the probability of getting a

(i) 3? (ii) 6? (iii) 5? (iv)1?Solution 4

Question 5

In a survey of 200 ladies, it was found that 142 like coffee, while 58 dislike it.

Find the probability that a lady chosen at random

(i) Likes coffee (ii) dislikes coffeeSolution 5

Question 6

The percentages of marks obtained by a student in six unit tests are given below :

Unit test	I	II	III	IV	V	VI
Percentageof marksobtained	53	72	28	46	67	59

A unit test is selected at random. What is the probability that the student gets more than 60% marks in the test?Solution 6

Number of tests in which he gets more than60% marks =2

Total numbers of tests =6

_{Required probability}

Question 7

On a particular day, at a crossing in a city, the various types of 240 vehicles going past during a time interval were observed as under:

Type of vehicles	Two-wheelers	Three-wheelers	Four -wheelers
Frequency	84	68	88

Out of these vehicles, one is chosen at random. What is the probability that the chosen vehicle is a two-wheeler?Solution 7

Question 8

On one page of a telephone directory, there are 200 phone numbers. The frequency distribution of their units digits is given below:

Units digit	0	1	2	3	4	5	6	7	8	9
Frequency	19	22	23	19	21	24	23	18	16	15

One of the numbers is chosen at random from the page. What is the probability that the units digit of the chosen number is (i) 5? (ii) 8?Solution 8

Question 9

The following table shows the blood groups of 40 students of the class.

Blood Group	A	B	O	AB
Number of students	11	9	14	6

One student of the class is chosen at random. What is the probability that the chosn student has blood group

(i) O? (ii) AB?Solution 9

Question 10

12 packets of salt, each marked 2 kg, actually contained the following weights (in kg) of salt:

1.950, 2.020, 2.060, 1.980, 2.030, 1.970,

2.040, 1.990, 1.985, 2.025, 2.000, 1.980.

Out of these packets, one packet is chosen at random.

What is the probability that the chosen packet contains more than 2 kg of salt?Solution 10

Total number of salt packets = 12

Number of packets containing more than 2 kg of salt = 5

Therefore, 

Probability that the chosen packet contains more than 2 kg of salt

Question 11

In a cricket match, a batsman hits a boundary 6 times out of 30 balls he plays. Find the probability that he did not hit a boundary.Solution 11

Total number of ball played = 30

Number of times boundary was hit = 6

⇒ Number of times boundary was not hit = 30 – 6 = 24

Therefore, 

Probability that the batsman did not hit the boundary

Question 12

An organisation selected 2400 families at random and surveyed them to determine a relationship between the income level and the number of vehicles in a family. The information gathered is listed in the table below:

Monthly income (in Rs.)	Number of vehicles per family
0	1	2	3 or more
Less than Rs.25000	10	160	25	0
Rs.25000 – Rs.30000	0	305	27	2
Rs.30000 – Rs.35000	1	535	29	1
Rs.35000 – Rs.40000	2	469	59	25
Rs.40000 or more	1	579	82	88

Suppose a family is chosen at random. Find the probability that the family chosen is

(i) earning Rs.25000 – Rs.30000 per month and owning exactly 2 vehicles.

(ii) earning Rs.40000 or more per month and owning exactly 1 vehicle.

(iii) earning less than Rs.25000 per month and not owning any vehicle.

(iv) earning Rs.35000 – Rs.40000 per month and owning 2 or more vehicles.

(v) owning not more than 1 vehicle.Solution 12

Question 13

The table given below shows the mark obtained by 30 students in a test.

Marks(Class interval)	1 – 10	11 – 20	21 – 30	31 – 40	41 – 50
Number of students(Frequency)	7	10	6	4	3

Out of these students, one is chosen at random. What is the probability that the marks of the chosen student

(i) are 30 or less?

(ii) are 31 or more?

(iii) lie in the interval 21-30?Solution 13

Question 14

The table given shows the ages of 75 teachers in a school.

Age (in years)	18 – 29	30 – 39	40 – 49	50 – 59
Number of teachers	3	27	37	8

A teacher from this school is chosen at random. What is probability that the selected teacher is

(i) 40 or more than 40 years old?

(ii) of an age lying between 30-39 years (including both)?

(iii) 18 years or more and 49 years or less?

(iv) 18 years or more old?

(v) above 60 years of age?

NOTE Here 18 – 29 means 18 or more but less than or equal to 29.Solution 14

Question 15

Following are the ages (in years) of 360 patients, getting medical treatment in a hospital:

Age(in years)	10 – 20	20 – 30	30 – 40	40 – 50	50 – 60	60 – 70
Number of patients	90	50	60	80	50	30

One of the patients is selected at random.

What is probability that his age is

(i) 30 years or more but less than 40 years?

(ii) 50 years or more but less than 70 years?

(iii) 10 years or more but less than 40 years?

(iv) 10 years or more?

(v) less than 10 years?Solution 15

Question 16

The marks obtained by 90 students of a school in mathematics out of 100 are given as under:

Marks	0-20	20-30	30-40	40-50	50-60	60-70	70 and above
No. of students	7	8	12	25	19	10	9

From these students, a student is chosen at random.

What is the probability that the chosen student

(i) get 20% or less marks? (ii) get 60% or more marks?Solution 16

Question 17

It is known that a box of 800 electric bulbs contains 36 defective bulbs. One bulb is taken at random out of the box. What is probability that the bulb chosen is nondefective?Solution 17

Total number of electric bulbs = 800

Number of defective bulbs = 36

⇒ Number of non-defective bulbs = 800 – 36 = 764

Hence, probability that the bulb chosen is non-defective

Question 18

Fill in the blanks.

(i) Probability of an impossible event = ….. .

(ii) Probability of a sure event = ….. .

(iii) Let E be an event. Then, P(not E) = ….. .

(iv) P(E) + P(not E) = ….. .

(v) ….. ≤ P(E) ≤ …. .Solution 18

Fill in the blanks.

(i) Probability of an impossible event = 0

(ii) Probability of a sure event = 1

(iii) Let E be an event. Then, P(not E) = 1 – P(E)

(iv) P(E) + P(not E) = 1

(v) 0 ≤ P(E) ≤ 1

Exercise MCQ

Question 1

Three angles of quadrilateral are 80^°, 95° and 112°. Its fourth angle is

(a) 78°

(b) 73°

(c) 85°

(d) 100°Solution 1

Question 2

The angles of a quadrilateral are in the ratio 3:4:5:6. The smallest of these angles is

(a) 45°

(b) 60°

(c) 36°

(d) 48°Solution 2

Question 3

In the given figure, ABCD is a parallelogram in which ∠BAD = 75° and ∠CBD = 60°. Then, ∠BDC =?

(a) 60°

(b) 75°

(c) 45°

(d) 50°

Solution 3

Question 4

ABCD is a rhombus such that ∠ACB = 50°. Then, ∠ADB = ?

(a) 40° 

(b) 25° 

(c) 65°  

(d) 130° Solution 4

Correct option: (a)

ABCD is a rhombus.

⇒ AD ∥ BC and AC is the transversal.

⇒ ∠DAC = ∠ACB  (alternate angles)

⇒ ∠DAC = 50° 

In ΔAOD, by angle sum property,

∠AOD + ∠DAO + ∠ADO = 180° 

⇒ 90° + ∠50° + ∠ADO = 180° 

⇒ ∠ADO = 40° 

⇒ ∠ADB = 40° Question 5

In which of the following figures are the diagonals equal?

1. Parallelogram
2. Rhombus
3. Trapezium
4. Rectangle

Solution 5

Question 6

If the diagonals of a quadrilateral bisect each other at right angles, then the figure is a

a. trapezium

b. parallelogram

c. rectangle

d. rhombusSolution 6

Question 7

The lengths of the diagonals of a rhombus are 16 cm and 12 cm. The length of each side of the rhombus is

1. 10 cm
2. 12 cm
3. 9cm
4. 8cm

Solution 7

Question 8

The length of each side of a rhombus is 10 cm and one of its diagonals is of length 16 cm. The length of the other diagonal is

Solution 8

Question 9

A diagonal of a rectangle is inclined to one side of the rectangle at 35°. The acute angle between the diagonals is

(a) 55° 

(b) 70° 

(c) 45° 

(d) 50° Solution 9

Correct option: (b)

∠DAO + ∠OAB = ∠DAB

⇒ ∠DAO + 35° = 90° 

⇒ ∠DAO = 55° 

ABCD is a rectangle and diagonals of a rectangle are equal and bisect each other.

OA = OD

⇒ ∠ODA = ∠DAO (angles opposte to equal sides are equal)

⇒ ∠ODA = 55° 

In DODA, by angle sum property,

∠ODA + ∠DAO + ∠AOD = 180° 

⇒ 55° + ∠55° + ∠AOD = 180° 

⇒ ∠AOD = 70° Question 10

If ABCD is a parallelogram with two adjacent angles ∠A = ∠B, then the parallelogram is a

1. rhombus
2. trapezium
3. rectangle
4. none of these

Solution 10

Question 11

In a quadrilateral ABCD, if AO and BO are the bisectors of ∠A and ∠B respectively, ∠C = 70° and ∠D = 30°. Then, ∠AOB =?

