NCERT Solutions for Class 9 Mathematics Chapter 13- Surface areas and volumes

Mathematics is a subject requiring high knowledge and skills. For one to be good at Mathematics, one must be thorough with the vital concepts and also be comfortable with calculations. Hence, students should know all the techniques associated with the study of Mathematics.

Class 9 Mathematics Chapter 13 is more of a kind of chapter of mensuration covered in lower classes. In order to score well in this chapter, one must be through with all the formulas mentioned in the chapter. The main topics covered in NCERT Class 9 Mathematics Chapter 13 include the total and lateral surface area of cube, cuboid, cylinder, cone and sphere. You will also be acquainted with how to find the volume of cubes, cuboids, cylinders, cones and spheres.

The chapter has been covered in detail in NCERT Solutions for Class 9 Mathematics Chapter 13. Students can find all the formulas compiled in quick notes format, which will facilitate them to quickly grasp the topics in this chapter. Students will also be able to apply these formulas as required in the problems once they refer to our NCERT Solutions.

All the students taking the guidance from Extramarks and following Extramarks resources have seen a lot of changes in their results and have turned out to be great academic performers. Thus, becoming confident and developing a strong personality. You can find all the study material required for your preparation on the Extramarks website.

Key Topics Covered In NCERT Solutions for Class 9 Mathematics Chapter 13

Mensuration has always been an important part of Geometry as it covers the detailed study of the calculation of areas of 2D and 3D shapes. The chapter surface areas and volumes are another name for mensuration. As the name suggests, the chapter covers all the formulas associated with cube, cuboid, cylinder, cone and sphere in calculating their surface areas and volumes. Thus, it is more of a formula learning-oriented chapter for the students.

Students should read and revise the formulas as much as they can in order to be thorough with the chapter. NCERT Solutions Class 9 Mathematics Chapter 13 has covered all the major points related to this chapter in a very precise manner aiding students to get rid of the fear of multiple formulas included in this chapter. Students can get NCERT Solutions for Class 9 Mathematics Chapter 13 from the Extramarks’ website.

NCERT Solutions for Class 9 Mathematics Chapter 13 requires students to use solutions, reasoning, thinking skills, and facilities in solving sums with a better approach.

Introduction

In this chapter, we will learn about shapes like cuboids, cubes, circular cylinders, circular cones, spheres, etc., and we will learn to find out their surface area and their volume. 

Surface area of cuboid and cube

In this part of the chapter, we will find out the surface area of the cuboid and cube. You can use the following formula for the surface area of the cuboid and the cube:

  • Surface area of cuboid = 2(lb + bh + hl)
  • Surface area of cube = 6a2

You can find multiple examples to practise and learn more about this section in our NCERT Solutions for Class 9 Mathematics Chapter 13, available on the Extramarks’ website.

Surface area of a right circular cylinder

In this part of the chapter, we will find out the surface area of the right circular cylinder using the formula:

  • Total surface area of right circular cylinder = 2πr(r + h)

Surface area of a right circular cone

In this part of the chapter, we will find out the surface area of the right circular cone. The total surface area of the right circular cone is given by:

  • Total Surface area of the right circular cone = πr(l + r) 

Surface area of a sphere and hemisphere.

In this part of the chapter, we will find out the surface area of the sphere and hemisphere. The formulas are as follows:

  • Surface area of sphere = 4πr2
  • Surface area of hemisphere = 2πr2.

For more details regarding the chapter and to find solutions to all the exercises given in the chapter in the NCERT Solutions for Class 9 Mathematics Chapter 13.

Volume of cuboid and cube

In this part of the chapter, we will find out the volume of the cuboid and cube. Their volume can be calculated by using the formula:

  • Volume of cuboid = lbh
  • Volume of cube = a3

Volume of cylinder

In this part of the chapter, we will find out the volume of the cylinder. The formula for the calculation of volume is

  • Volume of cylinder = πr2h

Volume of a right circular cone.

