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Chapter 20 – Surface Areas and Volume of A Right Circular Cone Exercise Ex. 20.1

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6Find its curved surface area of a cone with base radius 5.25 cm and slant height 10 cm.Solution 6

Question 7Find the total surface area of a cone, if its slant height is 21 m and diameter of its base is 24 m.Solution 7

Question 8

Solution 8

Question 9The curved surface area of a cone is 4070 cm² and its diameter is 70 cm. What is its slant height? (Use  = 22/7).Solution 9

Question 10

Solution 10

Question 11A joker’s cap is in the form of 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 11

Question 12Find the ratio of the curved surface area of two cones if their diameters of the bases are equal and slant heights are in the ratio 4:3.Solution 12

Question 13

Solution 13

Question 14

Solution 14

Question 15Curved 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 15(i)    Slant height of cone = 14 cm
    Let radius of circular end of cone be r.
    CSA of cone = 

    Thus, the radius of circular end of the cone is 7 cm.

(ii)    Total surface area of cone = CSA of cone + Area of base
                     =

Thus, the total surface area of the cone is 462 .

Question 16The slant height and base diameter of a conical tomb are 25 m and 14 m respectively. Find the cost of white-washing its curved surface at the rate of Rs 210 per 100 m².Solution 16

Question 17A conical tent is 10 m high and the radius of its base is 24 m. Find slant height of the tent. If the cost of 1 m² canvas is Rs 70, find the cost of the canvas required to make the tent,Solution 17

Question 18

Solution 18

Question 19

What length of tarpaulin 3 m wide will be required to make conical tent of height 8 m and base radius 6 m? Assume that the extra length of material that will be required for stitching margins and wastage in cutting is approximately 20 cm. (Use  = 3.14).

Solution 19Height (h) of conical tent = 8 m
    Radius (r) of base of tent = 6 m
    Slant height (l) of tent = 
    CSA of conical tent =  = (3.14  6  10)  = 188.4 

Let length of tarpaulin sheet required be L.
As 20 cm will be wasted so, effective length will be (L – 0.2 m)
Breadth of tarpaulin = 3 m
Area of sheet = CSA of tent
    [(L – 0.2 m)  3] m = 188.4 
    L – 0.2 m = 62.8 m
    L = 63 m

    Thus, the length of the tarpaulin sheet will be 63 m.Question 20

Solution 20

Question 21

A cylinder and a cone have equal radii of their bases and equal heights. If their curved surface areas are in the ratio 8 : 5, show that the radius of each to the height of each is 3:4.Solution 21

Question 22

Solution 22

Question 23

Solution 23

Chapter 20 – Surface Areas and Volume of A Right Circular Cone Exercise Ex. 20.2

Question 1Find the volume of the right circular cone with

(i) radius 6 cm, height 7 cm

(ii) radius 3.5 cm, height 12 cm

(iii) height 21 cm and slant height 28 cm.Solution 1(i)    Radius (r) of cone = 6 cm
       Height (h) of cone = 7 cm
       Volume of cone 

(ii)   Radius (r) of cone = 3.5 cm
       Height (h) of cone = 12 cm
       Volume of cone  
(iii)

Question 2Find the capacity in litres of a conical vessel with
(i)    radius 7 cm, slant height 25 cm
(ii)   height 12 cm, slant height 13 cm

Solution 2(i)    Radius (r) of cone = 7 cm
       Slant height (l) of cone = 25 cm
       Height (h) of cone  
       Volume of cone  
       Capacity of the conical vessel =  litres= 1.232 litres(ii)    Height (h) of cone = 12 cm
        Slant height (l) of cone = 13 cm        
        Radius (r) of cone 
        Volume of cone   = 314.28 cm³
        Capacity of the conical vessel = litres =  litres.

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

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

Question 10

Solution 10

Question 11A right angled triangle of which the sides containing the right angle are 6.3 cm and 10 cm in length, is made to turn round on the longer side. Find the volume of the solid, thus generated. Also, find its curved surface area.
Solution 11

Question 12

Solution 12

Question 13The volume of a right circular cone is 9856 cm³. If the diameter of the base is 28 cm, find

(i) height of the cone

(ii) slant height of the cone

(iii) curved surface area of the coneSolution 13(i)    Radius of cone =   =14 cm
       Let height of cone be h.
       Volume of cone = 9856 cm³
        h = 48 cm       Thus, the height of the cone is 48 cm. (ii)   Slant height (l) of cone  
       Thus, the slant height of the cone is 50 cm. (iii)    CSA of cone = rl = Question 14

A conical pit of top diameter 3.5 m is 12 m deep. What is its capacity in kilolitres?Solution 14Radius (r) of pit =
Depth (h) of pit = 12 m
Volume of pit =Capacity of the pit = (38.5  1) kilolitres = 38.5 kilolitres

Question 15Monica has a piece of canvas whose area is 551 m². She uses it to have a conical tent made, with a base radius of 7 m. Assuming that all the stitching margins and the wastage incurred while cutting, amounts to approximately 1 m², find the volume of the tent that can be made with it.Solution 15

Chapter 19 – Surface Areas and Volume of a Circular Cylinder Exercise Ex. 19.1

Question 1Curved surface area of a right circular cylinder is 4.4 m². If the radius of the base of the cylinder is 0.7 m, find its height.

Solution 1

Question 2In a hot 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 2 Height (h) of cylindrical pipe = Length of cylindrical pipe = 28 m
  Radius (r) of circular end of pipe =  cm = 2.5 cm = 0.025 m
  CSA of cylindrical pipe =  = 4.4 
    Thus, the area of radiating surface of the system is 4.4 m² or 44000 cm².

Question 3A cylindrical pillar is 50 cm in diameter and 3.5 m in height. Find the cost of painting the curved surface of the pillar at the rate of Rs.12.50 per m².Solution 3Height of the pillar = 3.5 m
Radius of the circular end of the pillar = cm = 25 cm  = 0.25 m
CSA of pillar =  = 
Cost of painting 1  area = Rs 12.50
Cost of painting 5.5  area = Rs (5.5 12.50) = Rs 68.75

Thus, the cost of painting the CSA of pillar is Rs 68.75.

Question 4It is required to make a closed cylindrical tank of height 1 m and base diameter 140 cm from a metal sheet. How many square meters of the sheet are required for the same?   Solution 4Height (h) of cylindrical tank = 1 m.
    Base radius (r) of cylindrical tank =  = 70 cm = 0.7 m
Area of sheet required = total surface area of tank = 

So, it will require 7.48 of metal sheet.

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

The inner diameter of a circular well is 3.5 m. It is 10 m deep. Find (i)    Its inner curved surface area,
 (ii)    The cost of plastering this curved surface at the rate of Rs 40 per m²

Solution 8Inner radius (r) of circular well = 1.75 m
Depth (h) of circular well = 10 m (i) Inner curved surface area = 

          = (44 x 0.25 x 10) 
          = 110 m²(ii) Cost of plastering 1 m² area = Rs 40                    
    Cost of plastering 110 m² area = Rs (110 x 40) = Rs 4400

Question 9The students of a Vidyalaya were asked to participate in a competition for making and decorating penholders in the shape of a cylinder with a base, using cardboard. Each penholder was to be of radius 3 cm and height 10.5 cm. The Vidyalaya was to supply the competitors with cardboard. If there were 35 competitors, how much cardboard was required to be bought for the competition?   Solution 9Radius of circular end of cylindrical penholder = 3 cm
Height of penholder = 10.5 cm
Surface area of 1 penholder = CSA of penholder + Area of base of     SA of 1 penholder =  + 
Area of cardboard sheet used by 1 competitor = 
    Area of cardboard sheet used by 35 competitors
 = 7920 cm2
    Thus, 7920 cm² of cardboard sheet will be required for the competition.

Question 10

Solution 10

Question 11

Twenty cylindrical pillars of the Parliament House are to be cleaned. If the diameter of each pillar is 0.50 m and height is 4 m. What will be the cost of cleaning them at the rate of Rs 2.50 per square metre?Solution 11

Question 12

Solution 12

Question 13

The total surface area of a hollow metal cylinder open at both ends of external radius 8 cm and height 10 cm is 338 cm². Taking r to be inner radius, obtain an equation in r and use it to obtain the thickness of the metal in the cylinder.Solution 13

Question 14

Find the lateral or curved surface area of a cylindrical petrol storage tank that is 4.2 m in diameter and 4.5 m high. How much steel was actually used, if  of the steel actually used was wasted in making the closed tank?Solution 14Height (h) cylindrical tank = 4.5 m
Radius (r) of circular end of cylindrical tank =m = 2.1m
    (i)    Lateral or curved surface area of tank = 
                                    = 
                                    = 59.4 m2            

    (ii)    Total surface area of tank = 2 (r + h)
              = 
              = 87.12 m²

    Let A m² steel sheet be actually used in making the tank.


Thus, 95.04  steel was used in actual while making the tank.    

Chapter 19 – Surface Areas and Volume of a Circular Cylinder Exercise Ex. 19.2

Question 1A soft drink is available in two packs –

(i) a tin can with a rectangular base of length 5 cm and width 4 cm, having a height of 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 1The tin can will be cuboidal in shape.Length (l) of tin can = 5 cm
Breadth (b) of tin can = 4 cm
Height (h) of tin can = 15 cm
Capacity of tin can = l  b  h = (5  4  15) cm³ = 300 cm³Radius (R) of circular end of plastic cylinder = Height (H) of plastic cylinder = 10 cmCapacity of plastic cylinder = R²H  ==385 cm³Thus, the plastic cylinder has greater capacity.
Difference in capacity = (385 – 300) cm³ = 85 cm³

Question 2The pillars of a temple are cylindrically shaped. If each pillar has a circular base of radius 20 cm and height 10m. How much concrete mixture would be required to build 14 such pillars?

Solution 2

Question 3The 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 3

Question 4If the lateral surface of a cylinder is 94.2 cm² and its height is 5 cm, then find
(i) radius of its base    (ii) its volume. (Use  = 3.14)Solution 4(i)    Height (h) of cylinder = 5 cm
        Let radius of cylinder be r.
        CSA of cylinder = 94.2 cm²
        2rh = 94.2 cm²
        (2  3.14  r  5) cm = 94.2 cm²
        r = 3 cm (ii)    Volume of cylinder = r²h = (3.14  (3)² 5) cm³ = 141.3 cm³

Question 5The capacity of a closed cylindrical vessel of height 1 m is 15.4 litres. How many square metres of metal sheet would be needed to make it?Solution 5Let radius of the circular ends of the cylinder be r.
Height (h) of the cylindrical vessel = 1 m
Volume of cylindrical vessel = 15.4 litres = 0.0154 m³

Total  Surface area of vessel = 2 r(r+h)

                                            Thus, 0.4708 m² of metal sheet would be needed to make the cylindrical vessel.   Question 6A 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 6Radius (r) of cylindrical bowl = cm = 3.5 cm
Height (h) up to which the bowl is filled with soup = 4 cmVolume of soup in 1 bowl = r²h= 
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 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11

Solution 11

Question 12

Solution 12

Question 13

Solution 13

Question 14

Solution 14

Question 15

Solution 15

Question 16

Solution 16

Question 17

Solution 17

Question 18

Solution 18

Question 19

A cylinderical container with diameter of base 56 cm contains sufficient water to submerge a rectangular solid of iron with dimensions 32 cm x 22 cm x 14 cm. Find the rise in the level of the water when the solid is completely submerged.Solution 19

Question 20

Solution 20

Question 21

Solution 21

Question 22

Solution 22

Question 23

Solution 23

Question 24

Solution 24

Question 25

Solution 25

Question 26

Solution 26

Question 27

A well with 14 m diameter is dug 8 m deep. The earth taken out of it has been evenly spread all around it to around it to a width of 21 m to form an embankment. Find the height of the embankment.Solution 27

Question 28

Solution 28

Question 29

Water flows out through a circular pipe whose internal diameter is 2 cm, at the rate of 6 meters per second into cylindrical tank. The water is collected in a cylindrical vessel radius of whose base is 60 cm. Find the rise in the level of water in 30 minutes?Solution 29

Question 30

Solution 30

Question 31

The sum of the radius of the base and height of a solid cylinder is 37 m. If the total surface area of the solid cylinder is 1628 m². Find the volume of the cylinder.Solution 31

Question 32A well with 10 m inside diameter 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 32

Chapter 18 – Surface Areas and Volume of a Cuboid and Cube Exercise Ex. 18.1

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4Mary wants to decorate her Christmas tree. She wants to place the tree on a wooden block covered with coloured paper with picture of Santa Claus on it. She must know the exact quantity of paper to buy for this purpose. If the box has length, breadth and height as 80 cm, 40 cm and 20 cm respectively. How many square sheets of paper of side 40 cm would she require?