1. 40°
2. 50°
3. 80°
4. 100°

Solution 11

Question 12

The bisectors of any adjacent angles of a parallelogram intersect at

1. 30°
2. 45°
3. 60°
4. 90°

Solution 12

Question 13

The bisectors of the angles of a parallelogram enclose a

1. rhombus
2. square
3. rectangle
4. parallelogram

Solution 13

Correct option: (c)

The bisectors of the angles of a parallelogram enclose a rectangle.Question 14

If bisectors of ∠A and ∠B of a quadrilateral ABCD intersect each other at P, of ∠B and ∠C at Q, of ∠C and ∠D at R and of ∠D and ∠A at S then PQRS is a

(a) rectangle

(b) parallelogram

(c) rhombus

(d) quadrilateral whose opposite angles are supplementarySolution 14

Correct option: (d)

In ΔAPB, by angle sum property,

∠APB + ∠PAB + ∠PBA = 180° 

In ΔCRD, by angle sum property,

∠CRD + ∠RDC + ∠RCD = 180° 

Now, ∠SPQ + ∠SRQ = ∠APB + ∠CRD

= 360° – 180° 

= 180° 

Now, ∠PSR + ∠PQR = 360° – (∠SPQ + ∠SRQ)

= 360° – 180° 

= 180° 

Hence, PQRS is a quadrilateral whose opposite angles are supplementary. Question 15

The figure formed by joining the mid-points of the adjacent sides of a quadrilateral is a

1. rhombus
2. square
3. rectangle
4. parallelogram

Solution 15

Question 16

The figure formed by joining the mid-points of the adjacent sides of a square is a

1. rhombus
2. square
3. rectangle
4. parallelogram

Solution 16

Question 17

The figure formed by joining the mid-points of the adjacent sides of a parallelogram is a

1. rhombus
2. square
3. rectangle
4. parallelogram

Solution 17

Question 18

The figure formed by joining the mid-points of the adjacent sides of a rectangle is a

1. rhombus
2. square
3. rectangle
4. parallelogram

Solution 18

Question 19

The figure formed by joining the mid-points of the adjacent sides of a rhombus is a

1. rhombus
2. square
3. rectangle
4. parallelogram

Solution 19

Question 20

The quadrilateral formed by joining the midpoints of the sides of a quadrilateral ABCD, taken in order, is a rectangle, if

(a) ABCD is a parallelogram

(b) ABCD is a rectangle

(c) diagonals of ABCD are equal

(d) diagonals of ABCD are perpendicular to each otherSolution 20

Correct option: (d)

In ΔABC, P and Q are the mid-points of sides AB and BC respectively.

In ΔADC, R and S are the mid-points of sides CD and AD respectively.

From (i) and (ii),

PQ ∥ RS and PQ = RS

Thus, in quadrilateral PQRS, a pair of opposite sides are equal are parallel.

So, PQRS is a parallelogram.

Let the diagonals AC and BD intersect at O.

Now, in ΔABD, P and S are the mid-points of sides AB and AD respectively.

Thus, in quadrilateral PMON, PM ∥ NO and PN ∥ MO.

⇒ PMON is a parallelogram.

⇒ ∠MPN = ∠MON (opposite angles of a parallelogram are equal)

⇒ ∠MPN = ∠BOA (since ∠BOA = ∠MON)

⇒ ∠MPN = 90° (since AC ⊥ BD, ∠BOA = 90°)

⇒ ∠QPS = 90° 

Thus, PQRS is a parallelogram whose one angle, i.e. ∠QPS = 90°.

Hence, PQRS is a rectangle if AC ⊥ BD. Question 21

The quadrilateral formed by joining the midpoints of the sides of a quadrilateral ABCD, taken in order, is a rhombus, if

(a) ABCD is a parallelogram

(b) ABCD is a rhombus

(c) diagonals of ABCD are equal

(d) diagonals of ABCD are perpendicular to each otherSolution 21

Correct option: (c)

In ΔABC, P and Q are the mid-points of sides AB and BC respectively.

In ΔBCD, Q and R are the mid-points of sides BC and CD respectively.

In ΔADC, S and R are the mid-points of sides AD and CD respectively.

In ΔABD, P and S are the mid-points of sides AB and AD respectively.

⇒ PQ ∥ RS and QR ∥ SP [From (i), (ii), (iii) and (iv)]

Thus, PQRS is a parallelogram.

Now, AC = BD (given)

⇒ PQ = QR = RS = SP [From (i), (ii), (iii) and (iv)]

Hence, PQRS is a rhombus if diagonals of ABCD are equal. Question 22

The figure formed by joining the midpoints of the sides of a quadrilateral ABCD, taken in order, is a square, only if

(a) ABCD is a rhombus

(b) diagonals of ABCD are equal

(c) diagonals of ABCD are perpendicular

(d) diagonals of ABCD are equal and perpendicularSolution 22

Correct option: (d)

In ΔABC, P and Q are the mid-points of sides AB and BC respectively.

In ΔBCD, Q and R are the mid-points of sides BC and CD respectively.

In ΔADC, S and R are the mid-points of sides AD and CD respectively.

In ΔABD, P and S are the mid-points of sides AB and AD respectively.

⇒ PQ || RS and QR || SP [From (i), (ii), (iii) and (iv)]

Thus, PQRS is a parallelogram.

Now, AC = BD (given)

⇒ PQ = QR = RS = SP [From (i), (ii), (iii) and (iv)]

Let the diagonals AC and BD intersect at O.

Now,

Thus, in quadrilateral PMON, PM || NO and PN || MO.

⇒ PMON is a parallelogram.

⇒ ∠MPN = ∠MON (opposite angles of a parallelogram are equal)

⇒ ∠MPN = ∠BOA (since ∠BOA = ∠MON)

⇒ ∠MPN = 90° (since AC ⊥ BD, ∠BOA = 90°)

⇒ ∠QPS = 90° 

Thus, PQRS is a parallelogram such that PQ = QR = RS = SP and ∠QPS = 90°.

Hence, PQRS is a square if diagonals of ABCD are equal and perpendicular. Question 23

If an angle of a parallelogram is two-third of its adjacent angle, the smallest angle of the parallelogram is

1. 108°
2. 54°
3. 72°
4. 81°

Solution 23

Question 24

If one angle of a parallelogram is 24° less than twice the smallest angle, then the largest angles of the parallelogram is

1. 68°
2. 102°
3. 112°
4. 136°

Solution 24

Question 25

If ∠A, ∠B, ∠C and ∠D of a quadrilateral ABCD taken in order, are in the ratio 3:7:6:4, then ABCD is a

1. rhombus
2. kite
3. trapezium
4. parallelogram

Solution 25

Question 26

Which of the following is not true for a parallelogram?

1. Opposite sides are equal.
2. Opposite angles are equal.
3. Opposite angles are bisected by the diagonals.
4. Diagonals bisect each other.

Solution 26

Question 27

If APB and CQD are two parallel lines, then the bisectors of ∠APQ, ∠BPQ, ∠CQP and ∠PQD enclose a

1. square
2. rhombus
3. rectangle
4. kite

Solution 27

Question 28

In the given figure, ABCD is a parallelogram in which ∠BDC = 45° and ∠BAD = 75°. Then, ∠CBD =?

1. 45°
2. 55°
3. 60°
4. 75°

Solution 28

Question 29

If area of a ‖gm with side ɑ and b is A and that of a rectangle with side ɑ and b is B, then

(a) A > B

(b) A = B

(c) A < B

(d) A ≥ BSolution 29

Question 30

In the given figure, ABCD is a ‖gm and E is the mid-point at BC, Also, DE and AB when produced meet at F. Then,

Solution 30

Question 31

P is any point on the side BC of a ΔABC. P is joined to A. If D and E are the midpoints of the sides AB and AC respectively and M and N are the midpoints of BP and CP respectively then quadrilateral DENM is

(a) a trapezium

(b) a parallelogram

(c) a rectangle

(d) a rhombusSolution 31

Correct option: (b)

In ΔABC, D and E are the mid-points of sides AB and AC respectively.

Hence, DENM is a parallelogram.Question 32

The parallel sides of a trapezium are ɑ and b respectively. The line joining the mid-points of its non-parallel sides will be

Solution 32

Question 33

In a trapezium ABCD, if E and F be the mid-points of the diagonals AC and BD respectively. Then, EF =?

Solution 33

Question 34

In the given figure, ABCD is a parallelogram, M is the mid-point of BD and BD bisects ∠B as well as ∠D. Then, ∠AMB=?

1. 45°
2. 60°
3. 90°
4. 30°

Solution 34

Question 35

In the given figures, ABCD is a rhombus. Then

(a) AC² + BD² = AB²

(b) AC² + BD² = 2AB²

(c) AC² + BD² = 4AB²

(d) 2(AC² + BD²)=3AB²

Solution 35

Question 36

In a trapezium ABCD, if AB ‖ CD, then (AC² + BD²) =?

(a) BC² + AD² + 2BC. AD

(b) AB² +CD² + 2AB.CD

(c) AB² + CD² + 2AD. BC

(d) BC² + AD² + 2AB.CD

Solution 36

Question 37

Two parallelogram stand on equal bases and between the same parallels. The ratio of their area is

1. 1:2
2. 2:1
3. 1:3
4. 1:1

Solution 37

Question 38

In the given figure, AD is a median of ΔABC and E is the mid-point of AD. If BE is joined and produced to meet AC in F, then AF =?

Solution 38

Question 39

The diagonals AC and BD of a parallelogram ABCD intersect each other at the point O such that ∠DAC = 30°and ∠AOB = 70°. Then, ∠DBC =?

1. 40°
2. 35°
3. 45°
4. 50°

Solution 39

Question 40

Three statement are given below:

1. In a ‖gm, the angle bisectors of two adjacent angles enclose a right angle.
2. The angle bisectors of a ‖gm form a rectangle.
3. The triangle formed by joining the mid-point of the sides of an isosceles triangle is not necessarily an isosceles.