In this part of the chapter, we will find out the Volume of a right cylinder cone using the formula:

  • Volume of a right cylinder cone =⅓* πr2h 

Volume of a sphere

In this part of the chapter, we will find out the volume of a sphere and volume of a hemisphere using the formula listed below:

  • Volume of a sphere = 4/3*πr3
  • Volume of a hemisphere = 2/3*πr3

Summary

In this chapter, we covered all about the 2D and 3D shapes and the formulas to calculate their areas

Surface area 

  • Surface area of cuboid = 2(lb + bh + hl)
  • Surface area of cube = 6a
  • Total Surface area of right circular cylinder = 2πr(r + h)
  • Total Surface area of right circular cone = πr(l + r) 
  • Surface area of sphere = 4πr2
  • Surface area of hemisphere = 2πr2

Volume

    • Volume of cuboid = lbh
    • Volume of cube = a3
  • Volume of cylinder = πr2h
  • Volume of a right cylinder cone =⅓* πr2h
  • Volume of a sphere = 4/3*πr3
  • Volume of a hemisphere = 2/3*πr3

The chapter is completely covered in detail in the NCERT Solutions for Class 9 Mathematics Chapter 13 available on the Extramarks’ website.

NCERT Solutions for Class 9 Mathematics Chapter 13: Exercise &  Solutions

Even if one is thorough with all the formulas included in the chapter surface areas and volumes, one must be able to solve the problems accurately till the last step and be able to get the right answers. This mastery comes with a lot of practice. For this reason, we have included solutions to all the exercises given in the NCERT textbook in our NCERT Solutions for Class 9 Mathematics Chapter 13 in a detailed manner. One can also find extra questions related to the chapter on our Extramarks’ website, a trusted source for all the NCERT-related study material.

You can find for exercise specific questions and solutions for NCERT Solutions for Class 9 Mathematics Chapter 13 by referring to the following links:

  •  Chapter 13: Exercise 13.1 Question and answers    
  •   Chapter 13: Exercise 13.2 Question and answers 
  •  Chapter 13: Exercise 13.3 Question and answers 
  •  Chapter 13: Exercise 13.4 Question and answers 
  •  Chapter 13: Exercise 13.5 Question and answers 
  •  Chapter 13: Exercise 13.6 Question and answers 
  •  Chapter 13: Exercise 13.7 Question and answers 
  •  Chapter 13: Exercise 13.8Question and answers 
  •  Chapter 13: Exercise 13.9Question and answers 

Along with NCERT Solutions for Class 9 Mathematics Chapter 13, students can explore NCERT Solutions on our Extramarks website for all primary and secondary classes.

  • NCERT Solutions Class 1
  • NCERT Solutions Class 2
  • NCERT Solutions Class 3
  • NCERT Solutions Class 4
  • NCERT Solutions Class 5 
  • NCERT Solutions Class 6
  • NCERT Solutions Class 7
  • NCERT Solutions Class 8
  • NCERT Solutions Class 9
  • NCERT solutions Class 10
  • NCERT solutions Class 11
  • NCERT solutions Class 12

NCERT Exemplar for Class 9 Mathematics 

Students must include NCERT Exemplar Class 9 Mathematics in their core study material as the book is designed by the subject matter experts. Thus, it helps students to get an idea of the types of questions they would have to face in their examinations. As a result, students find their examination papers easy after solving questions from the NCERT Exemplar.

The book is designed as per the latest CBSE syllabus. Students can promise themselves success in every examination they face after practising from this book. It has proved to be fruitful for the teachers too in framing their question papers, thus becoming an important resource for the students as well as teachers.

Learning to solve the problems in a well stipulated time is quite necessary to complete the entire examination in a given time frame. Hence, NCERT Exemplar provides quick tips and tricks to solve every sum, thereby making calculations easy. You can get NCERT Exemplar Class 9 Mathematics from the Extramarks’ website today.

Key Features of NCERT Solutions for Class 9 Mathematics Chapter 13

Students with good calculating skills always excel in Mathematics. Hence, in NCERT Solutions for Class 9 Mathematics Chapter 13, we have covered various aspects which help in making the calculations of the students strong. The key features are as follows: 

  • Multiple Questions are covered from all the topics and sub-topics of the chapter in the NCERT Solutions for Class 9 Mathematics Chapter 13, keeping in mind no concept remains untouched
  • All sets of questions right from basic to advanced, are covered noting the levels of different kinds of students
  • All the questions are covered as per the latest CBSE curriculum.
  • After completing the NCERT Solutions for Class 9 Mathematics Chapter 13, students will be able to distinguish between the areas and volumes of different figures easily and confidently.