Solution 4

Question 5

Solution 5

Question 6Three equal cubes are placed adjacently in a row. Find the ratio of total surface area of the new cuboid to that of the sum of the surface areas of the three cubes.Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9Hameed has built a cubical water tank with lid for his house, with each outer edge 1.5 m long. He gets the outer surface of the tank excluding the base, covered with square tiles of side 25 cm. Find how much he would spend for the titles, if the cost of tiles is Rs. 360 per dozen.Solution 9

Question 10

Solution 10

Question 11

Solution 11

Question 12

Ravish wanted to make a temporary shelter for his car, by making a box-like structure with tarpaulin that covers all the four sides and the top of the car (with the front face as a flap which can be rolled up). Assuming that the stitching margins are very small, and therefore negligible, how much tarpaulin would be required to make the shelter of height 2.5 m, with base dimensions 4 m x 3m?Solution 12Length of shelter = 4 m
Breadth of shelter = 3 m
Height of shelter = 2.5 m

The tarpaulin will be required for top and four sides of the shelter.
Area of Tarpaulin required = 2(lh + bh) + lb
= [2(4  2.5 + 3  2.5) + 4  3] m²
= [2(10 + 7.5) + 12] m²
= 47 m²

Question 13

An open box is made of wood 3 cm thick. Its external length, breadth and height are 1.48m, 1.16 m and 8.3 dm. Find the cost of painting the inner surface at Rs 50 per sq meter.Solution 13

Question 14

Solution 14

Question 15The paint in a certain container is sufficient to paint an area equal to 9.375 m². How many bricks of dimensions 22.5 cm × 10 cm × 7.5 cm can be painted out of this container?Solution 15Total surface area of one brick = 2(lb + bh + lh)
 = [2(22.5 × 10 + 10 × 7.5 + 22.5 × 7.5)]cm²
 = 2(225 + 75 + 168.75) 
          = (2 × 468.75) cm²
 = 937.5 cm²
Let n number of bricks be painted by the container.
Area of n bricks = 937.5n cm²
Area that can be painted by the container = 9.375 m² = 93750 cm²
 93750 = 937.5n
n = 100
Thus, 100 bricks can be painted out by the container.

Question 16

Solution 16

Question 17

The cost of preparing the walls of a room 12 m long at the rate of Rs 1.35 per square meter is Rs 340.20 and the cost of matting the floor at 85 paise per square meter is Rs 91.80. Find the height of the room.Solution 17

Question 18

Solution 18

Question 19A wooden bookshelf has external dimensions as follows: Height = 110 cm, Depth = 25 cm, Breadth = 85 cm. The thickness of the plank is 5 cm everywhere. The external faces are to be polished and the inner faces are to be painted. If the rate of polishing is 20 paise per cm² and the rate of painting is 10 paise per cm², find the total expenses required for polishing and painting the surface of the bookshelf. 

                                           Solution 19    External length (l) of bookshelf = 85 cm
    External breadth (b) of bookshelf = 25 cm
    External height (h) of bookshelf = 110 cm
    External surface area of shelf while leaving front face of shelf
                                                  = lh + 2 (lb + bh)
                                                  = [85  110 + 2 (85  25 + 25  110)] cm²
                                                  = 19100 cm²
    Area of front face = [85  110 – 75  100 + 2 (75  5)] cm²
                                                  = 1850 + 750 cm²
                                                  = 2600 cm²
    Area to be polished = (19100 + 2600) cm² = 21700 cm²
    Cost of polishing 1 cm² area = Rs 0.20
    Cost of polishing 21700 cm² area = Rs (21700  0.20) = Rs 4340    

    Now, length (l), breadth (b) height (h) of each row of bookshelf is 75 cm, 20 cm, and    30cm  respectively.
    Area to be painted in 1 row = 2 (l + h) b + lh
                                           = [2 (75 + 30)  20 + 75  30] cm²
                                           = (4200 + 2250) cm²
                                           = 6450 cm²
    Area to be painted in 3 rows = (3  6450) cm² = 19350 cm²
    Cost of painting 1 cm² area = Rs 0.10
    Cost of painting 19350 cm² area = Rs (19350  0.10) = Rs 1935    
    Total expense required for polishing and painting the surface of the bookshelf                                             = Rs(4340 + 1935) = Rs 6275

Chapter 18 – Surface Areas and Volume of a Cuboid and Cube Exercise Ex. 18.2

Question 1A cuboidal water tank is 6 m long, 5 m wide and 4.5 m deep. How, many litres of water can it holds? Solution 1Volume of tank = l  b  h = (6  5  4.5) m³ = 135 m³  It is given that:
 1 m³ = 1000 litres


 Thus, the tank can hold 135000 litres of water.

Question 2A cuboidal vessel is 10 m long and 8 m wide. How high must it be made to hold 380 cubic metres of a liquid?   Solution 2Let height of cuboidal vessel be h.
Length (l) of vessel = 10 m
Width (b) of vessel = 8 m
Volume of vessel = 380 m³
l  b  h = 380        
 10  8  h = 380
 h = 4.75

 Thus, the height of the vessel should be 4.75 m.    

Question 3Find the cost of digging a cuboidal pit 8 m long, 6 m broad and 3 m deep at the rate of Rs 30 per m³.

Solution 3Length (l) of the cuboidal pit = 8 m
Width (b) of the cuboidal pit = 6 m
 Depth (h) of the cuboidal pit = 3 m
Volume of the cuboidal pit = l  b  h = (8  6  3)  = 144 m³
Cost of digging 1 m³ = Rs 30
Cost of digging 144 m³ = Rs (144 30) = Rs 4320

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11

Solution 11

Question 12

Solution 12

Question 13

Solution 13

Question 14

Solution 14

Question 15

Solution 15

Question 16A village, having a population of 4000, requires 150 litres of water per head per day. It has a tank measuring 20 m × 15 m × 6 m. For how many days will the water of this tank last?Solution 16Length (l) of the cuboidal tank = 20 m
Breadth (b) of the cuboidal tank = 15 m
Height (h) of the cuboidal tank = 6 m

Capacity of tank = l × b × h    = (20 × 15 × 6) m³ = 1800 m³ = 1800000 litres

Water consumed by people of village in 1 day = 4000 × 150 litres = 600000 litres

Let water of this tank lasts for n days.
Water consumed by all people of village in n days = capacity of tank
n × 600000 = 1800000
n = 3    
Thus, the water of tank will last for 3 days.

Question 17

A child playing with building blocks, which are of the shape of the cubes, has built a structure as shown in fig. If the edge of each cube is 3 cm, find the volume of the structure built by the child

Solution 17

Question 18A godown measures 40 m  25 m  10 m. Find the maximum number of wooden crates each measuring 1.5 m  1.25 m  0.5 m that can be stored in the godown.    

Solution 18Length  of the godown = 40 m
    Breadth  of the godown = 25 m
    Height  of the godown = 10 m

Volume of godown = l₁ b₁ h₁ = (40  25  10)  = 10000 

    Length  of a wooden crate = 1.5 m
    Breadth  of a wooden crate = 1.25 m
    Height  of a wooden crate = 0.5 m

Volume of a wooden crate =  = (1.5  1.25  0.5) m3 = 0.9375 

Let n wooden crates be stored in the godown.
Volume of n wooden crates = volume of godown
0.9375  n = 10000

Thus, 10666 wooden crates can be stored in godown.

Question 19

A wall of length 10 m was to be built across an open ground. The height of the wall is 4 m and thickness of the wall is 24 cm. If this wall is to be built up with bricks whose dimensions are 24 cm x 12 cm x 8 cm, how many bricks would be required?Solution 19

Question 20

Solution 20

Question 21

Solution 21

Question 22A river 3 m deep and 40 m wide is flowing at the rate of 2 km per hour. How much water will fall into the sea in a minute?Solution 22Rate of water flow = 2 km per hour  
Depth (h) of river = 3 m
Width (b) of river = 40 m
Volume of water flowed in 1 min 
Thus, in 1 minute 4000  = 4000000 litres of water will fall into the sea.    

Question 23

Solution 23

Question 24

Solution 24

Question 25

Solution 25

Question 26

A rectangular container, whose base is a square of side 5 cm, stands on a horizontal table, and holds water upto 1 cm from the top. When a cube is placed in the water it is completely submerged, the water rises to the top and 2 cubic cm of water overflows. Calculate the volume of the cube and also the length of its edge.Solution 26

Question 27

Solution 27

Question 28

Solution 28

Chapter 17 – Heron’s Formula Exercise Ex. 17.1

Question 1

Find the area of the triangle whose sides are respectively 150 cm, 120 cm and 200 cm.Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5The perimeter of triangular field is 540 m and its sides are in the ratio 25:17:12. Find the area of the triangle.Solution 5The sides of the triangular field are in the ratio 25:17:12.
Let the sides of triangle be 25x, 17x, and 12x.
  Perimeter of this triangle = 540 m
             25x + 17x + 12x = 540 m
                         54x = 540 m
                            x = 10 m
  Sides of triangle will be 250 m, 170 m, and 120 m. Semi-perimeter (s) =  By Heron’s formula:

    So, area of the triangle is 9000 m².Question 6

The perimeter of right triangle is 300m. If its sides are in the ratio 3 : 5 : 7. Find the area of the triangle.Solution 6

Question 7

The perimeter of a triangular field is 240 dm. If two of its sides are 78 dm and 50 dm, find the length of the perpendicular on the side of length 50 dm from the opposite vertex.Solution 7

Question 8

A triangle has sides 35 cm, 54 cm and 61 cm long. Find its area. Also, find the smallest of its altitudes.Solution 8

Question 9

The lengths of the sides of a triangle are in the ratio 3 : 4 : 5 and its perimeter is 144 cm. Find the area of the triangle and the height corresponding to the longest side.Solution 9

Question 10

The perimeter of an isosceles triangle is 42 cm and its base is (3/2) times each of the equal sides. Find the length of each side of the triangle, area of the triangle and the height of the triangle.Solution 10

Question 11

Find the area of the shaded region in fig.12.12

Solution 11

Chapter 17 – Heron’s Formula Exercise Ex. 17.2

Question 1Find the area of a quadrilateral ABCD in which AB = 3 cm, BC = 4 cm, CD = 4 cm, DA = 5 cm and AC = 5 cm.Solution 1

For ABC
AC² = AB² + BC²
(5)² = (3)² + (4)²
So, ABC is a right angle triangle, right angled at point B.
Area of ABCFor ADC
Perimeter = 2s = AC + CD + DA = (5 + 4 + 5) cm = 14 cm
            s = 7 cm
By Heron’s formula
Area of triangle 

Area of ABCD = Area of ABC + Area of ACD
    = (6 + 9.166) cm² = 15.166 cm² = 15.2 cm² (approximately)

Question 2

Solution 2

Question 3

Solution 3

Question 4A park, in the shape of a quadrilateral ABCD, has  = 90^o, AB = 9 m, BC = 12 m, CD = 5 m and AD = 8 m. How much area does it occupy?Solution 4Let us join BD.
In BCD applying Pythagoras theorem
BD² = BC² + CD²
       = (12)² + (5)²
       = 144 + 25
BD² = 169
  BD = 13 m

Area of BCD

                  For ABD

                    By Heron’s formula Area of triangle  

                               Area of park = Area of ABD + Area of BCD
                         = 35.496 + 30 m²                         = 65.496 m²                         = 65. 5 m² (approximately)

Question 5

Find the area of a rhombus whose perimeter is 80 m and one of whose diagonal is 24 m.Solution 5

Question 6

A rhombus sheet, whose perimeter is 32 m and whose one diagonal is 10 m long, is painted on both sides at the rate of Rs 5 per m². Find the cost of painting.Solution 6

Question 7

Solution 7

Question 8

Find the area of a quadrilateral ABCD in which AB = 42 cm, BC = 21 cm, CD = 29 cm, DA = 34 cm and diagonal BD = 20 cm.Solution 8

Question 09

The adjacent sides of a parallelogram ABCD measure 34 cm and 20 cm, and the diagonal AC measures 42 cm. Find the area of the parallelogram.Solution 09

Question 10

Find the area of the blades of the magnetic compass shown in fig.

(Take √11 = 3.32)

Solution 10

Question 11

A triangle and a parallelogram have the same base and the same area. If the sides of the triangle are 13 cm, 14 cm and 15 cm and the parallelogram stands on the base 14 cm, find the height of the parallelogram.Solution 11

Question 12

Solution 12

Question 13

Solution 13

Question 14

A hand fan is made by stitching 10 equal size triangular strips of two different types of paper as shown in fig., The dimensions of equal strips are 25 cm, 25 cm and 14 cm. Find the area of each type of paper needed to make the hand fan.