Which is true?

1. I only
2. II only
3. I and II
4. II and III

Solution 40

Question 41

Three statements are given below:

 I. In a rectangle ABCD, the diagonal AC bisects ∠A as well as ∠C.

 II. In a square ABCD, the diagonal AC bisects ∠A as well as ∠C

 III. In a rhombus ABCD, the diagonal AC bisects ∠A as well as ∠C

Which is true?

1. I only
2. II and III
3. I and III
4. I and II

Solution 41

Question 42

In a quadrilateral PQRS, opposite angles are equal. If SR = 2 cm and PR = 5 cm then determine PQ. Solution 42

Since opposite angles of quadrilateral are equal, PQRS is a parallelogram.

⇒ PQ = SR (opposite sides of parallelogram are equal)

⇒ PQ = 2 cmQuestion 43

Diagonals of a parallelogram are perpendicular to each other. Is this statement true? Give reasons for your answer.Solution 43

The given statement is false.

Diagonals of a parallelogram bisect each other. Question 44

What special name can be given to a quadrilateral PQRS if ∠P + ∠S = 118°?Solution 44

In quadrilateral PQRS, ∠P and ∠S are adjacent angles.

Since the sum of adjacent angles ≠ 180°, PQRS is not a parallelogram.

Hence, PQRS is a trapezium. Question 45

All the angles of a quadrilateral can be acute. Is this statement true? Give reasons for your answer.Solution 45

The given statement is false.

We know that the sum of all the four angles of a quadrilateral is 360°.

If all the angles of a quadrilateral are acute, the sum will be less than 360°. Question 46

All the angles of a quadrilateral can be right angles. Is this statement true? Give reasons for your answer.Solution 46

The given statement is true.

We know that the sum of all the four angles of a quadrilateral is 360°.

If all the angles of a quadrilateral are right angles,

Sum of all angles of a quadrilateral = 4 × 90° = 360° Question 47

All the angles of a quadrilateral can be obtuse. Is this statement true? Give reasons for your answer.Solution 47

The given statement is false.

We know that the sum of all the four angles of a quadrilateral is 360°.

If all the angles of a quadrilateral are obtuse, the sum will be more than 360°. Question 48

Can we form a quadrilateral whose angles are 70°, 115°, 60° and 120°? Give reasons for your answer.Solution 48

We know that the sum of all the four angles of a quadrilateral is 360°.

Here,

70° + 115° + 60° + 120° = 365° ≠ 360° 

Hence, we cannot form a quadrilateral with given angles. Question 49

What special name can be given to a quadrilateral whose all angles are equal?Solution 49

A quadrilateral whose all angles are equal is a rectangle. Question 50

If D and E are respectively the midpoints of the sides AB and BC of ΔABC in which AB = 7.2 cm, BC = 9.8 cm and AC = 3.6 cm then determine the length of DE.Solution 50

D and E are respectively the midpoints of the sides AB and BC of ΔABC.

Thus, by mid-point theorem, we have

Question 51

In a quadrilateral PQRS, the diagonals PR and QS bisect each other. If ∠Q = 56°, determine ∠R.Solution 51

Since the diagonals PR and QS of quadrilateral PQRS bisect each, PQRS is a parallelogram.

Now, adjacent angles of parallelogram are supplementary.

⇒ ∠Q + ∠R = 180° 

⇒ 56° + ∠R = 180° 

⇒ ∠R = 124° Question 52

In the adjoining figure, BDEF and AFDE are parallelograms. Is AF = FB? Why or why not?

Solution 52

AFDE is a parallelogram

⇒ AF = ED …(i)

BDEF is a parallelogram.

⇒ FB = ED …(ii)

From (i) and (ii),

AF = FB Question 53

In each of the questions are question is followed by two statements I and II. The answer is

1. if the question can be answered by one of the given statements alone and not by the other;
2. if the question can be answered by either statement alone;
3. if the question can be answered by both the statements together but not by any one of the two;
4. if the question cannot be answered by using both the statement together.

Is quadrilateral ABCD a ‖gm?

1. Diagonal AC and BD bisect each other.
2. Diagonal AC and BD are equal.

The correct answer is : (a)/ (b)/ (c)/ (d).Solution 53

Correct option: (a)

If the diagonals of a quad. ABCD bisect each other, then the quad. ABCD is a parallelogram.

So, I gives the answer.

If the diagonals are equal, then the quad. ABCD is a parallelogram.

So, II gives the answer.Question 54

In each of the questions are question is followed by two statements I and II. The answer is

1. if the question can be answered by one of the given statements alone and not by the other;
2. if the question can be answered by either statement alone;
3. if the question can be answered by both the statements together but not by any one of the two;
4. if the question cannot be answered by using both the statement together

Is quadrilateral ABCD a rhombus?

1. Quad. ABCD is a ‖gm.
2. Diagonals AC and BD are perpendicular to each other.

The correct answer is: (a) / (b)/ (c)/ (d).Solution 54

Correct option: (c)

If the quad. ABCD is a ‖gm, it could be a rectangle or square or rhombus.

So, statement I is not sufficient to answer the question.

If the diagonals AC and BD are perpendicular to each other, then the ‖gm could be a square or rhombus.

So, statement II is not sufficient to answer the question.

However, if the statements are combined, then the quad. ABCD is a rhombus.Question 55

In each of the questions are question is followed by two statements I and II. The answer is

1. if the question can be answered by one of the given statements alone and not by the other;
2. if the question can be answered by either statement alone;
3. if the question can be answered by both the statements together but not by any one of the two;
4. if the question cannot be answered by using both the statement together

Is ‖gm ABCD a square?

1. Diagonals of ‖gm ABCD are equal.
2. Diagonals of ‖gm ABCD intersect at right angles.

The correct answer is: (a)/ (b)/ (c)/ (d).Solution 55

Correct option: (c)

If the diagonals of a ‖gm ABCD are equal, then ‖gm ABCD could either be a rectangle or a square.

If the diagonals of the ‖gm ABCD intersect at right angles, then the ‖gm ABCD could be a square or a rhombus.

However, if both the statements are combined, then ‖gm ABCD will be a square.Question 56

In each of the questions are question is followed by two statements I and II. The answer is

1. if the question can be answered by one of the given statements alone and not by the other;
2. if the question can be answered by either statement alone;
3. if the question can be answered by both the statements together but not by any one of the two;
4. if the question cannot be answered by using both the statement together

Is quad. ABCD a parallelogram?

1. Its opposite sides are equal.
2. Its opposite angles are equal.

The correct answer is: (a)/ (b)/ (c)/ (d)Solution 56

Correct option: (b)

If the opposite sides of a quad. ABCD are equal, the quadrilateral is a parallelogram.

If the opposite angles are equal, then the quad. ABCD is a parallelogram.Question 57

 Each question consists of two statements, namely, Assertion (A) and Reason (R). For selecting the correct answer, use the following code:

1. Both Assertion (A) and Reason (R) are true and Reason (R) is a correct explanation of Assertion (A).
2. Both Assertion (A) and Reason (R) are true but Reason (R) is not a correct explanation of Assertion (A)
3. Assertion (A) is true and Reason (R) is false.
4. Assertion (A) is false and Reason (R) is true.

Assertion (A)	Reason (R)
If three angles of a quadrilateral are 130°, 70° and 60°, then the fourth angle is 100°.	The sum of all the angle of a quadrilateral is 360°.

The correct answer is: (a)/ (b)/ (c)/ (d).Solution 57

Question 58

Each question consists of two statements, namely, Assertion (A) and Reason (R). for selecting the correct answer, use the following code:

1. Both Assertion (A) and Reason (R) are true and Reason (R) is a correct explanation of Assertion (A).
2. Both Assertion (A) and Reason (R) are true but Reason (R) is not a correct explanation of Assertion (A)
3. Assertion (A) is true and Reason (R) is false.
4. Assertion (A) is false and Reason (R) is true.

Assertion (A)	Reason (R)
ABCD is a quadrilateral in which P, Q, R and S are the mid-points of AB, BC, CD and DA respectively.Then, PQRS is a parallelogram.	The line segment joining the mid-points of any two sides of a triangle is parallel to the third side and equal to half of it.

The correct answer is: (a)/ (b)/ (c)/ (d).Solution 58

The Reason (R) is true and is the correct explanation for the Assertion (A).Question 59

Each question consists of two statements, namely, Assertion (A) and Reason (R). for selecting the correct answer, use the following code:

1. Both Assertion (A) and Reason (R) are true and Reason (R) is a correct explanation of Assertion (A).
2. Both Assertion (A) and Reason (R) are true but Reason (R) is not a correct explanation of Assertion (A)
3. Assertion (A) is true and Reason (R) is false.
4. Assertion (A) is false and Reason (R) is true.

Assertion (A)	Reason (R)
In a rhombus ABCD, the diagonal AC bisects ∠A as well as ∠C.	The diagonals of a rhombus bisect each other at right angles.

The correct answer is: (a)/ (b)/ (c)/ (d).Solution 59

Question 60

Each question consists of two statements, namely, Assertion (A) and Reason (R). for selecting the correct answer, use the following code:

1. Both Assertion (A) and Reason (R) are true and Reason (R) is a correct explanation of Assertion (A).
2. Both Assertion (A) and Reason (R) are true but Reason (R) is not a correct explanation of Assertion (A)
3. Assertion (A) is true and Reason (R) is false.
4. Assertion (A) is false and Reason (R) is true.

Assertion (A)	Reason (R)
Every parallelogram is a rectangle.	The angle bisectors of a parallelogram form a rectangle.