Q.1 A plastic box 1.5 m long, 1.25 m wide and 65 cm deep is to be made. It is opened at he top. Ignoring the thickness of the plastic sheet, determine:
(i) The area of the sheet required for making the box.
(ii) The cost of sheet for it, if a sheet measuring 1m2 costs ₹ 20.

Ans.

     Length of the box = 1.5 m    Breadth of the box = 1.25 m      Height of the box = 65 cm                           = 0.65 mi Surface area of open box= lb+2b+lh= 1.5×1.25+21.25+1.5×0.65= 1.875+3.575= 5.45 m2Thus, the area of the sheet required for making the box is 5.45 m2.ii Cost of 1m2 sheet = ₹ 20Cost of sheet = ₹ 20×5.45 = ₹109

Q.2 The length, breadth and height of a room are 5 m, 4 m and 3 m respectiv y. Find the cost of white washing the walls of the room and the ceiling at the rate of ₹ 7.50 per m2.

Ans.

The length of a room = 5 mThe breadth of a room = 4 mThe height of a room = 3 mSurface area of 4 walls and ceiling = lb+2b+lh                                          = 5×4+24+5×3                                          = 20+54                                          = 74 m2Rate of whitewashing of 1 m2 area = ₹ 7.50Rate of whitewashing the room = ₹ 7.50×74                                     = ₹ 555Thus, the cost of whitewashing the room is ₹ 555.

Q.3 The floor of a rectangular hall has a perimeter 250 m. If the cost of painting e four walls at the rate of Rs 10 per m2 is Rs 15000, fin the height of the hall.

Ans.

The perimeter of a rectangular hall = 250 m            Rate of painting the walls = ₹10 per m2     The cost of painting the 4 walls = ₹15000                      Let height of hall = h m          Area of 4 walls of the room = 2l+bh                                           = 250 h       Cost of painting the four walls = 10×250 h ₹ 15000 = 10×250 h                                         h = 150002500 = 6 mThus, the height of hall is 6 m.

Q.4 The paint in a certain container is sufficient to paint an area equal to 9.375 m. How many bricks of dimensions 22.5 cm×10 cm×7.5 cm can be painted out of this container?

Ans.

The area to paint by a certain container = 9.375 m2Dimensions of given brick = 22.5cm×10cm×7.5cm     Surface area of a brick = 222.5cm×10cm+10cm×7.5cm+7.5cm×22.5cm                              = 2225+75+168.25                              = 2468.75                              = 937.5 cm2                              = 937.510000 m2Number of bricks painted in given container = 9.375937.510000                                                     = 100Thus, 100 bricks can be painted out of this container.

Q.5 A cubical box has each edge 10 cm and another cuboidal box is 12.5 cm long, 10 wide and 8 cm high.

(i) Which box has the greater lateral surface area and by how much?

(ii) Which box has the smaller total surface area and by how much?

Ans.

(i)     Edge of a cubical box=10cmLateral surface area of cubical box     =4(10)2    =400cm2            Length of cuboidal box=12.5cm          Breadth of cuboidal box=10cm             Height of cuboidal box=8cmLateral surface area of cuboidal box    =2(l+b)h     =2(12.5+10)×8cm2     =360cm2cubical box has the greater lateral surface area.   Greater area of cubical box=400360     =40cm2(ii) Total surface area of cubical box     =6(side)2     =6(10)2    =600cm2Total surface area of cuboidal box    =2(lb+bh+hl)     =2(12.5×10+10×8+8×12.5)cm2     =2(125+80+100)cm2     =2(305)     =610cm2Total surface area of cuboidal box is greater than cubical box.Greater surface area of cubical box     =610600     =10cm2

Q.6 A small indoor greenhouse (herbarium) is made entirely of glass panes (including ase) held together with tape. It is 30 cm long, 25 cm wide and 25 cm high.

(i) What is the area of the glass?

(ii) How much of tape is needed for all the 12 edges?

Ans.

The length of greenhouse = 30 cm

The breadth of greenhouse = 25 cm
The height of greenhouse = 25 cm

(i) The area of the glass required
= 2(lb + bh + hl)
= 2(30 x 25 + 25 x 25
+ 25 x 30) cm2
= 4250 cm2

Thus, the area of the glass is 4250 cm2.