Solution 14

Chapter 16 – Constructions Exercise Ex. 16.1

Question 1

Draw a line segment of length 8.6 cm. Bisect it and measure the length of each part.Solution 1

Question 2

Draw a line segment AB of length 5.8 cm. Draw the perpendicular bisector of this segment.Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Draw a line segment AB and bisect it. Bisect one of the equal parts of obtain a line segment of length  (AB)Solution 6

Question 7

Solution 7

Chapter 16 – Constructions Exercise Ex. 16.2

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Using ruler and compasses only, draw a right angle.Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10(i)

Solution 10(i)

Question 10(ii)

Solution 10(ii)

Question 11(i)

Solution 11(i)

Question 11(ii)

Solution 11(ii)

Question 11(iii)

Solution 11(iii)

Question 11(iv)

Solution 11(iv)

Question 11(v)

Construct an angle of 15^o.Solution 11(v)

Question 11(vi)

Solution 11(vi)

Chapter 16 – Constructions Exercise Ex. 16.3

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Construct a right angled triangle whose perimeter is equal to 10 cm and one actute angle equal to 60^o.Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Page No 113:

Exercise 7A

Question 1:

Write each of the following in figures:
(i) Fifty-eight point six three
(ii) One hundred twenty four point four two five
(iii) Seven point seven six
(iv) Nineteen point eight
(v) Four hundred four point zero four four
(vi) Point one seven three
(vii) Point zero one five

ANSWER:

(i) 58.63
(ii) 124.425
(iii) 7.76
(iv) 19.8
(v) 404.044 
(vi) 0.173
(vii) 0.015

Page No 113:

Question 2:

Write the place value of each digit in each of the following decimals:
(i) 14.83
(ii) 275.269
(iii) 46.075
(iv) 302.459
(v) 5370.34
(vi) 186.209

ANSWER:

(i) In 14.83, we have:
Place value of 1 = 1 tens = 10
Place value of 4 = 4 ones = 4
Place value of 8 = 8 tenths = 810810
Place value of 3 = 3 hundredths = 31003100

(ii) In 275.269, we have:
Place value of 2 = 2 hundreds = 200
Place value of 7 = 7 tens = 70
Place value of 5 = 5 ones = 5
Place value of 2 = 2 tenths = 210210
Place value of 6 = 6 hundredths = 61006100
Place value of 9 = 9 thousandths = 9100091000

(iii) In 46.075, we have:
Place value of 4 = 4 tens = 40
Place value of 6 = 6 ones = 6
Place value of 0 = 0 tenths = 010010= 0
Place value of 7 = 7 hundredths = 71007100
Place value of 5 = 5 thousandths = 5100051000

(iv) In 302.459, we have:
Place value of 3 = 3 hundreds = 300
Place value of 0 = 0 tens = 0
Place value of 2 = 2 ones = 2
Place value of 4 = 4 tenths = 410410
Place value of 5 = 5 hundredths = 51005100
Place value of 9 = 9 thousandths = 9100091000

(v) In 5370.34, we have:
Place value of 5 = 5 thousands = 5000
Place value of 3 = 3 hundreds = 300
Place value of 7 = 7 tens = 70
Place value of 0 = 0 ones = 0
Place value of 3 = 3 tenths = 310310
Place value of 4 = 4 hundredths = 41004100

(vi) In 186.209, we have:
Place value of 1 = 1 hundreds = 100
Place value of 8 = 8 tens = 80
Place value of 6 = 6 ones = 6
Place value of 2 = 2 tenths = 210210
Place value of 0 = 0 hundredths = 0
Place value of 9 = 9 thousandths = 9100091000

Page No 113:

Question 3:

Write each of the following decimals in expanded form:
(i) 67.83
(ii) 283.61
(iii) 24.675
(iv) 0.294
(v) 8.006
(vi) 4615.72

ANSWER:

(i) 67.83
 = 6 tens + 7 ones + 8 tenths + 3 hundredths
 = 60 + 7 + 810 + 310060 + 7 + 810 + 3100

(ii) 283.61
 = 2 hundreds  + 8 tens + 3 ones + 6 tenths + 1 hundredths
 = 200 + 80 + 3 + 610  + 1100200 + 80 + 3 + 610  + 1100

(iii) 24.675
 = 2 tens + 4 ones + 6 tenths + 7 hundredths + 5 thousandths
 = 20 + 4 + 610 + 7100 + 5100020 + 4 + 610 + 7100 + 51000

(iv) 0.294
 = 2 tenths + 9 hundredths + 4 thousandths
 = 210  + 9100 + 41000210  + 9100 + 41000

(v) 8.006
 = 8 ones + 0 tenths + 0 hundredths  + 6 thousandths
 = 8 + 010 + 0100 + 610008 + 010 + 0100 + 61000

(vi) 4615.72
 = 4 thousands + 6 hundreds + 1 tens + 5 ones + 7 tenths + 2 hundredths
 = 4000 + 600 + 10 + 5 + 710 + 21004000 + 600 + 10 + 5 + 710 + 2100

Page No 113:

Question 4:

Write each of the following in decimals form:
(i) 40 + 6 + 710 + 910040 + 6 + 710 + 9100
(ii) 500 + 70 + 8 + 310 + 1100 + 61000500 + 70 + 8 + 310 + 1100 + 61000
(iii) 700 + 30 + 1 + 810 + 4100700 + 30 + 1 + 810 + 4100
(iv) 600 + 5 + 7100 + 91000600 + 5 + 7100 + 91000
(v) 800 + 5 + 810 + 61000800 + 5 + 810 + 61000
(vi) 30 + 9 + 4100 + 8100030 + 9 + 4100 + 81000

ANSWER:

(i) 40 + 6 + 710 + 910040 + 6 + 710 + 9100 = 46 + 0.7 + .09 = 46.79

(ii) 500 + 70 + 8 + 310 + 1100 + 61000500 + 70 + 8 + 310 + 1100 + 61000 = 578 + 0.3 + 0.01 + 0.006 = 578.316

(iii) 700 + 30 + 1 + 810 + 4100700 + 30 + 1 + 810 + 4100 = 731 + 0.8 + 0.04 = 731.84

(iv) 600 + 5 + 7100 + 91000600 + 5 + 7100 + 91000 = 605 + 0.07 + 0.009 = 605.079

(v) 800 + 5 + 810 + 61000800 + 5 + 810 + 61000 = 805 + 0.8 + 0.006 = 805.806

(vi) 30 + 9 + 4100 + 8100030 + 9 + 4100 + 81000 = 39 + 0.04 + 0.008 = 39.048

Page No 114:

Question 5:

Convert each of the following into like decimals:
(i) 7.5, 64.23, 0.074
(ii) 0.6, 5.937, 2.36, 4.2
(iii) 1.6, 0.07, 3.58, 2.9
(iv) 2.5, 0.63, 14.08, 1.637

ANSWER:

(i) Each of the numbers has maximum 3 decimal places. So, we convert them into numbers having three decimal places by annexing suitable number of zeroes to the extreme right of the decimal part.
  7.5 = 7.500
64.23 = 64.230
0.074 = 0.074

(ii) Each of the numbers has maximum 3 decimal places. So, we convert them into numbers having three decimal places by annexing suitable number of zeroes to the extreme right of the decimal part.
  0.6 = 0.600
5.937 = 5.937
2.36 = 2.360
4.2 = 4.200

(iii) Each of the numbers has maximum 2 decimal places. So, we convert them into numbers having three decimal places by annexing suitable number of zeroes to the extreme right of the decimal part.
  1.6 = 1.60
0.07 = 0.07
3.58 = 3.58
2.9 = 2.90

(iv) Each of the numbers has maximum 3 decimal places. So, we convert them into numbers having three decimal places by annexing suitable number of zeroes to the extreme right of the decimal part.
  2.5 = 2.500
0.63 = 0.630
14.08 = 14.080
1.637 = 1.637

Page No 114:

Question 6:

Fill in each of the place holders with the correct symbol > or <:
(i) 84.23      76.3584.23      76.35
(ii) 7.608      7.687.608      7.68
(iii) 8.34      8.438.34      8.43
(iv) 12.06      12.00612.06      12.006
(v) 3.85      3.8053.85      3.805
(vi) 0.97      1.070.97      1.07

ANSWER:

(i) 84.23 > 76.35
Since 84 is greater than 76, 84.23 is greater than 76.35. (Comparing the whole number parts)

(ii) 7.608 < 7.680
Since 8 is greater than 0 at the hundredths place, 7.608 is smaller than 7.680.

(iii) 8.34 < 8.43
Since 4 is greater than 3 at the tenths place, 8.34 is smaller than 8.43.

(iv) 12.06 > 12.006
Since 6 is greater than 0 at the hundredths place, 12.06 is greater than 12.006.

(v) 3.850 > 3.805
Since 5 is greater than 0 at the hundredths place, 3.850 is greater than 3.805.

(vi) 0.97 < 1.07
Since 1 is greater than 0, 0.97 is smaller than 1.07. (Comparing the whole number parts)

Page No 114:

Question 7:

Arrange the following decimals in ascending order:
(i) 5.8, 7.2, 5.69, 7.14, 5.06
(ii) 0.6, 6.6, 6.06, 66.6, 0.06
(iii) 6.54, 6.45, 6.4, 6.5, 6.05
(iv) 3.3, 3.303, 3.033, 0.33, 3.003

ANSWER:

(i) 5.8, 7.2, 5.69, 7.14, 5.06
 Converting the given decimals into like decimals:
    5.80, 7.20, 5.69, 7.14, 5.06
  Clearly, 5.06 < 5.69 < 5.80 < 7.14 < 7.20
  Hence, the given decimals can be arranged in the ascending order as follows:
          5.06, 5.69, 5.80, 7.14 and 7.2

(ii) 0.6, 6.6, 6.06, 66.6, 0.06
  Converting the given decimals into like decimals:
   0.60, 6.60, 6.06, 66.60, 0.06
  Clearly, 0.06 < 0.60 < 6.06 < 6.60 < 66.60
  Hence, the given decimals can be arranged in the ascending order as follows:
           0.06, 0.60, 6.06, 6.60 and 66.60

(iii) 6.54, 6.45, 6.4, 6.5, 6.05
Converting the given decimals into like decimals:
  6.54, 6.45, 6.40, 6.50, 6.05
  Clearly, 6.05 < 6.40 < 6.45 < 6.50 < 6.54
  Hence, the given decimals can be arranged in the ascending order as follows:
         6.05, 6.40, 6.45, 6.50 and 6.54

(iv) 3.3, 3.303, 3.033, 0.33, 3.003
Converting the given decimals into like decimals:
  3.300, 3.303, 3.033, 0.330, 3.003
  Clearly, 0.330 < 3.003 < 3.033 < 3.300 < 3.303
  Hence, the given decimals can be arranged in the ascending order as follows:
         0.33, 3.003, 3.033, 3.300 and 3.303

Page No 114:

Question 8:

Arrange the following decimals in descending order:
(i) 7.3, 8.73, 73.03, 7.33, 8.073
(ii) 3.3, 3.03, 30.3, 30.03, 3.003
(iii) 2.7, 7.2, 2.27, 2.72, 2.02, 2.007
(iv) 8.88, 8.088, 88.8, 88.08, 8.008

ANSWER:

(i) 7.3, 8.73, 73.03, 7.33, 8.073
 Converting each decimal into like decimals:
  7.300, 8.730, 73.030, 7.330, 8.073
 Clearly, 73.030 > 8.730 > 8.073 > 7.330 > 7.300
 Hence, the given decimals can be arranged in the descending order as follows:
   73.03, 8.73, 8.073, 7.33 and 7.3

(ii) 3.3, 3.03, 30.3, 30.03, 3.003
 Converting each decimal into like decimals:
   3.300, 3.030, 30.300, 30.030, 3.003
 Clearly, 30.300 > 30.030 > 3.300 > 3.030 > 3.003
 Hence, the given decimals can be arranged in the descending order as follows:
   30.3, 30.03, 3.3, 3.03 and 3.003

(iii) 2.7, 7.2, 2.27, 2.72, 2.02, 2.007
 Converting each decimal into like decimals:
   2.700, 7.200, 2.270, 2.720, 2.020, 2.007
 Clearly, 7.200 > 2.720 > 2.700 > 2.270 > 2.020 > 2.007
 Hence, the given decimals can be arranged in the descending order as follows:
      7.2, 2.72, 2.7, 2.27, 2.02 and 2.007

(iv) 8.88, 8.088, 88.8, 88.08, 8.008
 Converting each decimal into like decimals:
        8.880, 8.088, 88.800, 88.080, 8.008
 Clearly, 88.800 > 88.080 > 8.880 > 8.088 > 8.008
 Hence, the given decimals can be arranged in the descending order as follows:
        88.8, 88.08, 8.88, 8.088 and 8.008

Page No 118:

Exercise 7B

Question 1:

Convert each of the following into a fraction in its simplest form:
.9

ANSWER:

We have:
 .9 = 910910

Page No 118:

Question 2:

Convert each of the following into a fraction in its simplest form:
0.6

ANSWER:

We have:
 0.6 = 610 = 35610 = 35

Page No 118:

Question 3:

Convert each of the following into a fraction in its simplest form:
.08

ANSWER:

We have:
 0.08 = 8100 = 450 = 2258100 = 450 = 225
    

Page No 118:

Question 4:

Convert each of the following into a fraction in its simplest form:
0.15

ANSWER:

We have:
 0.15 = 15100 = 32015100 = 320

Page No 118:

Question 5:

Convert each of the following into a fraction in its simplest form:
0.48

ANSWER:

We have:
 0.48 = 48100 = 122548100 = 1225

Page No 118:

Question 6:

Convert each of the following into a fraction in its simplest form:
.053

ANSWER:

We have:
 0.053 = 531000 531000 

Page No 118:

Question 7:

Convert each of the following into a fraction in its simplest form:
0.125

ANSWER:

We have:
 0.125 = 1251000 = 25200 =540= 181251000 = 25200 =540= 18

Page No 118:

Question 8:

Convert each of the following into a fraction in its simplest form:
.224

ANSWER:

We have:
0.224 = 2241000 = 56250 = 281252241000 = 56250 = 28125

Page No 118:

Question 9:

Convert each of the following as a mixed-fraction:
6.4

ANSWER:

We have:
 6.4 = 6410 = 325 = 6256410 = 325 = 625

Page No 118:

Question 10:

Convert each of the following as a mixed-fraction:
16.5

ANSWER:

We have:
 16.5 = 16510 = 332 =161216510 = 332 =1612

Page No 118:

Question 11:

Convert each of the following as a mixed-fraction:
8.36

ANSWER:

We have:
 8.36 = 836100 = 20925 = 8925836100 = 20925 = 8925

Page No 118:

Question 12:

Convert each of the following as a mixed-fraction:
4.275

ANSWER:

We have:
 4.275 = 42751000 = 17140 = 4114042751000 = 17140 = 41140

Page No 118:

Question 13:

Convert each of the following as a mixed-fraction:
25.06

ANSWER:

We have:
 25.06 = 2506100 = 125350 =253502506100 = 125350 =25350

Page No 118:

Question 14:

Convert each of the following as a mixed-fraction:
7.004

ANSWER:

We have:
7.004 = 70041000 = 1751250 = 7125070041000 = 1751250 = 71250

Page No 118:

Question 15:

Convert each of the following as a mixed-fraction:
2.052

ANSWER:

We have:
2.052 = 20521000 = 513250 = 21325020521000 = 513250 = 213250

Page No 118:

Question 16:

Convert each of the following as a mixed-fraction:
3.108

ANSWER:

We have:
3.108 = 31081000 = 777250 = 32725031081000 = 777250 = 327250

Page No 118:

Question 17:

Convert each of the following into a decimal:
23102310

ANSWER:

We have:
2310 = 2310 = 2 + 0.3 = 2.32310 = 2310 = 2 + 0.3 = 2.3

Page No 118:

Question 18:

Convert each of the following into a decimal:
167100167100

ANSWER:

We have:
167100 = 167100 = 1 + 0.67 = 1.67167100 = 167100 = 1 + 0.67 = 1.67

Page No 118:

Question 19:

Convert each of the following into a decimal:
15891001589100

ANSWER:

We have:
1589100 = 1589100 = 15+ 0.89 = 15.891589100 = 1589100 = 15+ 0.89 = 15.89

Page No 118:

Question 20:

Convert each of the following into a decimal:
5413100054131000

ANSWER:

We have:
54131000 = 54131000 = 5 + 0.413 = 5.41354131000 = 54131000 = 5 + 0.413 = 5.413

Page No 118:

Question 21:

Convert each of the following into a decimal:
214151000214151000

ANSWER:

We have:
214151000 = 214151000 = 21 + 0.415 = 21.415214151000 = 214151000 = 21 + 0.415 = 21.415

Page No 118:

Question 22:

Convert each of the following into a decimal:
254254

ANSWER:

We have:
254 = 614 = 6 + 0.25 = 6.25254 = 614 = 6 + 0.25 = 6.25

Page No 118:

Question 23:

Convert each of the following into a decimal:
335335

ANSWER:

We have:
335= 3 + 0.6 = 3.6335= 3 + 0.6 = 3.6

Page No 118:

Question 24:

Convert each of the following into a decimal:
14251425

ANSWER:

We have:
 1425 = 1 + 0.16 = 1.16 1425 = 1 + 0.16 = 1.16

Page No 118:

Question 25:

Convert each of the following into a decimal:
5175051750

ANSWER:

We have:
51750 = 5 + 0.34 = 5.3451750 = 5 + 0.34 = 5.34

Page No 118:

Question 26:

Convert each of the following into a decimal:
12381238

ANSWER:

We have:
1238 = 12 + 0.375 = 12.3751238 = 12 + 0.375 = 12.375

Page No 118:

Question 27:

Convert each of the following into a decimal:
2194021940

ANSWER:

We have:
21940 = 2 + 0.475 = 2.47521940 = 2 + 0.475 = 2.475

Page No 118:

Question 28:

Convert each of the following into a decimal:
19201920

ANSWER:

We have:
1920 = 0.951920 = 0.95

Page No 118:

Question 29:

Convert each of the following into a decimal:
37503750

ANSWER:

We have:
3750 =  0.743750 =  0.74

Page No 118:

Question 30:

Convert each of the following into a decimal:
107250107250

ANSWER:

We have:
107250 = 0.428107250 = 0.428

Page No 118:

Question 31:

Convert each of the following into a decimal:
340340

ANSWER:

We have:
340 = 0.075340 = 0.075

Page No 118:

Question 32:

Convert each of the following into a decimal:
7878

ANSWER:

We have:
78 = 0.87578 = 0.875

Page No 118:

Question 33:

Using decimals, express
(i) 8 kg 640 g in kilograms
(ii) 9 kg 37 g in kilograms
(iii) 6 kg 8 g in kilograms

ANSWER:

(i) 8 kg 640 g in kilograms:
     8 kg + 640 gm = 8 kg + 64010006401000 kg
     8 kg + 0.640 kg = 8.640 kg

(ii) 9 kg 37 g in kilograms:
     9 kg + 37 gm = 9 kg + 371000371000 kg
     9 kg + 0.037 kg = 9.037 kg

(iii) 6 kg 8 g in kilograms:
      6 kg + 8 gm = 6 kg + 8100081000 kg
      6 kg + 0.008 kg = 6.008 kg

Page No 118:

Question 34:

Using decimals, express
(i) 4 km 365 m in kilometres
(ii) 5 km 87 m in kilometres
(iii) 3 km 6 m in kilometres
(iv) 270 m in kilometres
(v) 35 m in kilometres
(vi) 6 m in kilometres

ANSWER:

(i) 4 km 365 m in kilometres:
    4 km + 365 m = 4 km + 36510003651000 km     [Since 1 km = 1000 m]
    4 km + 0.365 km = 4.365 km

(ii) 5 km 87 m in kilometres:
    5 km + 87 m = 5 km + 871000871000 km     [Since 1 km = 1000 m]
    5 km + 0.087 km = 5.087 km

(iii) 3 km 6 m in kilometres:
     3 km + 6 m = 3 km + 6100061000 km      [Since 1 km = 1000 m]
     3 km + 0.006 km = 3.006 km

(iv) 270 m in kilometres:
      27010002701000 km = 0.270 km                     [Since 1 km = 1000 m]

(v) 35 m in kilometres:
      351000351000 km = 0.035 km                      [Since 1 km = 1000 m]

(vi) 6 m in kilometres:
       6100061000 km = 0.006 km                    [Since 1 km = 1000 m]

Page No 118:

Question 35:

Using decimals, express
(i) 15 kg 850 g in kilograms
(ii) 8 kg 96 g in kilograms
(iii) 540 g in kilograms
(iv) 8 g in kilograms

ANSWER:

(i) 15 kg 850 g in kilograms:
    15 kg + 850 gm = 15 kg + 85010008501000 kg        [Since 1 kg = 1000 gm]
    15 kg + 0.850 kg = 15.850 kg

(ii) 8 kg 96 g in kilograms:
     8 kg + 96 gm = 8 kg + 961000 961000 kg         [Since 1 kg = 1000 gm]
     8 kg + 0.096 kg = 8.096 kg

(iii) 540 g in kilograms:
       540 gm = 54010005401000 kg = 0.540 kg        [Since 1 kg = 1000 gm]

(iv) 8 g in kilograms:
       8 gm = 8100081000 kg = 0.008 kg           [Since 1 kg = 1000 gm]

Page No 118:

Question 36:

Using decimals, express
(i) Rs 18 and 25 paise in rupees
(ii) Rs 9 and 8 paise in rupees
(iii) 32 paise in rupees
(iv) 5 paise in rupees

ANSWER:

(i) Rs 18 and 25 paise in rupees:
    Rs 18 + 25 paise = Rs 18 + Rs 2510025100         [Since Re 1 = 100 paise]
    Rs 18 + Rs 0.25 = Rs 18.25

(ii) Rs 9 and 8 paise in rupees:
    Rs 9 + 8 paise = Rs 9 + Rs 81008100         [Since Re 1 = 100 paise]
    Rs 9 + Rs 0.08 = Rs 9.08

(iii) 32 paise in rupees:
      32 paise = Rs 3210032100 = Rs 0.32          [Since Re 1 = 100 paise]

(iv) 5 paise in rupees:
      5 paise = Rs 51005100 = Rs 0.05          [Since Re 1 = 100 paise]

Page No 120:

Exercise 7C

Question 1:

Add the following decimals:
9.6, 14.8, 37 and 5.9

ANSWER:

9.6, 14.8, 37 and 5.9
Converting the decimals into like decimals:
9.6, 14.8, 37.0 and 5.9
Let us write the given numbers in the column form.
Now, adding:
     9.6
   14.8
   37.0
     5.9
   67.3
Hence, the sum of the given numbers is 67.3.

Page No 120:

Question 2:

Add the following decimals:
23.7, 106.94, 68.9 and 29.5

ANSWER:

23.7, 106.94, 68.9 and 29.5
Converting the decimals into like decimals:
23.70, 106.94, 68.90 and 29.50
Let us write the given numbers in the column form.
Now, adding:
   23.70
 106.94
  68.90
  29.50
 229.04
Hence, the sum of the given numbers is 229.04.

Page No 120:

Question 3:

Add the following decimals:
72.8, 7.68, 16.23 and 0.7

ANSWER:

72.8, 7.68, 16.23 and 0.7
Converting the decimals into like decimals:
72.80, 7.68, 16.23 and 0.70
Let us write the given numbers in the column form.
Now, adding:
   72.80
     7.68
   16.23
     0.70
   97.41
Hence, the sum of the given numbers is 97.41.

Page No 120:

Question 4:

Add the following decimals:
18.6, 84.75, 8.345 and 9.7

ANSWER:

18.6, 84.75, 8.345 and 9.7
Converting the decimals into like decimals:
18.600, 84.750, 8.345 and 9.700
Let us write the given numbers in the column form.
Now, adding:
   18.600
   84.750
     8.345
     9.700
  121.395
Hence, the sum of the given numbers is 121.395.

Page No 120:

Question 5:

Add the following decimals:
8.236, 16.064, 63.8 and 27.53

ANSWER:

8.236, 16.064, 63.8 and 27.53
Converting the decimals into like decimals:
8.236, 16.064, 63.800 and 27.530
Let us write the given numbers in the column form.
Now, adding:
     8.236
   16.064
   63.800
   27.530  
  115.630
Hence, the sum of the given numbers is 115.630.

Page No 120:

Question 6:

Add the following decimals:
28.9, 19.64, 123.697 and 0.354

ANSWER:

28.9, 19.64, 123.697 and 0.354
Converting the decimals into like decimals:
28.900, 19.640, 123.697 and 0.354
Let us write the given numbers in the column form.
Now, adding:
    28.900
    19.640
  123.697
     0.354 
  172.591
Hence, the sum of the given numbers is 172.591.

Page No 120:

Question 7:

Add the following decimals:
4.37, 9.638, 17.007 and 6.8

ANSWER:

4.37, 9.638, 17.007 and 6.8
Converting the decimals into like decimals:
4.370, 9.638, 17.007 and 6.800
Let us write the given numbers in the column form.
Now, adding:
    4.370
    9.683
  17.007
    6.800 
  37.815  
Hence, the sum of the given numbers is 37.815.

Page No 120:

Question 8:

Add the following decimals:
14.5, 0.038, 118.573 and 6.84

ANSWER:

14.5, 0.038, 118.573 and 6.84
Converting the decimals into like decimals:
14.500, 0.038, 118.573 and 6.840
Let us write the given numbers in the column form.
Now, adding:
   14.500
     0.038
  118.573
     6.840 
  139.951 
Hence, the sum of the given numbers is 139.951.

Page No 120:

Question 9:

During three days of a week, a rickshaw puller earns Rs 32.60, Rs 56.80 and Rs 72 respectively. What is his total earning during these days?

ANSWER:

Earning on the 1st day of the week =  Rs 32.60
Earning on the 2nd day of the week = Rs 56.80
Earning on the 3rd day of the week = Rs 72.00
Total earning =                                 Rs 161.40

Page No 120:

Question 10:

A man purchases an almirah for Rs 11025, gives Rs 172.50 as its cartage and spends Rs 64.80 on its repair. How much does the almirah cost him?

ANSWER:

Cost of the almirah =          Rs 11025.00
Money spent on cartage =       Rs 172.50
Money spent on repair =           Rs 64.800
Total cost of the almirah =  Rs 11262.3    

Page No 120:

Question 11:

Ramesh covers 36 km 235 m by taxi, 4 km 85 m by rickshaw and 1 km 80 m on foot What is the total distance covered by him?

ANSWER:

Distance covered by the taxi =        36 km 235 m
Distance covered by the rickshaw =  4 km 085 m
Distance covered on foot =               1 km 080 m
Total distance covered =                41 km  400 m

Page No 120:

Question 12:

A bag contains 45 kg 80 g of sugar and the mass of the empty bag is 950 g. What is the mass of the bag containing this much of sugar?

ANSWER:

Weight of sugar in the bag =   45 kg 080 g
Weight of the empty bag =       0 kg 950 g
Total weight of the bag =         46 kg 030 g

Page No 120:

Question 13:

Ramu bought 2 m 70 cm cloth for his shirt and 2 m 60 cm cloth for his pyjamas. Find the total length of cloth bought by him.

ANSWER:

Length of cloth for his shirt =         2 m 70 cm
Length of cloth for his pyjamas =    2 m 60 cm
Total length of cloth bought =         5 m 30 cm

Page No 120:

Question 14:

Radhika bought 2 m 5 cm cloth for her salwar and 3 m 35 cm cloth for her shirt. Find the total length of cloth bought by her.