The correct answer is: (a)/ (b)/ (c)/ (d).Solution 60

Question 61

Each question consists of two statement, namely, Assertion (A) and Reason (R). for selecting the correct answer, use the following code:

1. Both Assertion (A) and Reason (R) are true and Reason (R) is a correct explanation of Assertion (A).
2. Both Assertion (A) and Reason (R) are true but Reason (R) is not a correct explanation of Assertion (A)
3. Assertion (A) is true and Reason (R) is false.
4. Assertion (A) is false and Reason (R) is true.

Assertion (A)	Reason (R)
The diagonals of a ‖gm bisect each other.	If the diagonals of a ‖gm are equal and intersect at right angles, then the parallelogram is a square.

The correct answer is: (a)/ (b)/ (c)/ (d).Solution 61

Question 62

Column I	Column II
(a) Angle bisectors of a parallelogram form a	(p) parallelogram
(b) The quadrilateral formed by joining the mid-points of the pairs of adjacent sides of a square is a	(q) rectangle
(c) The quadrilateral formed by joining the mid-points of the pairs of adjacent side of a rectangle is a	(r) square
(d) The figure formed by joining the mid-points of the pairs of adjacent sides of a quadrilateral is	(s) rhombus

The correct answer is:

(a) -…….,

(b) -…….,

(c) -…….,

(d)-…….Solution 62

Question 63

Column I	Column II
(a) In the given figure, ABCD is a trapezium in which AB = 10 cm and CD = 7cm. If P and Q are the mid-points of AD and BC respectively, then PQ =	(p) equal
(b) In the given figure, PQRS is a ‖gm whose diagonal intersect at O. If PR = 13 cm, then OR=	(q) at right angle
(c) The diagonals of a square are	(r) 8.5 cm
(d) The diagonals of a rhombus bisect each other	(s) 6.5 cm

The correct answer is:

(a) -…….,

(b) -…….,

(c) -…….,

(d)-…….Solution 63

Exercise Ex. 10B

Question 1

In the adjoining figure, ABCD is a parallelogram in which =72⁰. Calculate _,and .

Solution 1

Question 2

In the adjoining figure , ABCD is a parallelogram in which

 and . Calculate _.

Solution 2

Question 3

In the adjoining figure, M is the midpoint of side BC of parallelogram ABCD such that ∠BAM = ∠DAM. Prove that AD = 2CD.

Solution 3

ABCD is a parallelogram.

Hence, AD || BC.

⇒ ∠DAM = ∠AMB (alternate angles)

⇒ ∠BAM = ∠AMB (since ∠BAM = ∠DAM)

⇒ BM = AB (sides opposite to equal angles are equal)

But, AB = CD (opposite sides of a parallelogram)

⇒ BM = AB = CD ….(i)

Question 4

In a adjoining figure, ABCD is a parallelogram in which =60^o. If the parallelogram in which and meet DC at P, prove that (i) PB=90^o, (ii) AD=DP and PB=PC=BC, (iii) DC=2AD.

Solution 4

Question 5

_{In the adjoining figure, ABCD is a parallelogram in which}

_Calculate 

Solution 5

Question 6

_{In a ||gm ABCD , if , find the value of x and the measure of each angle of the parallelogram.}Solution 6

Question 7

_{If an angle of a parallelogram is four fifths of its adjacent angle, find the angles of the parallelogram .}Solution 7

Question 8

_{Find the measure of each angle of parallelogram , if one of its angles is  less than twice the smallest angle.}Solution 8

Question 9

_{ABCD is a parallelogram in which AB=9.5 cm and its parameter is 30 cm. Find the length of each side of the parallelogram.}Solution 9

Question 10

_{In each of the figures given below, ABCD is a rhombus. Find the value of x and y in each case.}

Solution 10

Question 11

_{The lengths of the diagonals of a rhombus are 24 cm and 18 cm respectively . Find the length of each side of the rhombus.}Solution 11

Question 12

_{Each side of a rhombus is 10 cm long and one of its diagonals measures 16 cm. Find the length of the other diagonal and hence find the area of the rhombus.}Solution 12

Question 13

_{In each of the figures given below, ABCD is a rectangle . Find the values of x and y in each case.}

Solution 13

Question 14

In a rhombus ABCD, the altitude from D to the side AB bisect AB. Find the angle of the rhombusSolution 14

Let the altitude from D to the side AB bisect AB at point P.

Join BD.

In ΔAMD and ΔBMD,

AM = BM (M is the mid-point of AB)

∠AMD = ∠BMD (Each 90°)

MD = MD (common)

∴ ΔAMD ≅ ΔBMD (by SAS congruence criterion)

⇒ AD = BD (c.p.c.t.)

But, AD = AB (sides of a rhombus)

⇒ AD = AB = BD

⇒ ΔADB is an equilateral triangle.

⇒ ∠A = 60° 

⇒ ∠C = ∠A = 60° (opposite angles are equal)

⇒ ∠B = 180° – ∠A = 180° – 60° = 120° 

⇒ ∠D = ∠B = 120° 

Hence, in rhombus ABCD, ∠A = 60°, ∠B = 120°, ∠C = 60° and ∠D = 120°.Question 15

_{In the adjoining figure , ABCD is square . A line segment CX cuts AB at X and the diagonal BD at O, such that . Find the value of x.}

*Back answer incorrectSolution 15

Question 16

In a rhombus ABCD show that diagonal AC bisect ∠A as well as ∠C and diagonal BD bisect ∠B as well as ∠D.Solution 16

In ΔABC and ΔADC,

AB = AD (sides of a rhombus are equal)

BC = CD (sides of a rhombus are equal)

AC = AC (common)

∴ ΔABC ≅ ΔADC (by SSS congruence criterion)

⇒ ∠BAC = ∠DAC and ∠BCA = ∠DCA (c.p.c.t.)

⇒ AC bisects ∠A as well as ∠C.

Similarly,

In ΔBAD and ΔBCD,

AB = BC (sides of a rhombus are equal)

AD = CD (sides of a rhombus are equal)

BD = BD (common)

∴ ΔBAD ≅ ΔBCD (by SSS congruence criterion)

⇒ ∠ABD = ∠CBD and ∠ADB = ∠CDB (c.p.c.t.)

⇒ BD bisects ∠B as well as ∠D.Question 17

In a parallelogram ABCD, points M and N have been taken on opposite sides AB and CD respectively such that AM = CN. Show that AC and MN bisect each other.

Solution 17

In ΔAMO and ΔCNO

∠MAO = ∠NCO (AB ∥ CD, alternate angles)

AM = CN (given)

∠AOM = ∠CON (vertically opposite angles)

∴ ΔAMO ≅ ΔCNO (by ASA congruence criterion)

⇒ AO = CO and MO = NO (c.p.c.t.)

⇒ AC and MN bisect each other.Question 18

In the adjoining figure, ABCD is a parallelogram. If P and Q are points on AD and BC respectively such that  and , prove that AQCP is a parallelogram.

Solution 18

Question 19

In the adjoining figure, ABCD is a parallelogram whose diagonals intersect each other at O.A line segment EOF is drawn to meet AB at E and DC at F . Prove that OE=OF.

Solution 19

Question 20

The angle between two altitudes of a parallelogram through the vertex of an obtuse angle of the parallelogram is 60°. Find the angles of the parallelogram.Solution 20

∠DCM = ∠DCN + ∠MCN

⇒ 90° = ∠DCN + 60° 

⇒ ∠DCN = 30° 

In ΔDCN,

∠DNC + ∠DCN + ∠D = 180° 

⇒ 90° + 30° + ∠D = 180° 

⇒ ∠D = 60° 

⇒ ∠B = ∠D = 60° (opposite angles of parallelogram are equal)

⇒ ∠A = 180° – ∠B = 180° – 60° = 120° 

⇒ ∠C = ∠A = 120° 

Thus, the angles of a parallelogram are 60°, 120°, 60° and 120°.Question 21

ABCD is rectangle in which diagonal AC bisect ∠A as well as ∠C. Show that (i) ABCD is square, (ii) diagonal BD bisect ∠B as well as ∠D.Solution 21

(i) ABCD is a rectangle in which diagonal AC bisects ∠A as well as ∠C.

⇒ ∠BAC = ∠DAC ….(i)

And ∠BCA = ∠DCA ….(ii)

Since every rectangle is a parallelogram, therefore

AB ∥ DC and AC is the transversal.

⇒ ∠BAC = ∠DCA (alternate angles)

⇒ ∠DAC = ∠DCA [From (i)]

Thus, in ΔADC,

AD = CD (opposite sides of equal angles are equal)

But, AD = BC and CD = AB (ABCD is a rectangle)

⇒ AB = BC = CD = AD

Hence, ABCD is a square.

(ii) In ΔBAD and ΔBCD,

AB = CD

AD = BC

BD = BD

∴ ΔBAD ≅ ΔBCD (by SSS congruence criterion)

⇒ ∠ABD = ∠CBD and ∠ADB = ∠CDB (c.p.c.t.)

Hence, diagonal BD bisects ∠B as well as ∠D.Question 22

In the adjoining figure, ABCD is a parallelogram in which AB is produced to E so that BE= AB. Prove that ED bisects BC.

Solution 22

Question 23

In the adjoining figure , ABCD is a parallelogram and E is the midpoint of side BC. If DE and AB when produced meet at F, prove that AF=2AB.

Solution 23

Question 24

Two parallel lines l and m are intersected by a transversal t. Show that the quadrilateral formed by the bisectors of interior angles is a rectangle.Solution 24

l ∥ m and t is a transversal.