(ii) Length of required tape = 4(l + b + h)
= 4(30 + 25 + 25)
= 320 cm
Thus, 320 cm tape is needed for all the 12 edges

Q.7 Shanti Sweets Stall was placing an order for making cardboard boxes for packing t ir sweets. Two sizes of boxes were required. The bigger of dimensions 25 cm × 20 cm × 5 cm and the smaller of dimensions 15 cm × 12 cm × 5 cm. For all the overlaps, 5% of the total surface area is required extra. If the cost of the cardboard is Rs 4 for 1000 cm2, find the cost of cardboard required for supplying 250 boxes of each kind.

Ans.

Total surface area of bigger box
= 2(lb + bh + hl)
= 2(25×20+ 20×5+ 5×25)
= 2 (725)
= 1450 cm2
Required extra surface area = 5% of 1450 cm2
=72.5 cm2
Total surface area of smaller box
= 2(lb + bh + hl)
= 2(15×12+12×5+ 5×15)
= 2 (315)
= 630 cm2
Required extra surface area = 5% of 630 cm2
=31.5 cm2
Total required cardboard for both types boxes
= 1450 + 72.5 + 630 + 31.5
= 2184 cm2
Total required cardboard for both types 250 each boxes

= 250 x 2184 cm2
= 546000 cm2
Cost of 1000 cm2 cardboard = Rs. 4
Cost of required cardboard = Rs. (4/1000) x 546000
= Rs. 2184

Q.8 Parveen wanted to make a temporary shelter for her car, by making a box-like ructure 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 × 3 m?

Ans.

Parveen wanted to make a temporary shelter for her 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 × 3 m?

Q.9 The curved surface area of a right circular cylinder of height 14 cm is 88 cm. Find the diameter of the base of the cylinder

Ans.

Height of the cylinder = 14 cmCurved surface area of the cylinder = 88 cm2                                     2πrh = 88 cm2                                         r = 88227×14                                            = 8888 = 1               Diameter of the cylinder = 2r                                             = 21 = 2 cm

Q.10 It is required to make a closed cylindrical tank of height 1 m and base diameter 0 cm from a metal sheet. How many square metres of the sheet are required for the same?

Ans.

 The height of closed cylinderical tank=1mThe base diameter of cylinderical tank=140cm=1.4m The base radius of cylinderical tank=1.42 =0.7m     Required sheet for cylinder=2×227×0.7×1+2×227(0.7)2=4.4m2+3.08=7.48m2Thus, 7.48 m2 sheet is required for closed cylinderical tank.

Q.11 A metal pipe is 77 cm long. The inner diameter of a cross section is 4 cm, the o er diameter being 4.4 cm (figure shown below) . Find its
(i) inner curved surface area,
(ii) outer curved surface area,
(iii) total surface area.

Ans.

           Length of metal pipe = 77 cmInner diameter of metal pipe = 4cmOuter diameter of metal pipe = 4.4cmiInner surface area of metal pipe                                    =π×inner diameter×h                                    =227×4×77                                    =968cm2iiOuter surface area of metal pipe                                    =π×Outer diameter×h                                    =227×4.4×77                                    =1064.8cm2

iii Tital surface area = Outer surface area+Inner surface area + 2πr12r22=1064.8+968+2×2272.22-22r1 = 4.42 = 2.2,  r2 = 42 = 2= 2032.8+5.28= 2038.08 cm2Thus, the total surface area of metal pipe is 2038.08 cm2.

Q.12 The diameter of a roller is 84 cm and its length is 120 cm. It takes 500 complete volutions to move once over to level a playground. Find the area of the playground in m2.

Ans.

The diameter of a roller = 84cm   The radius of a roller = 842                             =42cm      The length of roller = 120 cmArea of playground covered in one revolution                              =curved surface area of roller                              =2×227×42×120                              =31680cm2Area of playground covered in 500 revolutions                              =500×31680cm2                              =15840000cm2                              =1584000010000  m2   1cm2=110000m2                              =1584m2Thus, the area of plalyground is 1584m2.