ANSWER:

Length of cloth for her salwar =        2 m 05 cm
Length of cloth for her shirt =           3 m 35 cm
Total length of cloth bought =           5 m 40 cm

Page No 122:

Question 1:

Exercise 7D

Subtract:
27.86 from 53.74

ANSWER:

Let us write the numbers in the column form with the larger one at the top.
Now, subtracting:
   53.74
 – 27.86
   25.88
∴∴ 53.74 – 27.86 = 25.88

Page No 122:

Question 2:

Subtract:
64.98 from 103.87

ANSWER:

Let us write the numbers in the column form with the larger one at the top.
Now, subtracting:
 103.87
– 64.98
  38.89
∴∴ 103.87 – 64.98 = 38.89

Page No 122:

Question 3:

Subtract:
59.63 from 92.4

ANSWER:

Converting the given numbers into like decimals:
59.63 and 92.40
Let us write them in the column form with the larger number at the top.
Now, subtracting:
   92.40
 – 59.63
   32.77
∴∴ 53.74 – 27.86 = 32.77

Page No 122:

Question 4:

Subtract:
56.8 from 204

ANSWER:

Converting the given numbers into like decimals:
56.80 and 204.00
Let us write them in the column form with the larger number at the top.
Now, subtracting:
  204.00
 – 56.80
   147.2
∴∴ 204.00 – 56.80 = 147.2

Page No 122:

Question 5:

Subtract:
127.38 from 216.2

ANSWER:

Converting the given numbers into like decimals:
127.38 and 216.20
Let us write them in the column form with the larger number at the top.
Now, subtracting:
  216.20
– 127.38
    88.82
∴∴ 216.20 – 127.38 = 88.82

Page No 122:

Question 6:

Subtract:
39.875 from 70.68

ANSWER:

Converting the given numbers into like decimals:
39.875 and 70.680
Let us write them in the column form with the larger number at the top.
Now, subtracting:
  70.680
– 39.875
  30.805
∴∴ 70.680 – 39.875 = 30.805

Page No 122:

Question 7:

Subtract:
348.237 from 523.12

ANSWER:

Converting the given numbers into like decimals:
348.237 and 523.120
Let us write them in the column form with the larger number at the top.
Now, subtracting:
  523.120
– 348.237
  174.883
∴∴ 523.120 – 348.237 = 174.883

Page No 122:

Question 8:

Subtract:
458.573 from 600

ANSWER:

Converting the given numbers into like decimals:
458.573 and 600.000
Let us write them in the column form with the larger number at the top.
Now, subtracting:
  600.000
– 458.573
  141.427
∴∴ 600.000 – 458.573 =141.427

Page No 122:

Question 9:

Subtract:
149.456 from 206.321

ANSWER:

Let us write the numbers in the column form with the larger one at the top.
Now, subtracting:
  206.321
– 149.456
    56.865
∴∴ 206.321 – 149.456 = 56.865

Page No 122:

Question 10:

Subtract:
0.612 from 3.4

ANSWER:

Converting the given numbers into like decimals:
3.400 and 0.612
Let us write them in the column form with the larger number at the top.
Now, subtracting:
  3.400
– 0.612
  2.788
∴∴ 3.400 – 0.612 = 2.788

Page No 122:

Question 11:

Simplify:
37.6 + 72.85 − 58.678 − 6.09

ANSWER:

Converting the given decimals into like decimals, then adding and, finally, subtracting:
                         
        37.60                        58.678
     + 72.85                      +  6.090                
       110.45                        64.768 
                         
                                 
                                      110.450
                                   −-  64.768
                                        45.682    

(37.60 + 72.85) −- (58.678 + 6.090)
= 110.450 −- 64.768
= 45.682                     

Page No 122:

Question 12:

Simplify:
75.3 − 104.645 + 178.96 − 47.9

ANSWER:

Converting the given decimals into like decimals, then adding and, finally, subtracting:
                         
      178.96                        104.645
     + 75.30                      + 47.900              
      254.26                        152.545

(75.30 + 178.96) −- (104.645 + 47.900)
254.260 −- 152.545
101.715                           254.260
                                   −- 152.545
                                       101.715
                                 
                              

Page No 122:

Question 13:

Simplify:
213.4 − 56.84 − 11.87 − 16.087

ANSWER:

Converting the given decimals into like decimals, then adding and, finally, subtracting:
                         
       56.840
       11.870                     
     +16.087                                  
       84.797                 

(213.400) −- (56.840 + 11.870 + 16.087)
213.400 −- 84.797
128.603                          213.400
                                   −-  84.797
                                      128.603
                                 
                              

Page No 122:

Question 14:

Simplify:
76.3 − 7.666 − 6.77

ANSWER:

Converting the given decimals into like decimals, then adding and, finally, subtracting:
                         
         7.666                       
     +  6.770                            
       14.436                   

(76.300) −- (7.666 + 6.770)
= 76.300 −- 14.436
= 61.864                          

  76.300
-14.436
 61.864
                                 
                              

Page No 122:

Question 15:

What is to be added to 74.5 to get 91?

ANSWER:

In order to get the number that must be added to 74.5 to get 91, we must subtract 74.5 from 91.0.

         91.0
    −-  74.5
         16.5

Thus, 16.5 is the required number.

Page No 122:

Question 16:

What is to be subtracted from 7.3 to get 0.862?

ANSWER:

In order to get the number that must be subtracted from 7.300 to get .0862, we have to subtract 0.862 from 7.300.

         7.300
    −-  0.862
        6.438 

Thus, 6.438 is the required number.

Page No 122:

Question 17:

By how much should 23.754 be increased to get 50?

ANSWER:

In order to get the number by which 23.754 must be increased to get 50, we have to subtract 23.754 from 50.000.
 
      50.000
   −-23.754
      26.246

Page No 122:

Question 18:

By how much should 84.5 be decreased to get 27.84?

ANSWER:

In order to get the number by which 84.50 must be decreased to get 27.84, we have to subtract 27.84 from 84.50.
 
      84.50
   −-27.84
      56.66

Page No 122:

Question 19:

If the school bags of Neelam and Garima weigh 6 kg 80 g and 5 kg 265 g respectively, whose bag is heavier and by how much?

ANSWER:

Weight of Neelam’s school bag =       6080 g  {Converting into grams: 6 kg + 80 g = (6000 + 80) g =  6080 g}
Weight of Garima’s school bag =     −-5265 g  {Converting into grams: 5 kg + 265 g = (5000 + 265)g = 5265 g}
Difference of the weights of bags =      815 g
Thus, the weight of Neelam’s school bag is more than that of Garima’s school bag by 815 grams, i.e. by 0.815 kg.

Page No 122:

Question 20:

Kunal purchased a notebook for Rs 19.75, a pencil for Rs 3.85 and a pen for Rs 8.35 from a book shop. He gave a 50-rupee note to the shopkeeper. What amount did he get back?

ANSWER:

Cost of the notebook =  Rs 19.75
Cost of the pencil =        Rs 3.85
Cost of the pen =        + Rs 8.35
Total cost payable =      Rs 31.95

Total money paid =      Rs 50.00
Total money spent = −-Rs 31.95
Balance =                   Rs 18.05 

Thus, Kunal got back Rs 18.05 from the shopkeeper.

Page No 122:

Question 21:

Sunita purchased 5 kg 75 g of fruits and 3 kg 465 g of vegetables, and put them in a bag. If this bag with these contents weighs 9 kg, find the weight of the empty bag.

ANSWER:

Weight of the fruits =                                  5 kg 075 g
Weight of the vegetables =                       + 3 kg 465 g
Total weight of the contents of the bag =      8 kg 540 g

Total weight of the bag with its contents =    9 kg 000 g
Total weight of the contents of the bag =  −- 8 kg 540 g
Weight of the empty bag =                           0 kg 460 g
Thus, the weight of the empty bag is 460 grams.

Page No 123:

Question 22:

The distance between Reeta’s house and her office is 14 km. She covers 10 km 65 m by scooter, 3 km 75 m by bus and the rest on foot. How much distance does she cover by walking?

ANSWER:

Converting into metres:
10 km 65 m = (10 + 0.065) m = 10.065 m
3 km 75 m = (3 + 0.075) m = 3.075 m

Distance covered by the scooter =                             10.065 km
Distance covered by the bus =                                 + 3.075 km
Total distance covered by the bus and the scooter =   13.140 km

Total distance between the house and the office =       14.000 km
Total distance covered by the bus and the scooter = −-13.140 km
Distance covered on foot =                                          0.860 km

∴∴ Distance covered by walking = 0.860 km = 860 metres

Page No 123:

Exercise 7E

Question 1:

Mark (✓) against the correct answer in each of the following:
710=?710=?
(a) 7.1
(b) 1.7
(c) 0.7
(d) 0.07

ANSWER:

(c) 0.7

710710 = 7 tenths = 0.7

Page No 123:

Question 2:

Mark (✓) against the correct answer in each of the following:
5100=?5100=?
(a) 5.1
(b) 5.01
(c) 0.5
(d) 0.05

ANSWER:

(d) 0.05
 51005100 = 5 hundredths = 0.05

Page No 123:

Question 3:

Mark (✓) against the correct answer in each of the following:
91000=?91000=?
(a) 0.0009
(b) 0.009
(c) 9.001
(d) none of these

ANSWER:

(b) 0.009
9100091000 = 9 thousandths = 0.009

Page No 123:

Question 4:

Mark (✓) against the correct answer in each of the following:
161000=?161000=?
(a) 0.016
(b) 0.16
(c) 0.0016
(d) 1.006

ANSWER:

(a) 0.016
161000161000 = 16 thousandths = 0.016

Page No 123:

Question 5:

Mark (✓) against the correct answer in each of the following:
1341000=?1341000=?
(a) 13.4
(b) 1.34
(c) 0.134
(d) 0.0134

ANSWER:

(c) 0.134
 13410001341000 = 134 thousandths = 0.134

Page No 123:

Question 6:

Mark (✓) against the correct answer in each of the following:
217100=?217100=?
(a) 2.17
(b) 2.017
(c) 0.217
(d) 21.7

ANSWER:

(a) 2.17
 21710017100 = 2 + 1710017100 = 2 + 0.17 = 2.17

Page No 123:

Question 7:

Mark (✓) against the correct answer in each of the following:
43100=?43100=?
(a) 4.3
(b) 4.03
(c) 4.003
(d) 43.10

ANSWER:

(b) 4.03
 431003100 = 4 + 31003100 = 4 + 0.03 = 4.03

Page No 123:

Question 8:

Mark (✓) against the correct answer in each of the following:
6.25 = ?
(a) 612612
(b) 614614
(c) 62126212
(d) none of these

ANSWER:

b) 614614614
6.25 = 6 + 0.25 = 6 + 2510025100 = 6 + 1414 = 614614

Page No 123:

Question 9:

Mark (✓) against the correct answer in each of the following:
625=?625=?
(a) 2.4
(b) 0.24
(c) 0.024
(d) none of these

ANSWER:

(b) 0.24
     625625 =  0.24
                                  
            25) 60 (0.24
               −-50
                  100
              −- 100
                   0  

Page No 123:

Question 10:

Mark (✓) against the correct answer in each of the following:
478=?478=?
(a) 4.78
(b) 4.87
(c) 4.875
(d) none of these

ANSWER:

(c) 4.875
478478 =  4 + 7878 = 4 + 0.875 = 4.875

Page No 123:

Question 11:

Mark (✓) against the correct answer in each of the following:
24.8 = ?
(a) 24452445
(b) 24252425
(c) 24152415
(d) none of these

ANSWER:

(a)  24452445
24.8 = 24 + 0.8 = 24 + 810810 = 24 + 4545 = 24452445

Page No 124:

Question 12:

Mark (✓) against the correct answer in each of the following:
2125=?2125=?
(a) 2.4
(b) 2.04
(c) 2.004
(d) none of these

ANSWER:

(b) 2.04
21252125 = 2 + 125125 = 2 + 0.04 = 2.04

Page No 124:

Question 13:

Mark (✓) against the correct answer in each of the following:
2+310+41002+310+4100
(a) 2.304
(b) 2.403
(c) 2.34
(d) none of these

ANSWER:

(c) 2.34
2 + 310 + 4100310 + 4100 = 2 + 0.3 + 0.04 = 2.34

Page No 124:

Question 14:

Mark (✓) against the correct answer in each of the following:
2+6100=?2+6100=?
(a) 2.006
(b) 2.06
(c) 2.6
(d) none of these

ANSWER:

(b) 2.06
2 + 61006100 = 2 + 0.06 = 2.06

Page No 124:

Question 15:

Mark (✓) against the correct answer in each of the following:
4100+710000=?4100+710000=?
(a) 0.47
(b) 0.407
(c) 0.0407
(d) none of these

ANSWER:

(c) 0.0407
4100 + 7100004100 + 710000 = 0.04 + 0.0007 = 0.0407

Page No 124:

Question 16:

The correct expanded from of 2.06 is
(a) (2 × 10) + (6 × 110)2 × 10 + 6 × 110
(b) (2 × 1) + (6 ×110)2 × 1 + 6 ×110
(c) (2 × 1) + (6 × 1100)2 × 1 + 6 × 1100
(d) none of these

ANSWER:

(c) (2 × 1) + (6 × 1100)(2 × 1) + (6 × 1100)

2.06 =  2 + 61006100 = (2 × 1) + (6 × 1100)(2 × 1) + (6 × 1100)2 × 1) + (6 × 1(2 × 1) + (6 (2 × 1) + (6 

Page No 124:

Question 17:

Amoung 2.6, 2.006, 2.66 and 2.08, the largest number is
(a) 2.006
(b) 2.08
(c) 2.6
(d) 2.66

ANSWER:

(d) 2.66
Converting the given decimals into like decimals:
2.600, 2.006, 2.660 and 2.080
Among the given decimals, 2.660 is the largest.

Page No 124:

Question 18:

Which of the following is the correct order?
(a) 2.2 < 2.02 < 2.002 < 2.222
(b) 2.002 < 2.02 < 2.2 < 2.222
(c) 2.02 < 2.22 < 2.002 < 2.222
(d) none of these

ANSWER:

(b) 2.002 < 2.02 < 2.2 < 2.222

Converting the given decimals into like decimals:
2.002, 2.020, 2.200, 2.222

∴∴ 2.002 < 2.02 < 2.2 < 2.222

Page No 124:

Question 19:

Which is larger: 2.1 or 2.055?
(a) 2.1
(b) 2.055
(c) cannot be compared

ANSWER:

(a) 2.1
If we convert the given decimals into like decimals, we get 2.100 and 2.055.
At tenths place, 1 is greater than 0. Thus, 2.100 is greater than 2.055.