⇒ ∠APR = ∠PRD (alternate angles)

⇒ ∠SPR = ∠PRQ (PS and RQ are the bisectors of ∠APR and ∠PRD)

Thus, PR intersects PS and RQ at P and R respectively such that ∠SPR = ∠PRQ i.e., alternate angles are equal.

⇒ PS ∥ RQ

Similarly, we have SR ∥ PQ.

Hence, PQRS is a parallelogram.

Now, ∠BPR + ∠PRD = 180° (interior angles are supplementary)

⇒ 2∠QPR + 2∠QRP = 180° (PQ and RQ are the bisectors of ∠BPR and ∠PRD)

⇒ ∠QPR + ∠QRP = 90° 

In ΔPQR, by angle sum property,

∠PQR + ∠QPR + ∠QRP = 180° 

⇒ ∠PQR + 90° = 180° 

⇒ ∠PQR = 90° 

Since PQRS is a parallelogram,

∠PQR = ∠PSR

⇒ ∠PSR = 90° 

Now, ∠SPQ + ∠PQR = 180° (adjacent angles in a parallelogram are supplementary)

⇒ ∠SPQ + 90° = 180° 

⇒ ∠SPQ = 90° 

⇒ ∠SRQ = 90° 

Thus, all the interior angles of quadrilateral PQRS are right angles.

Hence, PQRS is a rectangle.Question 25

K, L, M and N are point on the sides AB, BC, CD and DA respectively of a square ABCD such that AK = BL = CM = DN. Prove that KLMN is a square.Solution 25

AK = BL = CM = DN (given)

⇒ BK = CL = DM = AN (i)(since ABCD is a square)

In ΔAKN and ΔBLK,

AK = BL (given)

∠A = ∠B (Each 90°)

AN = BK [From (i)]

∴ ΔAKN ≅ ΔBLK (by SAS congruence criterion)

⇒ ∠AKN = ∠BLK and ∠ANK = ∠BKL (c.p.c.t.)

But, ∠AKN + ∠ANK = 90° and ∠BLK + ∠BKL = 90° 

⇒ ∠AKN + ∠ANK + ∠BLK + ∠BKL = 90° + 90° 

⇒ 2∠AKN + 2∠BKL = 180° 

⇒ ∠AKN + ∠BKL = 90° 

⇒ ∠NKL = 90° 

Similarly, we have

∠KLM = ∠LMN = ∠MNK = 90° 

Hence, KLMN is a square.Question 26

A is given . if lines are drawn through A, B, C parallel respectively to the sides BC, CA and AB forming  , as shown in the adjoining figure, show that 

Solution 26

Question 27

In the adjoining figure, is a triangle and through A,B, C lines are drawn , parallel respectively to BC, CA and AB, intersecting at P, Q and R. prove that the perimeter of  is double the perimeter of  .

Solution 27

Exercise Ex. 10A

Question 1

Three angles of a quadrilateral are 75°, 90° and 75°. Find the measure of the fourth angle.Solution 1

Let the measure of the fourth angle = x° 

For a quadrilateral, sum of four angles = 360° 

⇒ x° + 75° + 90° + 75° = 360° 

⇒ x° = 360° – 240° 

⇒ x° = 120° 

Hence, the measure of fourth angle is 120°. Question 2

The angle of the quadrilateral are in the ratio 2:4:5:7. Find the angles.Solution 2

Question 3

In the adjoining figure , ABCD is a trapezium in which AB || DC. If =55⁰ and = 70⁰, find  _and .

Solution 3

Since AB || DC

Question 4

In the adjoining figure , ABCD is a square andis an equilateral triangle . Prove that

(i)AE=BE, (ii) =15⁰

Solution 4

Given:

Question 5

In the adjoining figure , BMAC and DNAC. If BM=DN, prove that AC bisects BD.

Solution 5

Question 6

In given figure, ABCD is a quadrilateral in which AB=AD and BC= DC. Prove that (i) AC bisects  and , (ii) BE=DE,

(iii) 

Solution 6

Question 7

In the given figure , ABCD is a square and _PQR=90⁰. If PB=QC= DR, prove that (i) QB=RC, (ii) PQ=QR, (iii) QPR=45⁰.

Solution 7

Question 8

If O is a point with in a quadrilateral ABCD, show that OA+OB+OC+OD> AC+BD.Solution 8

Given: O is a point within a quadrilateral ABCD

Question 9

In the adjoining figure, ABCD is a quadrilateral and AC is one of the diagonals .Prove that:

(i)AB+BC+CD+DA> 2AC

(ii)AB+BC+CD>DA

(iii)AB+BC+CD+DA>AC+BD

Solution 9

Given: ABCD is a quadrilateral and AC is one of its diagonals.

Question 10

Prove that the sum of all the angles of a quadrilateral is 360^0.Solution 10

Given: ABCD is a quadrilateral.

Exercise Ex. 10C

Question 1

P, Q, R and S are respectively the midpoints of the sides AB, BC, CD and DA of a quadrilateral ABCD. Show that

(i) PQ ∥ AC and PQ = 

(ii) PQ ∥ SR

(iii) PQRS is a parallelogram.

Solution 1

(i) In ΔABC, P and Q are the mid-points of sides AB and BC respectively.

(ii) In ΔADC, R and S are the mid-points of sides CD and AD respectively.

From (i) and (ii), we have

PQ = SR and PQ ∥ SR

(iii) Thus, in quadrilateral PQRS, one pair of opposite sides are equal and parallel.

Hence, PQRS is a parallelogram. Question 2

A square is inscribed in an isosceles right triangle so that the square and the triangle have one angle common. Show that the vertex of the square opposite the vertex of the common angle bisect the hypotenuse.Solution 2

Let ΔABC be an isosceles right triangle, right-angled at B.

⇒ AB = BC

Let PBSR be a square inscribed in ΔABC with common ∠B.

⇒ PB = BS = SR = RP

Now, AB – PB = BC – BS

⇒ AP = CS ….(i)

In ΔAPR and ΔCSR

AP = CS  [From (i)

∠APR = ∠CSR (Each 90°)

PR = SR (sides of a square)

∴ ΔAPR ≅ ΔCSR (by SAS congruence criterion)

⇒ AR = CR (c.p.c.t.)

Thus, point R bisects the hypotenuse AC.Question 3

In the adjoining figure , ABCD is a ||gm in which E and F are the midpoints of AB and CD respectively . If GH is a line segment that cuts AD, EF, and BC at G, P and H respectively, prove that GP = PH.

Solution 3

Question 4

M and N are points on opposites sides AD and BC of a parallelogram ABCD such that MN passes through the point of intersection O of its diagonals AC and BD. show that MN is bisected at O.Solution 4

In ΔAOM and ΔCON

∠MAO = ∠OCN  (Alternate angles)

AO = OC (Diagonals of a parallelogram bisect each other)

∠AOM = ∠CON  (Vertically opposite angles)

∴ ΔAOM ≅ ΔCON  (by ASA congruence criterion)

⇒ MO = NO (c.p.c.t.)

Thus, MN is bisected at point O. Question 5

In the adjoining figure, PQRS is a trapezium in which PQ ∥ SR and M is the midpoint of PS. A line segment MN ∥ PQ meets QR at N. Show that N is the midpoint of QR.

Solution 5

Construction: Join diagonal QS. Let QS intersect MN at point O.

PQ ∥ SR and MN ∥ PQ

⇒ PQ ∥ MN ∥ SR

By converse of mid-point theorem a line drawn, through the mid-point of any side of a triangle and parallel to another side bisects the third side. 

Now, in ΔSPQ

MO ∥ PQ and M is the mid-point of SP

So, this line will intersect QS at point O and O will be the mid-point of QS.

Also, MN ∥ SR

Thus, in ΔQRS, ON ∥ SR and O is the midpoint of line QS.

So, by using converse of mid-point theorem, N is the mid-point of QR.Question 6

In a parallelogram PQRS, PQ = 12 cm and PS = 9 cm. The bisector of ∠P meets SR in M. PM and QR both when produced meet at T. Find the length of RT.Solution 6

PM is the bisector of ∠P.

⇒ ∠QPM = ∠SPM ….(i)

PQRS is a parallelogram.

∴ PQ ∥ SR and PM is the transversal.

⇒ ∠QPM = ∠MS (ii)(alternate angles)

From (i) and (ii),

∠SPM = ∠PMS ….(iii)

⇒ MS = PS = 9 cm (sides opposite to equal angles are equal)

Now, ∠RMT = ∠PMS (iv)(vertically opposite angles)

Also, PS ∥ QT and PT is the transversal.

∠RTM = ∠SPM

⇒ ∠RTM = ∠RMT

⇒ RT = RM (sides opposite to equal angles are equal)

RM = SR – MS = 12 – 9 = 3 cm

⇒ RT = 3 cmQuestion 7

In the adjoining figure , ABCD is a trapezium in which AB|| DC and P, Q are the mid points of AD and BC respectively. DQ and AB when produced meet at E. Also, AC and PQ intersect at R. Prove that (i) DQ=QE, (ii) PR||AB, (iii) AR=RC.