Q.13 A cylindrical pillar is 50 cm in diameter and 3.5 m in height. Find the cost of p nting the curved surface of the pillar at the rate of Rs 12.50 per m2.

Ans.

Diameter of a cylindrical pillar = 50 cm                                     = 50100  m                                     = 12 m    Radius of a cylindrical pillar = 14 m    Height of a cylindrical pillar = 3.5 m   Curved surface area of pillar = 2πrh                                     = 2×227×14×3.5                                     = 112 m2    Cost of painting of 1m2 area = Rs.12.50  Cost of painting of 112 m2 area = Rs.12.50×112                                      = Rs.68.75Therefore, the cost of painting the Curved surface area of the pillar is Rs 68.75.

Q.14 Curved surface area of a right circular cylinder is 4.4 m2. If the ra us of the base of the cylinder is 0.7 m, find its height.

Ans.

Curved surface area of right circular cylinder = 4.4 m2   Radius of the base of right circular cylinder = 0.7 mThen,                                          2πrh = 4.4 m2                                      2×227×0.7×h = 4.4 m2                                                     h=4.44.4m                                                       =1mThus, the height of right circular cylinder is 1 m.

Q.15 The inner diameter of a circular well is 3.5 m. It is 10 m deep. Find</str g>
(i) its inner curved surface area,
(ii) the cost of plastering this curved surface at the rate of Rs 40 per m2.

Ans.

The inner radius of a circular well = 3.52 m       The depth of a circular well = 10 mi Inner curved surface area of well = 2πrh                                            = 2×227×3.52×10                                            = 110 m2ii Cost of plastering 1 m2 of well = ₹ 40Cost of plastering 110 m2 of well = ₹ 40×110                                        = ₹ 4400Thus, the cost of plastering the inner curved surface of thewell at the rate of ₹ 40 per m2 is ₹4400.

Q.16 In a hot water heating system, there is a cylindrical pipe of length 28 m and dia ter 5 cm. Find the total radiating surface in the system.

Ans.

Length of hot water heating system=28mDiameter of hot water heating system=5cmRadius of hot water heating system=5100mTotal radiating surface of the system=2πrh                                              =2×227×5100×28                                              =8.8m2Thus, the total radiating surface area of the system is 8.8 m2.

Q.17 Find
( the lateral or curved surface area of a closed cylindrical petrol storage tank that is 4.2 m in diameter and 4.5 m high.
(ii) how much steel was actually used, if (1/12) of the steel actually used was wasted in making the tank.

Ans.

The diameter of cylinderical petrol tank = 4.2 m    The radius of cylinderical petrol tank = 2.1 mThe height of The cylinderical petrol tank = 4.5 mi  The curved surface area of petrol tank = π×diameter×h                                                   = 227×4.2×4.5                                                    = 59.4 m2Thus, the lateral or curved surface area of a closed cylindricalpetrol storage tank 59.4 m2.ii Total surface area of closed cylinderical tank                                   = 2πrh+r                                    = 2×227×2.14.5+2.1                                    = 44×0.3×6.6                                    = 87.12 m2Let actually used steel in making tank = x m2Wasted steel in making cylinderical tank = 112 of xUsed steel in making cylinderical tank = xx12                                  87.12 m2 = 11x12                                            x = 1211×87.12                                               = 95.04  m2Thus, actually used steel in making cylinderical tank is 95.04 m2.

Q.18 In Figure, you see the frame of lampshade. It is to be covered with a decorativ cloth. The frame has a base diameter of 20 cm and height of 30 cm. A margin of 2.5 cm is to be given for folding it over the top and bottom of the frame. Find how much cloth is required for covering the lampshade.

Ans.

Diameter of the base of the frame = 20 cmRadius of the base of the frame = 10 cm               Height of the frame = 30 cm                 Length of margin = 2.5 cmHeight of the frame with margin = 2.5+30+2.5                                       = 35 cmCloth required for covering the lampshade       = 2πrh         = 2×227×10×35         = 2200 cm2Thus, 2200 cm2 cloth is required for covering the lampshade.

Q.19 The students of a Vidyalaya were asked to participate in a competition for makin 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?

Ans.