Page No 124:

Question 20:

Mark (✓) against the correct answer in each of the following:
1 cm = ?
(a) 0.1 m
(b) 0.01 m
(c) 0.001 m
(d) none of these

ANSWER:

(b) 0.01 m
1 m = 100 cm
∴∴ 1 cm = 11001100m = 0.01 m   

Page No 124:

Question 21:

Mark (✓) against the correct answer in each of the following:
2 m 5 cm = ?
(a) 2.5 m
(b) 2.05 m
(c) 2.005 m
(d) 0.25 m

ANSWER:

(b) 2.05 m
2 m 5 cm = (2 + 51005100) m = (2 + 0.05) m = 2.05 m

Page No 124:

Question 22:

Mark (✓) against the correct answer in each of the following:
2 kg 8 g = ?
(a) 2.8 kg
(b) 2.08 kg
(c) 2.008 kg
(d) none of these

ANSWER:

(c) 2.008 kg
1 kg = 1000 g
∴∴ 2 kg 8 g = 2 kg + 8100081000 kg = (2 + 0.008) kg = 2.008 kg

Page No 124:

Question 23:

Mark (✓) against the correct answer in each of the following:
2 kg 56 g = ?
(a) 2.56 kg
(b) 2.056 kg
(c) 2.560 kg
(d) none of these

ANSWER:

(b) 2.056 kg
2 kg + 56 g = (2 + 561000561000) kg = (2 + 0.056) kg = 2.056 kg

Page No 124:

Question 24:

Mark (✓) against the correct answer in each of the following:
2 km 35 m = ?
(a) 2.35 km
(b) 2.350 km
(c) 2.035 km
(d) none of these

ANSWER:

(c) 2.035 km
1 km = 1000 m
∴∴ 2 km 35 m  = (2 + 351000351000) km = (2 + 0.035) km = 2.035 km

Page No 124:

Question 25:

Mark (✓) against the correct answer in each of the following:
0.4 + 0.004 + 4.4 = ?
(a) 4.444
(b) 5.2
(c) 4.804
(d) 5.404

ANSWER:

(c) 4.804
0.4 + 0.004 + 4.4
Converting into like decimals and then adding:
   4.400
   0.004
+ 0.400
   4.804

Page No 124:

Question 26:

Mark (✓) against the correct answer in each of the following:
3.5 + 4.05 − 6.005 = ?
(a) 1.545
(b) 1.095
(c) 1.6
(d) none of these

ANSWER:

(a) 1.545
Converting into like decimals:
3.500 + 4.050 − 6.005
   3.500
+ 4.050
   7.550

   7.550
− 6.005
   1.545

Page No 124:

Question 27:

Mark (✓) against the correct answer in each of the following:
6.3 − 2.8 = ?
(a) 0.35
(b) 3.5
(c) 3.035
(d) none of these

ANSWER:

(b) 3.5

   6.3
− 2.8
   3.5

Page No 124:

Question 28:

Mark (✓) against the correct answer in each of the following:
5.01 − 3.6 = ?
(a) 4.65
(b) 1.95
(c) 1.41
(d) none of these

ANSWER:

(c) 1.41
Converting into like decimals and then subtracting:
   5.01
− 3.60
   1.41

Page No 125:

Question 29:

Mark (✓) against the correct answer in each of the following:
2 − 0.7 = ?
(a) 1.3
(b) 1.5
(c) 2.03
(d) none of these

ANSWER:

(a) 1.3
Converting into like decimals and then subtracting:
   2.0
− 0.7
   1.3

Page No 125:

Question 30:

Mark (✓) against the correct answer in each of the following:
1.1 − 0.3 = ?
(a) 0.8
(b) 0.08
(c) 8
(d) none of these

ANSWER:

(a) 0.8
Converting into like decimals and then subtracting:
   1.1
− 0.3
   0.8

Page No 126:

Exercise 7F

Question 1:

Convert 458458 into a decimal fraction.

ANSWER:

458458 = 4 + 5858 =  4 + 0.625 = 4.625

Page No 126:

Question 2:

Express 105 cm into metres using decimals.

ANSWER:

1 cm = 11001100 m
∴∴ 105 cm = 105100105100 m = (1 + 51005100) m = (1 + 0.05) m = 1.05 m

Page No 126:

Question 3:

Express 6 km 5 m as km using decimals.

ANSWER:

1 m = 1100011000 km
  ∴∴ 5 m = 5100051000 km = 0.005 km
6 km 5 m = (6 + 0.005) km = 6.005 km

Page No 126:

Question 4:

Express 8 m as kilometre using decimals.

ANSWER:

1 m = 1100011000 km
∴∴8 m = 8100081000 km = 0.008 km

Page No 126:

Question 5:

Add 26.4, 163.05, 8.75 and 5.6.

ANSWER:

Converting into like decimals and then adding:
    26.40
  163.05
     8.75
+   5.60
  203.80

Page No 126:

Question 6:

Subtract 0.528 from 3.2.

ANSWER:

Converting into like decimals and then subtracting:
    3.200
−- 0.528
    2.672

Page No 126:

Question 7:

What is to be added to 63.5 to get 71?

ANSWER:

We subtract 63.5 from 71 in order to find the number that is to be added to 63.5 to make it 71.
    Converting into like decimals and then subtracting:  
     71.0
 −- 63.5
      7.5

Page No 126:

Question 8:

What is to be subtracted from 13 to get 5.4?

ANSWER:

We subtract 5.4 from 13 in order to find the number that is to be subtracted from 13 to get 5.4.
Converting into like decimals and then subtracting:
      13.0
   −-  5.4
       7.6 
Thus, we need to subtract 7.6 from 13 to get 5.4.

Page No 126:

Question 9:

Arrange the following decimals in descending order:
6.5, 6.05, 6.54, 6.4 and 6.45

ANSWER:

6.5, 6.05, 6.54, 6.4 and 6.45
Converting into like decimals:
6.50, 6.05, 6.54, 6.40 and 6.45
Clearly, 6.54 > 6.50 > 6.45 > 6.40 > 6.05
Hence, the given decimals can be arranged in the descending order in the following manner:
 6.54, 6.50, 6.45, 6.40 and 6.05

Page No 126:

Question 10:

Convert each of the following into a fraction in simplest form:
(i) .4
(ii) .35
(c) 0.08
(iv) 0.075

ANSWER:

(i) 0.4 = 410 = 25410 = 25

(ii) 0.35 = 35100 = 72035100 = 720

(iii) 0.08 = 8100 = 2258100 = 225

(iv) 0.075 = 751000 = 340751000 = 340

Page No 126:

Question 11:

Mark (✓) against the correct answer in each of the following:
325=?325=?
(a) 1.2
(b) 0.12
(c) 0.012
(d) none of these

ANSWER:

(b) 0.12
 325325 = 0.12
                  
      25)  30 (0.12
         −- 25
               50
           −- 50  
                 0  

Page No 126:

Question 12:

Mark (✓) against the correct answer in each of the following:
61000=?61000=?
(a) 6.001
(b) 0.0006
(c) 0.006
(d) 0.06

ANSWER:

(c) 0.006
 6100061000 = 6 thousandths = 0.006

Page No 126:

Question 13:

Mark (✓) against the correct answer in each of the following:
23100=?23100=?
(a) 2.003
(b) 2.03
(c) 2.3
(d) none of these

ANSWER:

(b) 2.03
2310023100 = 2 + 31003100 = 2 + 0.03 = 2.03

Page No 126:

Question 14:

Mark (✓) against the correct answer in each of the following:
The place value of 3 in 16.534 is
(a) 310310
(b) 31003100
(c) 3100031000
(d) 3

ANSWER:

(b) 31003100
 In 16.534, place value of 3 = 3 hundredths = 0.03 = 31003100

Page No 126:

Question 15:

Mark (✓) against the correct answer in each of the following:
478=?478=?
(a) 4.78
(b) 4.87
(c) 4.875
(d) none of these

ANSWER:

(c) 4.875
478478 = 4 + 7878 = 4 + 0.875 = 4.875

Page No 126:

Question 16:

Mark (✓) against the correct answer in each of the following:
5.01 − 3.6 = ?
(a) 4.65
(b) 1.95
(c) 1.41
(d) none of these

ANSWER:

(c) 1.41
Converting into like decimals:
     5.01
 −  3.60
     1.41

Page No 126:

Question 17:

Mark (✓) against the correct answer in each of the following:
3.5 + 4.05 − 6.005 = ?
(a) 1.545
(b) 1.095
(c) 1.6
(d) none of these

ANSWER:

(a) 1.545

(3.50 + 4.05) −- 6.005
 7.550 −- 6.005
Converting into like decimals and then adding:
      3.50
   + 4.05
      7.55
Converting into like decimals and then subtracting:
     7.550
−-  6.005
     1.545

Page No 126:

Question 18:

Mark (✓) against the correct answer in each of the following:
4100+710000=?4100+710000=?
(a) 0.47
(b) 0.407
(c) 0.0407
(d) none of these

ANSWER:

(c) 0.0407

  4100 + 7100004100 + 710000 = 0.04  + 0.0007
Converting into like decimals:
0.0400 + 0.0007 = 0.0407

Page No 126:

Question 19:

Mark (✓) against the correct answer in each of the following:
Among 2.6, 2.006, 2.66 and 2.08, the largest number is
(a) 2.006
(b) 2.08
(c) 2.6
(d) 2.66

ANSWER:

(d) 2.66
Converting all the given decimals into like decimals:
2.600, 2.006, 2.660 and 2.080
Among these given like decimals, 2.660 is the largest.

Page No 126:

Question 20:

Fill in the blanks.
(i) 1 m = …… km
(ii) 10 ml = …… 1
(iii) 16 kg 5 g = …… kg
(iv) 2 m 8 cm = …… m
(v) 3.02, 4.75, 1.63 are examples of …… decimals.

ANSWER:

(i) 1 m = 1100011000 km = 0.001 km

(ii) 10 mL = 11001100 1 = 0.01 L

(iii) 16 kg 5 g = (16 + 0.005) kg = 16.005 kg
(iv) 2 m 8 cm = (2 + 0.08) m = 2.08 m
(v) 3.02, 4.75, 1.63 are examples of like decimals.

Page No 127:

Question 21:

Write ‘T’ for true and ‘F’ for false for each of the statements given below:
(i) 3.02 < 3.2.
(ii) 3 g = 0.003 kg.
(iii) 3411000=3.410.3411000=3.410.
(iv) 6.2 and 6.200 are equivalent decimals.
(v) 2.3, 3.41, 4.53, 5.61 are examples of like decimals.

ANSWER:

(i) True
This is because after converting them into like decimals, we get 3.20 > 3.02, which is true.

(ii) True
1 g = 1100011000 kg
∴∴ 3 g = 3100031000 = 0.003 kg

(iii) False
34110003411000 = 0.341
This is because it is equal to 341 thousandths.

(iv) True

(v) False
In this case, every decimal should have 2 decimals in order to be like decimals. But, 2.3 has only 1 decimal.

Chapter 15 – Circles Exercise Ex. 15.1

Question 1

Fill in the blanks:

(i) All points lying inside/outside a circle are called …… points/ … points.

(ii) Circles having the same centre and different radii are called … circles.

(iii) A point whose distance from the centre of a circle is greater than its radius lies in … of the circle.

(iv) A continuous piece of a circle is … of the circle.

(v) The longest chord of a circle is a … of the circle.

(vi) An arc is a … when its ends are the ends of a diameter.

(vii) Segment of a circle is the region between an arc and … of the circle.

(viii) A circle divides the plane, on which it lies, in …. parts.Solution 1

(i) interior/exterior

(ii) concentric

(iii) the exterior

(iv) arc

(v) diameter

(vi) semi-circle

(vii) centre

(viii) threeQuestion 2

Write the truth value (T/F) of the following with suitable reasons:

(i) A circle is a plane figure.

(ii) Line segment joining the centre to any point on the circle is a radius of the circle.

(iii) If a circle is divided into three equal arcs each is a major arc.

(iv) A circle has only finite number of equal chords.

(v) A chord of a circle, which is twice as long is its radius is a diameter of the circle.

(vi) Sector is the region between the chord and its corresponding arc.

(vii) The degree measure of an arc is the complement of the central angle containing the arc.

(viii) The degree measure of a semi-circle is 180^o.Solution 2

(i) T

(ii) T

(iii) T

(iv) F

(v) T

(vi) T

(vii) F

(viii) T

Chapter 15 – Circles Exercise Ex. 15.2

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Give a method to find the centre of a given circle.Solution 4

Steps of construction:

(1) Take three point A, B and C on the given circle.

(2) Join AB and BC.

(3) Draw the perpendicular bisectors of chord AB and BC which interesect each other at O.

(4) Point O will be the required circle because we know that the perpendicular bisector of a chord always passes through the centre.Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11

The lengths of two parallel chords of a circle are 6 cm and 8 cm. If the smaller chord is at distance 4 cm from the centre, what is the distance of the other chord form the centre?Solution 11

                                              Distance of smaller chord AB from centre of circle = 4 cm.
OM = 4 cm

In OMB

In ONDOD=OB=5cm             (radii of same circle)

 So, distance of bigger chord from centre is 3 cm.Question 12

Solution 12

Question 13

Solution 13

Question 14

Prove that two different circles cannot intersect each other at more than two points.Solution 14

Suppose two different circles can intersect each other at three points then they will pass through the three common points but we know that there is one and only one circle with passes through three non-collinear points, which contradicts our supposition.

Hence, two different circles cannot intersect each other at more than two points.Question 15Two chords AB and CD of lengths 5 cm and 11cm respectively of a circle are parallel to each other and are on opposite sides of its centre. If the distance between AB and CD is 6 cm, find the radius of the circle.Solution 15Draw OM  AB and ON  CD. Join OB and OD 

                     (Perpendicular from centre bisects the chord)

Let ON be x, so OM will be 6 – x
In MOB

In NOD

 We have OB = OD             (radii of same circle)
So, from equation (1) and (2) 

From equation (2) 

So, radius of circle is found to be  cm.