Solution 7

Question 8

In the adjoining figure, AD is a medium of  _{and DE|| BA. Show that BE is also a median of .}

Solution 8

Question 9

_{In the adjoining figure , AD and BE are the medians of  and DF|| BE. Show that .}

Solution 9

Question 10

_{Prove that the line segments joining the middle points of the sides of a triangle divide it into four congruent triangles.}Solution 10

Question 11

_{In the adjoining figure, D,E,F are the midpoints of the sides BC, CA, and AB respectively, of . Show that .}

Solution 11

Question 12

Show that the quadrilateral formed by joining the mid points of the pairs of adjacent sides of a rectangle is a rhombus.Solution 12

Question 13

Show that the quadrilateral formed by joining the mid points of the pairs of adjacent sides of a rhombus is a rectangle.Solution 13

Question 14

Show that the quadrilateral formed by joining the mid points of the pairs of adjacent sides of a square is a square.Solution 14

Question 15

Prove that the line segment joining the midpoints of opposite sides of a quadrilateral bisect each other.Solution 15

Question 16

The diagonals of a quadrilateral ABCD are equal. Prove that the quadrilateral formed by joining the midpoints of its sides is a rhombus.Solution 16

In ΔABC, P and Q are the mid-points of sides AB and BC respectively.

In ΔBCD, Q and R are the mid-points of sides BC and CD respectively.

In ΔADC, S and R are the mid-points of sides AD and CD respectively.

In ΔABD, P and S are the mid-points of sides AB and AD respectively.

⇒ PQ || RS and QR || SP [From (i), (ii), (iii) and (iv)]

Thus, PQRS is a parallelogram.

Now, AC = BD (given)

⇒ PQ = QR = RS = SP  [From (i), (ii), (iii) and (iv)]

Hence, PQRS is a rhombus.Question 17

The diagonals of a quadrilateral ABCD are perpendicular to each other. Prove that the quadrilateral formed by joining the midpoints of its sides is a rectangle.Solution 17

In ΔABC, P and Q are the mid-points of sides AB and BC respectively.

In ΔADC, R and S are the mid-points of sides CD and AD respectively.

From (i) and (ii),

PQ || RS and PQ = RS

Thus, in quadrilateral PQRS, a pair of opposite sides are equal are parallel.

So, PQRS is a parallelogram.

Let the diagonals AC and BD intersect at O.

Now, in ΔABD, P and S are the mid-points of sides AB and AD respectively.

Thus, in quadrilateral PMON, PM || NO and PN || MO.

⇒ PMON is a parallelogram.

⇒ ∠MPN = ∠MON (opposite angles of a parallelogram are equal)

⇒ ∠MPN = ∠BOA (since ∠BOA = ∠MON)

⇒ ∠MPN = 90° (since AC ⊥ BD, ∠BOA = 90°)

⇒ ∠QPS = 90° 

Thus, PQRS is a parallelogram whose one angle, i.e. ∠QPS = 90°.

Hence, PQRS is a rectangle. Question 18

The midpoints of the sides AB, BC, CD and DA of a quadrilateral ABCD are joined to form a quadrilateral. If AC = BD and AC ⊥ BD then prove that the quadrilateral formed is a square.Solution 18

In ΔABC, P and Q are the mid-points of sides AB and BC respectively.

In ΔBCD, Q and R are the mid-points of sides BC and CD respectively.

In ΔADC, S and R are the mid-points of sides AD and CD respectively.

In ΔABD, P and S are the mid-points of sides AB and AD respectively.

⇒ PQ || RS and QR || SP [From (i), (ii), (iii) and (iv)]

Thus, PQRS is a parallelogram.

Now, AC = BD (given)

⇒ PQ = QR = RS = SP [From (i), (ii), (iii) and (iv)]

Let the diagonals AC and BD intersect at O.

Now,

Thus, in quadrilateral PMON, PM || NO and PN || MO.

⇒ PMON is a parallelogram.

⇒ ∠MPN = ∠MON (opposite angles of a parallelogram are equal)

⇒ ∠MPN = ∠BOA (since ∠BOA = ∠MON)

⇒ ∠MPN = 90° (since AC ⊥ BD, ∠BOA = 90°)

⇒ ∠QPS = 90° 

Thus, PQRS is a parallelogram such that PQ = QR = RS = SP and ∠QPS = 90°.

Hence, PQRS is a square.

Exercise MCQ

Question 1

Out of the following given figures which are on the same base but not between the same parables?

Solution 1

Question 2

In which of the following figures, you find polynomials on the same base and between the same parallels?

Solution 2

Question 3

The median of a triangle divides it into two

1. triangles of equal area
2. congruent triangles
3. isosceles triangle
4. right triangles

Solution 3

Question 4

The area of quadrilateral ABCD in the given figure is

1. 57 cm²
2. 108 cm²
3. 114 cm²
4. 195 cm²

Solution 4

Question 5

The area of trapezium ABCD in the given figure is

1. 62 cm²
2. 93 cm²
3. 124 cm²
4. 155 cm²

Solution 5

Question 6

In the given figure, ABCD is a ∥gm in which AB = CD = 5 cm and BD ⊥ DC such that BD = 6.8 cm, Then the area of ‖gm ABCD = ?

1. 17 cm²
2. 25 cm²
3. 34 cm²
4. 68 cm²

Solution 6

Question 7

In the given figure, ABCD is a ∥gm in which diagonals Ac and BD intersect at O. If ar(‖gm ABCD) is 52 cm², then the ar(ΔOAB)=?

1. 26 cm²
2. 18.5 cm²
3. 39 cm²
4. 13 cm²

Solution 7

Question 8

In the given figure, ABCD is a ∥gm in which DL ⊥ AB, If AB = 10 cm and DL = 4 cm, then the ar(‖gm ABCD) = ?

1. 40 cm²
2. 80 cm²
3. 20 cm²
4. 196 cm²

Solution 8

Question 9

The area of ∥gm ABCD is

(a) AB × BM

(b) BC × BN

(c) DC × DL

(d) AD × DL

Solution 9

Correct option: (c)

Area of ∥gm ABCD = Base × Height = DC × DL Question 10

Two parallelograms are on equal bases and between the same parallels. The ratio of their areas is

(a) 1 : 2

(b) 1 : 1

(c) 2 : 1

(d) 3 : 1Solution 10

Correct option: (b)

Parallelograms on equal bases and between the same parallels are equal in area. Question 11

In the given figure, ABCD and ABPQ are two parallelograms and M is a point on AQ and BMP is a triangle.

(a) true

(b) false

Solution 11

Correct option: (a)

ΔBMP and parallelogram ABPQ are on the same base BP and between the same parallels AQ and BP.

Parallelograms ABPQ and ABCD are on the same base AB and between the same parallels AB and PD.

Question 12

The midpoints of the sides of a triangle along with any of the vertices as the fourth point makes a parallelogram of area equal to

(a) 

(b) 

(c) 

(d) ar(ΔABC)

Solution 12

Correct option: (a)

ΔABC is divided into four triangles of equal area.

A(parallelogram AFDE) = A(ΔAFE) + A(DFE)

Question 13

The lengths of the diagonals of a rhombus are 12 cm and 16 cm. The area of the rhombus is

1. 192 cm²
2. 96 cm²
3. 64 cm²
4. 80 cm²

Solution 13

Question 14

Two parallel sides of a trapezium are 12 cm and 8 cm long and the distance between them is 6.5 cm. The area of the trapezium is

1. 74 cm²
2. 32.5 cm²
3. 65 cm²
4. 130 cm²

Solution 14

Question 15

In the given figure ABCD is a trapezium such that AL ⊥ DC and BM ⊥ DC. If AB = 7 cm, BC = AD = 5 cm and AL = BM = 4 cm, then ar(trap. ABCD)= ?

1. 24 cm²
2. 40 cm²
3. 55 cm²
4. 27.5 cm²

Solution 15

Question 16

In a quadrilateral ABCD, it is given that BD = 16 cm. If AL ⊥ BD and CM ⊥ BD such that AL = 9 cm and CM = 7 cm, then ar(quad. ABCD) = ?

1. 256 cm²
2. 128 cm²
3. 64 cm²
4. 96 cm²

Solution 16

Question 17

ABCD is a rhombus in which ∠C = 60°.

Then, AC : BD = ?

Solution 17

Question 18

In the given figure ABCD and ABFE are parallelograms such that ar(quad. EABC) = 17cm² and ar(‖gm ABCD) = 25 cm². Then, ar(ΔBCF) = ?

1. 4 cm²
2. 4.8 cm²
3. 6 cm²
4. 8 cm²

Solution 18

Question 19

ΔABC and ΔBDE are two equilateral triangles such that D is the midpoint of BC. Then, ar(ΔBDE) : ar(ΔABC) = ?

(a) 1 : 2

(b) 1 : 4

(c) 

(d) 3 : 4Solution 19

Question 20

In a ‖gm ABCD, if P and Q are midpoints of AB and CD respectively and ar(‖gm ABCD) = 16 cm², then ar(‖gm APQD) = ?

1. 8 cm²
2. 12 cm²
3. 6 cm²
4. 9 cm²

Solution 20

Question 21

The figure formed by joining the midpoints of the adjacent sides of a rectangle of sides 8 cm and 6 cm is a

1. rectangle of area 24 cm²
2. square of area 24 cm²
3. trapezium of area 24 cm²
4. rhombus of area 24 cm²

Solution 21

Question 22

In ΔABC, if D is the midpoint of BC and E is the midpoint of AD, then ar(ΔBED) = ?

Solution 22

Question 23

The vertex A of ΔABC is joined to a point D on BC. If E is the midpoint of AD, then ar(ΔBEC) = ?

Solution 23

Question 24

In ΔABC, it is given that D is the midpoint of BC; E is the midpoint of BD and O is the midpoint of AE. Then, ar(ΔBOE) = ?

Solution 24

Question 25

If a triangle and a parallelogram are on the same base and between the same parallels, then the ratio of the area of the triangle to the area of the parallelogram is

1. 1 : 2
2. 1 : 3
3. 1 : 4
4. 3 : 4

Solution 25

Question 26

In the given figure ABCD is a trapezium in which AB ‖ DC such that AB = a cm and DC = b cm, If E and F are the midpoints of AD and BC respectively. Then, ar(ABFE) : ar(EFCD) = ?