Radius of penholder = 3 cmHeight of penholder = 10.5 cmTotal surface area of a penholder = 2πrh+πr2                                         = πr2h+r                                         = 227×32×10.5+3                                         = 667×24 cm2  Cardboard required for one penholder =667×24 cm2Cardboard required for 35 penholders  =667×24 ×35 cm2                                                 = 7920 cm2Thus, the cardboard required for 35 penholders for competition  was 7920 cm2.

Q.20 Diameter of the base of a cone is 10.5 cm and its slant height is 10 cm. Find its curved surface area.

Ans.

Curved surface area of con

e = (22/7) x (10.5/2) x 10
= 165 cm2
Thus, the curved surface area of cone is 165 cm2.

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

Ans.

Diameter of the base of cone = 24 m
Radius of the base of cone = 12 m
Slant height of the cone = 21 m

Total surface area of cone =

π

r(l + r)
= (22/7) x 12(21 + 12)
= (22/7) x 12 x 33

= 1244.57 cm2
Thus, total surface area of the cone is 1244.57 cm2.

Q.22 A conical tent is 10 m high and the radius of its base is 24 m. Find

(i) slant height of the tent.

(ii) cost of the canvas required to make the tent, if the cost of 1 m2 canvas is Rs 70.

Ans.

  Height of conical tent=10m Radius of conical tent=24 m(i)Slant height of the conical tent=(10)2+(24)2=100+576=676=26m(ii)  Curved surface area of tent=πrl=227×24×26m2 The cost of 1m2 canvas=Rs.70The cost of the canvas required to make the tent=Rs.70×227×24×26=Rs.137280Thus,the cost of the canvas required to make the tent is Rs.137280.

Q.23 The 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.

Ans.

The slant height of conical tomb=25mThe base diameter of conical tomb=14m The base radius of conical tomb=7mCurved surface area of conical tomb=πrl=227×7×25=550m2 Rate of white-washing 100 m2 area=Rs.210  Rate of white-washing 1 m2 area=Rs.210100Rate of white-washing 550 m2 area=Rs.210100×550=Rs.1155Thus, the cost of white-washing the conical tomb is Rs. 1155.

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

Ans.

Radius of jokers cap=7 cm Height of jokers cap=24 cmSlant height of joker’s cap=72+242=49+576=625=25cmCurved surface area of cap=πrl=227×7×25=550cm2Area of the sheet required for 10 caps=10×550cm2=5500cm2Thus, the area of required sheet for 10 caps is 5500 cm2.

Q.25

A bus stop is barricaded from the remaining part ofthe road, by using 50 hollow cones made of recycled cardboard. Each cone has abase diameter of 40 cm and height 1 m. If the outer side of each of the cones is to bepainted and the cost of painting is Rs 12 per m 2 , what will be the cost of painting allthese cones? (Use π=3.14 and take 1.04 = 1.02) MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@6CE0@

Ans.

Diameter of hollow cone=40cmRadius of hollow cone=20cm =0.2mHeight of hollow cone=1 mSlant height of hollow cone=12+(0.2)2=1+0.04=1.04m=1.02m(approx.)  C.S.A. of hollow cone=πrl=3.14×0.2×1.02=0.64056m2C.S.A. of hollow 50 cones=50×0.64056m2=32.028m2Cost of paintingof 1 m2area=Rs.12Cost of paintingof 50cones=Rs.12×32.028=Rs.384.336=Rs.384.34(approx)Thus, the cost of painting 50 cones is Rs.384.34.

Q.26 Find the surface area of a sphere of radius:

(i) 10.5 cm (ii) 5.6 cm (iii) 14 cm

Ans.

Surface area of a sphere=4πr2(i) Surface area of a sphere=4π10.52cm2                                  =4×227×10.5×10.5cm2                                  =4×22×1.5×10.5cm2                                  =1386cm2iiSurface area of a sphere=4π5.62cm2                                    =4×227×5.6×5.6 cm2                                    =4×22×0.8×5.6 cm2                                    =394.24cm2(iii) Surface area of a sphere=4π142cm2                                  =4×227×14×14cm2                                  =4×22×2×14cm2                                  =2464cm2

Q.27 Find the surface area of a sphere of diameter:

(i) 14 cm (ii) 21 cm (iii) 3.5 m

Ans.