Chapter 15 – Circles Exercise Ex. 15.3

Question 1

Three girls Ishita, Isha and Nisha are playing a game by standing on a circle of radius 20 m drawn in a park. Ishita throws a ball to Isha, Isha to Nisha, Nisha to Ishita. If the distance between Ishita and Isha and between Isha and Nisha is 24 m each, what is the distance between Ishita and Nisha?Solution 1

Question 2

A circular park of radius 40 m is situated in a colony. Three boys Ankur, Amit and Anand are sitting at equal distance on its boundary each having a toy telephone in his hands to talk to each other. Find the length of the string of each phone.Solution 2

Chapter 15 – Circles Exercise Ex. 15.4

Question 1

In fig., O is the centre of the circle. If ∠APB = 50°, find ∠AOB and ∠OAB.

Solution 1

Question 2

In fig., O is the centre of the circle. Find ∠BAC.

Solution 2

Question 3(i)

Solution 3(i)

Question 3(ii)

Solution 3(ii)

Question 3(iii)

Solution 3(iii)

Question 3(iv)

Solution 3(iv)

Question 3(v)

Solution 3(v)

Question 3(vi)

Solution 3(vi)

Question 3(vii)

If O is the centre of the circle. Find the value of x in the following figure:

Solution 3(vii)

Question 3(viii)

Solution 3(viii)

Question 3(ix)

If O is the centre of the circle. Find the value of x in the following figure:

Solution 3(ix)

Question 3(x)

If O is the centre of the circle. Find the value of x in the following figure:

Solution 3(x)

Question 3(xi)

If O is the centre of the circle. Find the value of x in the following figure:

Solution 3(xi)

Question 3(xii)

Solution 3(xii)

Question 4

Solution 4

Question 5

In fig., O is the centre of the circle, BO is  the bisector of ∠ABC. Show that AB = AC.

Solution 5

Question 6

In fig., O and O’ are centres of two circles intersecting at B and C. ABD is straight line, find x.

Solution 6

Question 7

In fig., if ∠ACB = 40°, ∠DPB = 120°, find ∠CBD.

Solution 7

Question 8

A chord of a circle is equal to the radius of the circle. Find the angle subtended by the chord at a point on the minor arc and also at a point on the major arc.Solution 8

Question 9

In fig., it is given given that O is the centre of the circle and ∠AOC = 150°. Find ∠ABC.

Solution 9

Question 10

In fig., O is the centre of the circle, prove that ∠x = ∠y + ∠z.

Solution 10

Question 11

in fig., O is the centre of a circle and PQ is a diameter. If ∠ROS = 40°, find. ∠RTS.

Solution 11

Chapter 15 – Circles Exercise Ex. 15.5

Question 1

In fig., ΔABC is an equilateral triangle. Find m∠BEC.

Solution 1

Question 2

In fig., ΔPQR is an isosceles triangle with PQ = PR and m∠PQR = 35°. find m∠QSR and m∠QTR.

Solution 2

Question 3

In fig., O is the centre of the circle. If ∠BOD = 160°, find the values of x and y.

Solution 3

Question 4

In fig., ABCD is a cyclic qudrilateral. If ∠BCD = 100° and ABD = 70°, find ∠ADB.

Solution 4

Question 5

If ABCD is a cyclic quadrilateral in which AD ∥ BC. Prove that ∠B = ∠C.

Solution 5

Question 6

In fig., O is the centre of the circle. find ∠CBD.

Solution 6

Question 7

In fig., AB and CD are diameters of a circle with centre O. If ∠OBD = 50°, find ∠AOC.

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11

In fig., O is the centre of the circle and DAB = 50. calculate the values of x and y.

Solution 11

Question 12

In fig., if ∠BAC = 60°, and ∠BCA = 20°, find ∠ADC.

Solution 12

Question 13

In fig., if ABC is an equilateral triangle. Find ∠BDC and ∠BEC.

Solution 13

Question 14

In fig., O is the centre of the circle. If ∠CEA = 30°, find the values of x, y and z.

Solution 14

Question 15

In fig., ∠BAD = 78°, ∠DCF = x° and DEF = y° find the values of x and y.

Solution 15

Question 16

Solution 16

Question 17

In fig., ABCD is cyclic qudrilateral. Find the value of x.

Solution 17

Question 18(i)

Solution 18(i)

Question 18(ii)

Solution 18(ii)

Question 18(iii)

Solution 18(iii)

Question 19

Solution 19

Question 20

Solution 20

Question 21

Solution 21

Question 22

Solution 22

Question 23

In fig., ABCD is cyclic quadrilaterial in which AC an BD are its diagonals. If ∠DBC = 55° and ∠BAC = 45°, find ∠BCD.

Solution 23

Question 24

Solution 24

Question 25

Prove that the centre of the circle circumscribing the cyclic rectangle ABCD is the point of intersection of its diagonals.Solution 25

Let O be the centre of the circle circumscribing the cyclic rectangle ABCD. Since ABC = 90^o and AC is a chord of the circle, so, AC is a diameter of the circle. Similarly, BD is a diameter.

Hence, point of intersection of AC and BD is the centre of the circle.Question 26

Solution 26

Question 27

Solution 27

Question 28

Solution 28

Question 29

Prove that the line segment joining the mid-point of the hypotenuse of a right triangle to its opposite vertex is half of the hypotenuse.Solution 29

Chapter 14 – Areas of Parallelograms and Triangles Exercise Ex. 14.1

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 two parallels:

Solution 1

(i) ΔAPB and trapezium ABCD are on the same base AB and between the same parallels AB and CD.

(ii) Parallelograms ABCD and APQD are on the same base AD and between the same parallels AD and BQ.

(iii) Parallelogram ABCD and ΔPQR are between the same parallels AD and BC but they are not on the same base.

(iv) ΔQRT and parallelogram PQRS are on the same base QR and between the same parallels QR and PS.

(v) Parallelogram PQRS and trapezium SMNR are on the same base SR but they are not between the same parallels.

(vi) Parallelograms PQRS, AQRD, BQRC are between the same parallels. Also, parallelograms PQRS, BPSC and APSD are between the same parallels.

Chapter 14 – Areas of Parallelograms and Triangles Exercise Ex. 14.2

Question 1In the given figure, ABCD is parallelogram, AE  DC and CF  AD. If AB = 16 cm. AE = 8 cm and CF = 10 cm, find AD.  

Solution 1In parallelogram ABCD, CD = AB = 16 cm     [Opposite sides of a parallelogram are equal]
We know that,
Area of parallelogram = Base x corresponding attitude
Area of parallelogram ABCD = CD x AE = AD x CF  
16 cm x 8 cm = AD x 10 cmAD =  cm = 12.8 cm.Thus, the length of AD is 12.8 cm.Question 2

In Q. No. 1, if AD = 6 cm, CF = 10 cm, and AE = 8, find AB.Solution 2

Question 3

Solution 3

Question 4

Solution 4

Chapter 14 – Areas of Parallelograms and Triangles Exercise Ex. 14.3

Question 1

In fig., compute the area of quadrilateral ABCD.

Solution 1

Question 2

In the fig., PQRS is a square and T and U are, respectively, the mid-points of PS and QR. Find the area of ΔOTS if PQ = 8 cm.

Solution 2

Question 3

Compute the area of trapezium PQRS in fig.

Solution 3

Question 4

In fig., ∠AOB = 90, AC = BC, OA = 12 cm and OC = 6.5 cm. find the area of ΔAOB

Solution 4

Question 5

In fig., ABCD is a trapezium in which AB = 7 cm, AD = BC = 5 cm, DC = x cm, and distance between AB and DC is 4 cm. Find the value of x and area of trapezium ABCD.

Solution 5

Question 6

Solution 6

Question 7

In fig., ABCD is a trapezium in which AB ∥ DC. PRove that ar (ΔAOD) = ar (ΔBOC)

Solution 7

Question 8

Solution 8

Question 9

In fig., ABC and ABD are two triangles on the base Ab. If the line segment CD is bisected by AB at O, show that ar (Δ ABC) = ar (Δ ABD).

Solution 9

Question 10

Solution 10

Question 11

Solution 11

Question 12

Solution 12

Question 13

Solution 13

Question 14

Solution 14

Question 15

In fig., D and E are two points on BC such that BD = DE = EC. Show that ar (ΔABD) = ar (ΔADE) = ar(ΔAEC).

Solution 15

Draw a line l through A parallel to BC.

Given that, BD = DE = EC.

We observe that the triangles ABD, ADE and AEC are on the equal bases and between the same parallels l and BC. Therefore, their areas are equal.

Hence, ar (ΔABD) = ar (ΔADE) = ar(ΔAEC).Question 16

Solution 16

Question 17

Solution 17

Question 18

Solution 18

Question 19

Solution 19

(i) 
(ii) 

(iii)

Question 20

Solution 20

Question 21

In fig., CD ∥ AE and CY ∥ BA.

(i) Name a triangle equal in area of ΔCBX 

(ii) Prove that ar (ΔZDE) = ar (ΔCZA)

(iii) Prove that ar (BCZY) = ar (ΔEDZ)

Solution 21

Question 22

In fig., PSDA is a parallelogram in which PQ = QR = RS and AP ∥ BQ ∥CR. Prove that ar(Δ PQE) = ar (Δ CFD).

Solution 22

Question 23

In fig., ABCD is a trapezium in which AB DC and DC = 40 cm and AB = 60 cm. If X and Y are, respectively, the mid – points of AD and BC, prove that:

(i) XY = 50 cm

(ii) DCYX is a trapezium

(iii) ar (trap. DCYX) = (9/11)ar (trap.(XYBA)

Solution 23

Question 24

Solution 24

Question 25

In fig., X and Y are the mid-points of AC and AB respectively, QP ∥ BC and CYQ and BXP are straight lines. Prove that ar(Δ ABP) = ar (Δ ACQ)

Solution 25

Question 26

In fig., ABCD and AEFD are two parallelograms. Prove that

(i) PE = FQ

(ii) ar(ΔAPE) : ar(ΔPFA) = ar Δ(QFD) : ar (ΔPFD)

(iii) ar(ΔPEA) = ar (ΔQFD)

Solution 26

Question 27

In fig. ABCD is a ∥^gm. O is any point on AC. PQ ∥ AB and LM ∥ AD. Prove that ar(∥^gm DLOP) = ar (∥^gm BMOQ).

Solution 27

Question 28

Solution 28

Question 29

In fig., ABC and BDE are two equilateral triangles such that D is the mid-point of BC. AE intersects BC in F. Prove that

(i) ar (BDE) = ar (ABC)

(ii) ar(BDE) = ar(BAE)

(iii) ar (BFE) = ar(AFD)

(iv) ar(ABC) = 2 ar(BEC)

(v) ar (FED) = ar(AFC)

(vi) ar(BFE) = 2 ar (EFD)

Solution 29

Given, ABC and BDE are two equilateral triangles.

Let AB = BC = CA = x. Then, BD = = DE = BE

(i) We have,

ar(ABC) = x²

ar (BDE) = 

ar(BDE) = ar (ABC)

(ii) It is given that triangles ABC and BED are equilateral triangles.

ACB = DBE = 60^o

BE||AC(Since, alternate angles are equal)

Triangles BAE and BEC are on the same base BE and between the same parallels BE and AC.

ar (BAE) = ar(BEC)

ar (BAE) =2 ar (BDE)

[ ED is a median of EBC ar(BEC) = 2ar(BDE)]

ar (BDE) = ar(BAE)

(iii) Since ABC and BDE are equilateral triangles.

ABC = 60^o and BDE = 60^o

ABC = BDE

AB||DE(Since, alternate angles are equal)

Triangles BED and AED are on the same base ED and between the same parallels AB and DE.

ar (BED) = ar(AED)

ar (BED) ar(EFD) = ar(AED) ar(EFD)

ar(BEF) = ar(AFD)

(iv) Since ED is a median of BEC

ar (BEC) = 2 ar (BDE)

ar (BEC) = ar (ABC)[From (i), ar (BDE) = ar (ABC)]

ar(BEC) = ar (ABC)

ar (ABC) = 2 ar (BEC)

(v) Let h be the height of vertex E, corresponding to the side BD in triangle BDE.

Let H be the height of vertex A, corresponding to the side BC in triangle ABC.

From part (i),

ar(BDE) = ar (ABC)

From part (iii),

ar (BFE) = ar (AFD)

(vi) ar (AFC) = ar (AFD) + ar (ADC)

= ar (BFE) + ar (ABC)

(Using part (iii); and AD is the median of ABC)

= ar (BFE) + 4 ar (BDE)(Using part (i))

= ar (BFE) + 2 ar (BDE) (2)

Now, from part (v),

ar (BFE) = 2ar (FED) (3)

ar (BDE) = ar (BFE) + ar (FED)

= 2 ar (FED) + ar (FED)

= 3 ar (FED) (4)

From (2), (3) and (4), we get,

ar (AFC) = 2ar (FED) + 2 3 ar (FED) = 8 ar (FED)

Hence, ar (FED) = ar(AFC)Now, fromQuestion 30

If fig., ABC is a right triangle right angled at A, BCED, ACFG and ABMN are square on the sides BC, CA and AB respectively. Line segment AX DE meets BC at Y. Show that

(i) MBC ABD

(ii) ar (BYXD) = 2ar(MBC)

(iii) ar(BYXD) = ar(ABMN)

(iv) FCB ACE

(v) ar(CYXE) = 2ar (FCB)

(vi) ar(CYXE) = ar (ACFG)

(vii) ar(BCED) = ar (ABMN) + ar (ACFG)

Solution 30

(i) In MBC and ABD, we have

MB = AB

BC = BD

And MBC = ABD

[MBC and ABD are obtained by adding ABC to a right angle]

So, by SAS congruence criterion, we have

MBC ABD

ar (MBC) = ar(ABD) (1)

(ii) Clearly, triangle ABD and rectangle BYXD are on the same base BD and between the same parallels AX and BD.

ar(ABD) = ar (rect. BYXD)

ar (rect. BYXD) = 2 ar(ABD)

ar (rect. BYXD) = 2 ar (MBC)…(2)

[ ar (ABD) = ar (MBC), from (1)]

(iii) Since triangle MBC and square MBAN are on the same base MB and between the same parallels MB and NC.