1. a : b
2. (a + 3b) : (3a + b)
3. (3a + b) : (a + 3b)
4. (2a + b) : (3a + b)

Solution 26

Question 27

ABCD is a quadrilateral whose diagonal AC divides it into two parts, equal in area, then ABCD is

1. a rectangle
2. a ‖gm
3. a rhombus
4. all of these

Solution 27

Question 28

In the given figure, a ‖gm ABCD and a rectangle ABEF are of equal area, Then,

1. perimeter of ABCD = perimeter of ABEF
2. perimeter of ABCD < perimeter of ABEF
3. perimeter of ABCD > perimeter of ABEF
4. 

Solution 28

Question 29

In the given figure, ABCD is a rectangle inscribed in a quadrant of a circle of radius 10 cm. If AD = cm, then area of the rectangle is

1. 32 cm²
2. 40 cm²
3. 44 cm²
4. 48 cm²

Solution 29

Question 30

Which of the following is a false statement?

1. A median of a triangle divides it into two triangles of equal areas.
2. The diagonals of a ∥gm divide it into four triangles of equal areas.
3. In a ΔABC, if E is the midpoint of median AD, then ar(ΔBED) =

4. In a trap. ABCD, it is given that AB ‖ DC and the diagonals AC and BD intersect at O. Then, ar(ΔAOB) = ar(ΔCOD).

Solution 30

Question 31

Which of the following is a false statement?

1. If the diagonals of a rhombus are 18 cm and 14 cm, then its area is 126 cm².
2. 
3. A parallelogram and a rectangle on the same base and between the same parallels are equal in area.
4. If the area of a ‖gm with one side 24 cm and corresponding height h cm is 192 cm², then h = 8 cm.

Solution 31

Question 32

Look at the statements given below:

1. A parallelogram and a rectangle on the same base and between the same parallels are equal in area.
2. In a ‖gm ABCD, it is given that AB = 10 cm. The altitudes DE on AB and BF on AD being 6 cm and 8 cm respectively, then AD = 7.5 cm.

3. 

Which is true?

1. I only
2. II only
3. I and II
4. II and III

Solution 32

Question 33

Each question consists of two statements, namely, Assertion (A) and Reason (R). For selecting the correct answer, use the following code:

1. Both Assertion (A) and Reason (R) are true and Reason (R) is a correct explanation of Assertion (A).
2. Both Assertion (A) and Reason (R) are true but Reason (R) is not a correct explanation of Assertion (A).
3. Assertion (A) is true and Reason (R) is false.
4. Assertion (A) is false and Reason (R) is true.

Assertion (A)	Reason (R)
In a trapezium ABCD we have AB ‖ DC and the diagonals AC and BD intersect at O.Then, ar(ΔAOD) = ar(ΔBOC)	Triangles on the same base and between the same parallels are equal in areas.

The correct answer is: (a) / (b) / (c) / (d).Solution 33

Question 34

Each question consists of two statements, namely, Assertion (A) and Reason (R). For selecting the correct answer, use the following code:

1. Both Assertion (A) and Reason (R) are true and Reason (R) is a correct explanation of Assertion (A).
2. Both Assertion (A) and Reason (R) are true but Reason (R) is not a correct explanation of Assertion (A).
3. Assertion (A) is true and Reason (R) is false.
4. Assertion (A) is false and Reason (R) is true.

Assertion (A)	Reason (R)
If ABCD is a rhombus whose one angle is 60°, then the ratio of the lengths of its diagonals is .	Median of a triangle divides it into two triangles of equal area.

The correct answer is: (a) / (b) / (c) / (d).Solution 34

Question 35

Each question consists of two statements, namely, Assertion (A) and Reason (R). For selecting the correct answer, use the following code:

1. Both Assertion (A) and Reason (R) are true and Reason (R) is a correct explanation of Assertion (A).
2. Both Assertion (A) and Reason (R) are true but Reason (R) is not a correct explanation of Assertion (A).
3. Assertion (A) is true and Reason (R) is false.
4. Assertion (A) is false and Reason (R) is true.

Assertion (A)	Reason (R)
The diagonals of a ‖gm divide it into four triangles of equal area.	A diagonal of a ‖gm divides it into two triangles of equal area.

The correct answer is: (a) / (b) / (c) / (d).Solution 35

Question 36

Each question consists of two statements, namely, Assertion (A) and Reason (R). For selecting the correct answer, use the following code:

1. Both Assertion (A) and Reason (R) are true and Reason (R) is a correct explanation of Assertion (A).
2. Both Assertion (A) and Reason (R) are true but Reason (R) is not a correct explanation of Assertion (A).
3. Assertion (A) is true and Reason (R) is false.
4. Assertion (A) is false and Reason (R) is true.

Assertion (A)	Reason (R)
The area of a trapezium whose parallel sides measure 25 cm and 15 cm respectively and the distance between them is 6 cm, is 120 cm2.

The correct answer is: (a) / (b) / (c) / (d).Solution 36

Question 37

Each question consists of two statements, namely, Assertion (A) and Reason (R). For selecting the correct answer, use the following code:

1. Both Assertion (A) and Reason (R) are true and Reason (R) is a correct explanation of Assertion (A).
2. Both Assertion (A) and Reason (R) are true but Reason (R) is not a correct explanation of Assertion (A).
3. Assertion (A) is true and Reason (R) is false.
4. Assertion (A) is false and Reason (R) is true.

Assertion (A)	Reason (R)
In the given figure, ABCD is a ‖gm in which DE ⊥ AB and BE ⊥ AD. If AB = 16 cm, DE = 8cm and BF = 10cm, then AD is 12 cm.	Area of a ‖gm = base × height.

The correct answer is: (a) / (b) / (c) / (d).Solution 37

Exercise Ex. 11A

Question 2

In the adjoining figure, show that ABCD is a parallelogram. Calculate the area of ||gm ABCD.

Solution 2

Question 3

In a parallelogram ABCD, it is being given that AB= 10 cm and the altitudes corresponding to the sides AB and AD are DL =6 cm and BM =8cm, respectively Find AD.

Solution 3

Question 5

Find the area of the trapezium whose parallel sides are 9 cm and 6 cm respectively and the distance between these sides is 8 cm.Solution 5

Question 6(ii)

Calculatethe area of trapezium PQRS , givenin Fig.(ii)

Solution 6(ii)

Question 6(i)

Calculate the area of quadrilateral ABCD givenin Fig (i)

Solution 6(i)

Question 8

BD is one of the diagonals of a quad. ABCD . If , show that 

Solution 8

Question 10

In the adjoining figure , ABCD is a quadrilateral in which diag. BD =14cm . If  such that AL=8 cm. and CM =6 cm, find the area of quadrilateral ABCD.

Solution 10

Question 13

In the adjoining figure , ABCD is a trapezium in which AB||DC and its diagonals AC and BD intersects at O. prove that 

Solution 13

Question 14

_{In the adjoining figure , DE||BC. Prove that} 

Solution 14

Question 15

Prove that the median divides a triangle into two triangles of equal area.Solution 15

Question 16

Show that the diagonal divides a parallelogram into two triangles of equal area.Solution 16

Question 20

_{In adjoining figure, the diagonals AC and BD of a quadrilateral ABCD intersect at O. If BO=OD, prove that}

Solution 20

Question 21

The vertex A of  _{is joined to a point D on the side BC. The mid-point of AD is E. prove that}

Solution 21

Question 22

_{D is the mid-point of side BC of   and E is the midpoint of BD. If O is midpoint of AE, prove that} 

Solution 22

Question 24

_{In the adjoining figure , ABCD is a quadrilateral. A line through D , parallel to AC , meets BC produced in P. prove that} 

Solution 24

Question 25

In the adjoining figure ,  _{are on the same base BC with A and D on opposite sides of BC such that . Show that BC bisects AD.}

Solution 25

Question 28

_{P, Q, R, S are respectively the midpoints of the sides AB, BC, CD and DA of ||gm ABCD. Show that PQRS is a parallelogram and also that}

_{Ar(||gm PQRS)=x ar (||gm ABCD)}

Solution 28

Question 30

The base BC of   _{is divided at D such that .}

_{Prove that .}Solution 30

Question 32

_{The given figure shows the pentagon ABCDE. EG, drawn parallel to DA, meetsBA producedat G ,andCF, drawnparallel to DB, meets AB produced at F.}

_Showthat 

Solution 32

Question 34

_{In the adjoining figure , the point D divides the side BC of   in the ratio m:n. Prove that} 

Solution 34

Exercise Ex. 11

Question 1

Which of the following figures lie on the same base and between the same parallels. In such a case, write the common base and the two parallels.

Fig (i)

Fig (ii)

Fig (iii)

Fig (iv)

Fig (v)

Fig (vi)

Solution 1

Following figures lie on the same base and between the same parallels:

Figure (i): No

Figure (ii): No

Figure (iii): Yes, common base – AB, parallel lines – AB and DE

Figure (iv): No

Figure (v): Yes, common base – BC, parallel lines – BC and AD

Figure (vi): Yes, common base – CD, parallel lines – CD and BPQuestion 4

Find the area of a figure formed by joining the midpoints of the adjacent sides of a rhombus with diagonals 12 cm and 16 cm.Solution 4

Question 7

In the adjoining figure, ABCD is a trapezium in which AB ∥ DC; AB = 7 cm; AD = BC = 5 cm and the distance between AB and DC is 4 cm. Find the length of DC and hence, find the area of trap. ABCD.