Surface area of a sphere=4πr2iDiameter of sphere=14cm        Radius of sphere=7cmSurface area of a sphere=4π72                               =4×227×7×7                                 =616cm2iiDiameter of sphere=21cm        Radius of sphere=212cmSurface area of a sphere=4π2122                                 =4×227×212×212                                  =1386cm2iiiDiameter of sphere=3.5  m            Radius of sphere=3.52  mSurface area of a sphere=4π3.52                                 =4×227×3.52×3.52                                  =38.5m2

Q.28 Find the total surface area of a hemisphere of radius 10 cm. (Use

π

= 3.14)

Ans.

Total Surface area of a hemisphere = 3πr2                                            = 3π102 cm2                                            = 3×3.14×10×10 cm2                                            = 942 cm2

Q.29 The radius of a spherical balloon increases from 7 cm to 14 cm as air is being pumped into it. Find the ratio of surface areas of the balloon in the two cases.

Ans.

Surface area of a sphere=4πr2 When radius of balloon is 7 cm.Surface area of ballon=4π(7)2=4×227×7×7=616cm2 When radius of balloon is 14 cm.Surface area of ballon=4π(14)2=4×227×14×14=2464cm2The ratio of surface areas of balloon in the two cases=616cm22464cm2=14Thus, the ratio of surface areas of the balloon in the two cases is 1:4.

Q.30 A hemispherical bowl made of brass has inner diameter 10.5 cm. Find the cost of tin-plating it on the inside at the rate of Rs 16 per 100 cm2.

Ans.

 The inner radius of bowl=10.52cmInner surface area of bowl=2πr2=2×227×10.52×10.52=173.25cm2Cost of tin-plating 100 cm2=Rs.16Cost of tin-plating 1 cm2=Rs.16100 Cost of tin-plating bowl=Rs.16100×173.25=Rs.27.72Thus, the cost of tin-platting the inner surface area of bowl is Rs. 27.72.

Q.31 Find the radius of a sphere whose surface area is 154 cm2.

Ans.

Let radius of sphere be r cm.Surface area of sphere=154cm24πr2=154cm24×227×r2=154cm2r2=154×74×22cm2=7×74cm2r=72cmThus, the radius of sphere is 3.5 cm.

Q.32 The diameter of the moon is approximately one fourth of the diameter of the earth. Find the ratio of their surface areas.

Ans.

Surface area of a sphere=4πr2Let radius of the earth be R.Let radius of the moon be R4 . [Since, diameter of the moon is14of diameter of the earth.]Surface area of the earth=4πR2 Surface area of the moon=4π(R4)2The ratio of surface areas of the moon and the earth=4π(R4)24πR2=116Thus, the ratio of surface areas of the moon and the earth is 1:16.

Q.33 A hemispherical bowl is made of steel, 0.25 cm thick. The inner radius of the bowl is 5 cm. Find the outer curved surface area of the bowl.

Ans.

Inner radius of hemispherical bowl = 5 cm
Thickness of hemispherical bowl = 0.25 cm

Outer radius of hemispherical bowl = 5 + 0.25
= 5.25 cm
So, outer curved surface area of bowl = 2

π

r2
=2(22/7)( 5.25)2
= 173.25 cm2.
Thus, the outer curved surface area of the bowl is 173.25 cm2.

Q.34 A right circular cylinder just encloses a sphere of radius r (see Figure). Find


(i) surface area of the sphere,
(ii) curved surface area of the cylinder,
(iii) ratio of the areas obtained in (i) and (ii).

Ans.

Radius of sphere is r.(i) Surface area of sphere= 4πr2(ii) Height of cylinder=Diameter of sphere        =2r        Radius  (R) of cylinder=rCurved surface area of cylinder        =2πRh        =2πr(2r)        =4πr2(iii)Ratioof the surface areas of sphere and cylinder        =4πr24πr2        =1Thus,theratioof the surface areas of sphere and cylinder is 1:1.

Q.35 A matchbox measures 4 cm × 2.5 cm × 1.5 cm. What will be the volume of a packet containing 12 such boxes?

Ans.

Volume of a matchbox = 4 cm × 2.5 cm × 1.5 cm

= 15 cm3

Volume of a packet = Volume of 12 matchboxes
= 12 x 15 cm3
= 180 cm3

Thus, the volume of a packet containing 12 matchboxes is 180 cm3.