2 ar (MBC) = ar (MBAN) (3)

From (2) and (3), we have

ar (sq. MBAN) = ar(rec. BYXD)

(iv) In triangles FCB and ACE, we have

FC = AC

CB = CE

And, FCB = ACE

[FCB and ACE are obtained by adding ACB to a right angle]

So, by SAS congruence criterion, we have

FCB ACE

(v) We have,

FCB ACE

ar (FCB) = ar (ACE)

Clearly, ACE and rectangle CYXE are on the same base CE ad between the same parallels CE and AX.

2 ar (ACE) = ar (CYXE)

2 ar (FCB) = ar (CYXE) (4)

(vi) Clearly, FCB and rectangle FCAG are on the same base FC and between the same parallels FC and BG.

2ar (FCB) = ar(FCAG) (5)

From (4) and (5), we get

ar(CYXE) = ar (ACFG)

(vii) Applying Pythagoras theorem in ACB, we have

BC² = AB² + AC²

Chapter 13 – Quadrilaterals Exercise Ex. 13.1

Question 1

Solution 1

Question 2

Solution 2

Question 3The angles of quadrilateral are in the ratio 3: 5: 9: 13, Find all the angles of the quadrilateral.Solution 3Let the common ratio between the angles is x. So, the angles will be 3x, 5x, 9x and 13x respectively.
Since the sum of all interior angles of a quadrilateral is 360^o.
 3x + 5x + 9x + 13x = 360^o
30x = 360^o
    x = 12^o
Hence, the angles are
3x = 3  12 = 36^o
5x = 5  12 = 60^o
9x = 9  12 = 108^o
13x = 13  12 = 156^o

Question 4

Solution 4

Chapter 13 – Quadrilaterals Exercise Ex. 13.2

Question 1

Two opposite angles of a parallelogram are (3x – 2)° and (50 – x)°. Find the measure of each of the parallelogram.Solution 1

Question 2

If an angle of a parallelogram is two-third of its adjacent angle, find the angles of the parallelogram.Solution 2

Question 3

Find the measure of all the angles of a parallelogram, if one angle is 24^o less than twice the smallest angle.Solution 3

Question 4

The perimeter of a parallelogram is 22 cm. If the longer side measures 6.5 cm what is the measure of the shorter side?Solution 4

Question 5

In a parallelogram ABCD, ∠D = 135°, determine the measures of ∠A and ∠B.Solution 5

Question 6

ABCD is a parallelogram in which ∠A = 70. Compute ∠B, ∠C and ∠D.Solution 6

Question 7

In fig., ABCD is a parallelogram in which ∠DAB = 75° and ∠DBC = 60°. Compute ∠CDB and ∠ADB.

Solution 7

Question 8

Solution 8

i. F

ii. T

iii. F

iv. F

v. T

vi. F

vii. F

viii. TQuestion 9

In fig., ABCD is a parallelogram in which ∠A = 60°. If the bisectors of ∠A and ∠b meet at P, prove that AD = DP, PC = BC and DC = 2AD.

Solution 9

Question 10

In fig., ABCD is a parallelogram and E is the mid-point of side BC. IF DE and AB when produced meet at F, prove that AF = 2AB.

Solution 10

Chapter 13 – Quadrilaterals Exercise Ex. 13.3

Question 1

In a parallelogram ABCD, determine sum of angles ∠C and ∠D.Solution 1

C and D are cosecutive interior angles on the same side of the transversal CD. Therefore, 

C + D = 180^oQuestion 2

In a parallelogram ABCD, if ∠B = 135°, determine the measures of its other angles.Solution 2

Question 3

ABCD is a square. AC and BD intersect at O. State the measure of AOB.Solution 3

Since, diagonals of a square bisect each other at right angle. Therefore, AOB = 90^oQuestion 4

Solution 4

Question 5

The sides AB and CD of a parallelogram ABCD are bisected at E and F. Prove that EBFD is a parallelogram.Solution 5

Question 6

P and Q are the points of trisection of the diagonal BD of a parallelogram ABCD. Prove that CQ is parallel to AP. Prove also that AC bisects PQ.Solution 6

Question 7

ABCD is a square E, F, G and H are points on AB, BC, CD and DA respectively, such that AE = BF = CG = DH. Prove that EFGH is square.Solution 7

Question 8

ABCD is a rhombus, EABF is a straight line such that EA = AB = BF. Prove that ED and FC when produced meet at right angles.Solution 8

Question 9

ABCD is a parallelogram, AD is produced to E so that DE = DC and EC produced meets AB produced in F. Prove that BF = BC.Solution 9

Chapter 13 – Quadrilaterals Exercise Ex. 13.4

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

In fig., triangle ABC is right-angled at B. Give that AB = 9 cm, AC = 15 cm and D, E are the mid-points of the sides AB and AC respectively, calculate

(i) The length of BC

(ii) The area of ADE.

Solution 7

Question 8

In fig., M, N, and P are the mid-points of AB, AC and BC respectively. If MN = 3 cm, Np = 3.5 cm and MP = 2.5 cm, Calculate BC, AB and AC.

Solution 8

Question 9

In fig., AB = AC and CP ∥ BA and AP is the bisector of exterior ∠CAD of ΔABC. Prove that (i) ∠PAC = ∠BCA (ii) ABCP is a parallelogram.

Solution 9

Question 10

Solution 10

Question 11

Solution 11

Question 12Show that the line segments joining the mid-points of the opposite sides of a quadrilateral bisect each other.Solution 12

                                          Let ABCD is a quadrilateral in which P, Q, R and S are mid-points of sides AB, BC, CD and DA respectively.
Join PQ, QR, RS, SP and BD.
In ABD, S and P are mid points of AD and AB respectively.
So, By using mid-point theorem, we can say that
SP || BD and SP =   BD             … (1)  
Similarly in BCD
QR || BD and QR =  BD               … (2)
From equations (1) and (2), we have
SP || QR and SP = QR  
As in quadrilateral SPQR one pair of opposite sides are equal and parallel to
each other.
So, SPQR is a parallelogram.Since, diagonals of a parallelogram bisect each other.
Hence, PR and QS bisect each other.

Question 13

Fill in the blanks to make the following statements correct:

(i) The triangle formed by joining the mid-points of the sides of an isosceles traingle is ______.

(ii) The triangle formed by joining the mid-points of the sides of a right triangle is ______ .

(iii) The figure formed by joining the mid-points of consecutive sides of a quadrilateral is ______ .Solution 13

(i) isosceles

(ii) right triangle

(iii) parallelogramQuestion 14

Solution 14

Question 15

In fig., BE ⊥ AC. AD is any line from A to BC interesting BE in H. P, Q and R are respectively the mid-points of AH, AB and BC. Prove that ∠PQR = 90°.

Solution 15

Question 16

Solution 16

Question 17

In fig., ABCD is a parallelogram in which P is the mid-point of DC and Q is a point on AC such that CQ = (1/4)AC. If PQ produced meets BC at R, prove that R is a mid-point of BC.

Solution 17

Question 18

In fig., ABCD and PQRC are rectangle and Q is the mid-point of AC. Prove that 

i. DP = PC ii. PR = (1/2) AC

Solution 18

Question 19

ABCD is a parallelogram, E and F are the mid-points of AB and CD respectively. GH is any line intersecting AD, EF and BC and G, P and H respectively. Prove that GP = PH.Solution 19

Question 20

BM and CN are perpendiculars to a line passing through the vertex A of a triangle ABC. If L is the mid-point of BC, prove that LM = LN.Solution 20

Chapter 12 – Congruent Triangles Exercise Ex. 12.1

Question 1

In fig., the sides BA and CA have been produced such that BA = AD and CA = AE.

Prove that segment DE || BC

Solution 1

Question 2

Solution 2

Question 3

Prove that the medians of an equilateral triangle are equal.Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

The vertical angle of an isosceles triangle is 100^o. Find its base angles.Solution 6

Question 7

 In fig., AB = Ac and ∠ACD = 105°, find ∠BAC. 

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

In fig., AB =AC and DB = DC, find the ratio ∠ABD = ∠ACD. 

Solution 10

Question 11

Determine the measure of each of the equal angles of a right-angled isosceles triangle.

                                                                                   OR

ABC is a right-angled triangle in which A = 90^o and AB = AC. Find B and C.Solution 11

Question 12

Solution 12

Question 13

AB is a line segment. P and Q are points on opposite sides of AB such that each of them is equidistant from the points A and B (See fig.). Show that the line PQ is perpendicular bisector of AB.

Solution 13

Chapter 12 – Congruent Triangles Exercise Ex. 12.2

Question 1

Solution 1

Question 2

In fig., it is given RT = TS, ∠1 = 2∠2 and ∠4 = 2∠3 prove that ΔRBT ≅ ΔSAT.

Solution 2

Question 3

Solution 3

Chapter 12 – Congruent Triangles Exercise Ex. 12.3

Question 1

In two right triangles one side and acute angle of one are equal to the corresponding side and angle of the other. Prove that the triangles are congruent.Solution 1

Let ABC and DEF be two right triangles.

Question 2

Solution 2

Question 3

Solution 3

Question 4Show that the angles of an equilateral triangle are 60^o each.Solution 4

 Let us consider that ABC is an equilateral triangle.
So, AB = BC = AC
Now, AB = AC⇒ ∠C = ∠B         (angles opposite to equal sides of a triangle are equal)

We also have
AC = BC    
⇒ ∠B = ∠A             (angles opposite to equal sides of a triangle are equal)

So, we have
∠A = ∠B = ∠C
    Now, in ΔABC
∠A + ∠B + ∠C = 180^o
⇒ ∠A + ∠A + ∠A = 180^o
⇒ 3∠A = 180^o
⇒ ∠A = 60^o
⇒ ∠A = ∠B = ∠C = 60^o
Hence, in an equilateral triangle all interior angles are of 60^o.

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Chapter 12 – Congruent Triangles Exercise Ex. 12.4

Question 1

In fig., it is given that Ab = CD and AD = BC. prove that ΔADC ≅ ΔCBA

Solution 1

Question 2

Solution 2

Chapter 12 – Congruent Triangles Exercise Ex. 12.5

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

In fig., AD ⊥ CD and CB ⊥ CD. If AQ = BP an DP = CQ, prove that ∠DAQ = ∠CBP.

Solution 4

Question 5

Which of the following statements are True (T) and which are False (f):

(i) Sides opposite to equal angles of a triangle may be unequal.

(ii) Angles opposite to equal sides of a triangle are equal.

(iii) The measure of each angle of an equilaterial triangle is 60^o.

(iv) If the altitude from one vertex of a triangle bisects the opposite side, then the triangle may be isoscles.

(v) The bisectors of two equal angles of a traingle are equal.

(vi) If the bisector of the vertical angle of a triangle bisects the base, then the triangle may be isosceles.

(vii) The two altitudes corresponding to two equal sides of a triangle need not be equal.

(viii) If any two sides of a right triangle are respectively equal to two sides of other right triagnle, then the two triangles are congruent.

(ix) Two right triangles are congruent if hypotenuse and a side of one triangle are respectively equal to the hypotenuse and a side of the other triangle.Solution 5

(i) False

(ii) True

(iii) True

(iv) False

(v) True

(vi) False

(vii) False

(viii) False

(ix) TrueQuestion 6

Solution 6

(i) equal

(ii) equal

(iii) equal

(iv) BC

(v) AC

(vi) equal to

(vii) EFDQuestion 7

Solution 7

Chapter 12 – Congruent Triangles Exercise Ex. 12.6

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Is it possible to draw a triangle with sides of length 2cm, 3cm and 7 cm?Solution 4

Here, 2 + 3 < 7

Hence, it is not possible because triangle can be drawn only if the sum of any two sides is greater than third side.Question 5

Solution 5

Question 6

Solution 6

Question 7

In fig., prove that:

i. CD + DA + AB + BC > 2AC

ii. CD + DA + AB > BC

Solution 7

Question 8

Which of the following statements are true (T) and which are false (F)?

(i) Sum of the three sides of a triangle is less than the sum of its three altitudes.

(ii) Sum of any two sides of a triangle is greater than twice the median drawn to the third side.

(iii) Sum of any two sides of a triangle is greater than the third side.

(iv) Difference of any two sides of a triangle is equal to the third side.

(v) If two angles of a triangle are unequal, then the greater angle has the larger side opposite to it.

(vi) Of all the line segments that can be drawn from a point to a line not containing it, the perpendicular line segment is the shortest one.Solution 8

(i) False

(ii) True

(iii) True

(iv) False

(v) True

(vi) TrueQuestion 9

Solution 9

(i) largest

(ii) less

(iii) greater

(iv) smaller

(v) less

(vi) greaterQuestion 10

Solution 10

Question 11

Solution 11