Solution 7

Question 9

M is the midpoint of the side AB of a parallelogram ABCD. If ar(AMCD) = 24 cm², find ar(Δ ABC).Solution 9

Construction: Join AC

Diagonal AC divides the parallelogram ABCD into two triangles of equal area.

⇒ A(ΔADC) = A(ΔABC) ….(i)

ΔADC and parallelogram ABCD are on the same base CD and between the same parallel lines DC and AM.

Since M is the mid-point of AB,

A(AMCD) = A(ΔADC) + A(ΔAMC)

Question 11

If P and Q are any two points lying respectively on the sides DC and AD of a parallelogram ABCD then show that ar(ΔAPB) = ar(ΔBQC).Solution 11

Since ΔAPB and parallelogram ABCD are on the same base AB and between the same parallels AB and DC, we have

Similarly, ΔBQC and parallelogram ABCD are on the same base BC and between the same parallels BC and AD, we have 

From (i) and (ii),

A(ΔAPB) = A(ΔBQC) Question 12

In the adjoining figure, MNPQ and ABPQ are parallelograms and T is any point on the side BP. Prove that

Solution 12

(i) Parallelograms MNPQ and ABPQ are on the same base PQ and between the same parallels PQ and MB.

(ii) ΔATQ and parallelogram ABPQ are on the same base AQ and between the same parallels AQ and BP. 

Question 17

In the adjoining figure, ABC and ABD are two triangles on the same base AB. If line segment CD is bisected by AB at O, show that

ar(Δ ABC) = ar(Δ ABD).

Solution 17

We know that median of a triangle divides it into two triangles of equal area.

Now, AO is the median of ΔACD.

⇒ A(ΔCOA) = A(ΔDOA) ….(i)

And, BO is the median of ΔBCD.

⇒ A(ΔCOB) = A(ΔDOB) ….(ii)

Adding (i) and (ii), we get

A(ΔCOA) + A(ΔCOB) = A(ΔDOA) + A(ΔDOB)

⇒ A(ΔABC) = A(ΔABD)Question 18

D and E are points on sides AB and AC respectively of Δ ABC such that ar(ΔBCD) = ar(ΔBCE). Prove that DE ∥ BC.Solution 18

Since ΔBCD and ΔBCE are equal in area and have a same base BC.

Therefore,

Altitude from D of ΔBCD = Altitude from E of ΔBCE

⇒ ΔBCD and ΔBCE are between the same parallel lines.

⇒ DE ∥ BC Question 19

P is any point on the diagonal AC of parallelogram ABCD. Prove that ar(ΔADP) = ar(ΔABP).Solution 19

Construction: Join BD.

Let the diagonals AC and BD intersect at point O.

Diagonals of a parallelogram bisect each other.

Hence, O is the mid-point of both AC and BD.

We know that the median of a triangle divides it into two triangles of equal area.

In ΔABD, OA is the median.

⇒ A(ΔAOD) = A(ΔAOB) ….(i)

In ΔBPD, OP is the median.

⇒ A(ΔOPD) = A(ΔOPB) ….(ii)

Adding (i) and (ii), we get

A(ΔAOD) + A(ΔOPD) = A(ΔAOB) + A(ΔOPB)

⇒ A(ΔADP) = A(ΔABP)Question 23

In a trapezium ABCD, AB ∥ DC and M is the midpoint of BC. Through M, a line PQ ∥ AD has been drawn which meets AB in P and DC produced in Q, as shown in the adjoining figure. Prove that ar(ABCD) = ar(APQD).

Solution 23

In ΔMCQ and ΔMPB,

∠QCM = ∠PBM (alternate angles)

CM = BM (M is the mid-point of BC) 

∠CMQ = ∠PMB (vertically opposite angles)

∴ ΔMCQ ≅ ΔMPB

⇒ A(ΔMCQ) = A(ΔMPB)

Now,

A(ABCD) = A(APQD) + A(DMPB) – A(ΔMCQ)

⇒ A(ABCD) = A(APQD)Question 26

ABCD is a parallelogram in which BC is produced to P such that CP = BC, as shown in the adjoining figure. AP intersect CD at M. If ar(DMB) = 7 cm², find the area of parallelogram ABCD.

Solution 26

In ΔADM and ΔPCM,

∠ADM = ∠PCM (alternate angles)

AD = CP (AD = BC = CP) 

∠AMD = ∠PMC (vertically opposite angles)

∴ ΔADM ≅ ΔPCM

⇒ A(ΔADM) = A(ΔPCM)

And, DM = CM (c.p.c.t.)

⇒ BM is the median of ΔBDC.

⇒ A(ΔDMB) = A(ΔCMB) 

⇒ A(ΔBDC) = 2 × A(ΔDMB) = 2 × 7 = 14 cm²

Now,

A(parallelogram ABCD) = 2 × A(ΔBDC) = 2 × 14 = 28 cm² Question 27

In a parallelogram ABCD, any point E is taken on the side BC. AE and DC when produced meet at a point M. Prove that

ar(ΔADM) – ar(ABMC)Solution 27

Construction: Join AC and BM

Let h be the distance between AB and CD.

Question 29

In a triangle ABC, the medians BE and CF intersect at G. Prove that ar(ΔBCG) = ar(AFGE).Solution 29

Construction: Join EF

Since the line segment joining the mid-points of two sides of a triangle is parallel to the third side,

FE ∥  BC

Clearly, ΔBEF and ΔCEF are on the same base EF and between the same parallel lines.

∴ A(ΔBEF) = A(ΔCEF)

⇒ A(ΔBEF) – A(ΔGEF) = A(ΔCEF) – A(ΔGEF)

⇒ A(ΔBFG) = A(ΔCEG) …(i)

We know that a median of a triangle divides it into two triangles of equal area.

⇒ A(ΔBEC) = A(ΔABE)

⇒ A(ΔBGC) + A(ΔCEG) = A(quad. AFGE) + A(ΔBFG)

⇒ A(ΔBGC) + A(ΔBFG) = A(quad. AFGE) + A(ΔBFG) [Using (i)]

⇒ A(A(ΔBGC) = A(quad. AFGE)Question 31

Solution 31

ΔDBC and ΔEBC are on the same base and between the same parallels.

⇒ A(ΔDBC) = A(ΔEBC) ….(i)

BE is the median of ΔABC.

Question 33

In the adjoining figure, CE ∥ AD and CF ∥ BA. Prove that ar(ΔCBG) = ar(ΔAFG).

Solution 33

ΔBCF and ΔACF are on the same base CF and between the same parallel lines CF and BA.

∴ A(ΔBCF) = A(ΔACF)

⇒ A(ΔBCF) – A(ΔCGF) = A(ΔACF) – A(ΔCGF)

⇒ A(ΔCBG) = A(ΔAFG)Question 35

In a trapezium ABCD, AB ∥ DC, AB = a cm, and DC = b cm. If M and N are the midpoints of the nonparallel sides, AD and BC respectively then find the ratio of ar(DCNM) and ar(MNBA).

Solution 35

Construction: Join DB. Let DB cut MN at point Y.

M and N are the mid-points of AD and BC respectively.

⇒ MN ∥ AB ∥ CD

In ΔADB, M is the mid-point of AD and MY ∥ AB.

∴ Y is the mid-point of DB.

Similarly, in ΔBDC,

Now, MN = MY + YN

Construction: Draw DQ ⊥ AB. Let DQ cut MN at point P.

Then, P is the mid-point of DQ.

i.e. DP = PQ = h (say)

Question 36

ABCD is a trapezium in which AB ∥ DC, AB = 16 cm and DC = 24 cm. If E and F are respectively the midpoints of AD and BC, prove that Solution 36

Construction: Join AC. Let AC cut EF at point Y.

E and F are the mid-points of AD and BC respectively.

⇒ EF ∥ AB ∥ CD

In ΔADC, E is the mid-point of AD and EY ∥ CD.

∴ Y is the mid-point of AC.

Similarly, in ΔABC,

Now, EF = EY + YF

Construction: Draw AQ ⊥ DC. Let AQ cut EF at point P.

Then, P is the mid-point of AQ.

i.e. AP = PQ = h (say)

Question 37

In the adjoining figure, D and E are respectively the midpoints of sides AB and AC of ΔABC. If PQ ∥ BC and CDP and BEQ are straight lines then prove that ar(ΔABQ) = ar(ΔACP).

Solution 37

Since D and E are the mid-points of AB and AC respectively,

DE ∥ BC ∥ PQ

In ΔACP, AP ∥ DE and E is the mid-point of AC.

⇒ D is the mid-point of PC (converse of mid-point theorem)

In ΔABQ, AQ ∥ DE and D is the mid-point of AB.

⇒ E is the mid-point of BQ (converse of mid-point theorem)

From (i) and (ii),

AP = AQ

Now, ΔACP and ΔABQ are on the equal bases AP and AQ and between the same parallels BC and PQ.

⇒ A(ΔACP) = A(ΔABQ)Question 38

In the adjoining figure, ABCD and BQSC are two parallelograms. Prove that ar(ΔRSC) = ar(ΔPQB).

Solution 38

In ΔRSC and ΔPQB,

∠CRS = ∠BPQ (RC ∥ PB, corresponding angles)

∠RSC = ∠PQB (RC ∥ PB, corresponding angles)

SC = QB (opposite sides of a parallelogram BQSC)

∴ ΔRSC ≅ ΔPQB (by AAS congruence criterion)

⇒ A(ΔRSC) = A(ΔPQB)