Q.36 A cuboidal water tank is 6 m long, 5 m wide and 4.5 m deep. How many litres of water can it hold? (1 m3 = 1000 l)

Ans.

Volume of cuboidal water tank = 6 x 5 x 4.5 m3
= 135 m3
Since, 1 m3 = 1000 litres
So, volume of water in tank = 135 x 1000 litres
= 135000 litres

Q.37 A cuboidal vessel is 10 m long and 8 m wide. How high must it be made to hold 380 cubic metres of a liquid?

Ans.

 Volume of a liquid = 380 m3 Length of cuboidal vessel = 10 mBreadth of cuboidal vessel = 8 mLet height of vessel be h m.Then, volume of cuboidal vessel=l×b×h380 m3=10×8×hh=38080=4.75mThus, cuboidal vessel should be 4.75 m high.

Q.38 Find 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.

Ans.

Volume of cuboidal pit = 8 x 6 x 3
= 144 m3
Cost of digging 1 m3 pit = Rs. 30
Then, the cost of digging cuboidal pit
= Rs. 30 x 144
= Rs. 4320
Thus, the cost of digging a cuboidal pit is Rs. 4320.

Q.39 The capacity of a cuboidal tank is 50000 litres of water. Find the breadth of the tank, if its length and depth are respectively 2.5 m and 10 m.

Ans.

Capacity of cuboidal tank = 50,000 litres
Length of cuboidal tank = 2.5 m
Depth of cuboidal tank = 10 m
Let breadth of cuboidal tank = p m
Since, 1000 litres = 1 m3
So, 50,000 litres = 50 m3
Then, Volume of tank = lbh
50 m3= 2.5 x p x 10
p = 2 m

Thus, the breadth of the tank is 2 m.

Q.40 A village, having a population of 4000, approximately 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?

Ans.

Volume of water tank = 20 m × 15 m × 6 m
= 1800 m3
= 1800000 litres
[Since, 1 m3 = 1000 litres]
Water require for a person = 150 litres
Water require for 4000 persons
= 150 x 4000 litres
= 600000 litres
Number of days to last the water of this tank
= (1800000/600000) days
= 3 days

Thus, the water of tank will last in 3 days in a village of 4000 people.

Q.41 A godown measures 40 m × 25 m × 15 m. Find the maximum number of wooden crates each measuring 1.5 m × 1.25 m × 0.5 m that can be stored in the godown.

Ans.

Number of wooden crates in godown=Volume of godownVolume of wooden crates=40 m × 25 m × 15 m1.5 m × 1.25 m × 0.5 m=400 m × 2500 m × 150 m15 m × 125 m × 5 m=80×20×10=16000Thus, 16000 wooden crates can be stored in the godown.

Q.42 A solid cube of side 12 cm is cut into eight cubes of equal volume. What will be the side of the new cube? Also, find the ratio between their surface areas.

Ans.

  Side of solid cube=12 cmNumber of cubes of equal volume= 8Side of each cube of equal volume=Volume of solid cube8=12×12×128=216cm3Side of each cube of equal volume=2163=6cmRatio of sufaces area of both cubes=4(12)24(6)2=12×126×6=4:1 Thus, the required ratio of surface areas of both cubes is 4:1.

Q.43 A part of a 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?

Ans.

Speed of river = 2 km per hour                   =  2000/60 m per minute                  = 1003 m per minThen,Volume of flowing water in one minute        = 10033 40        = 4000 m3Thus, 4000 m3 water will fall into the sea in a minute.

Q.44 The circumference of the base of a cylindrical vessel is 132 cm and its height is 25 cm. How many litres of water can it hold? (1000 cm3=1 l )

Ans.

Circumference of the base of a cylindrical vessel=132cm          2πr=132cm                  2×227×r=132cm    r=132×744    r=21cm      Height of cylinderical vessel=25cmVolume of water in the vessel=πr2h        =227×(21)2×25        =34650cm3        =346501000litres    [1cm3=11000  litres]        =34.65litres      Thus, the cyliderical vessel can hold 34.65 litres water.

Q.45 The inner diameter of a cylindrical wooden pipe is 24 cm and its outer diameter is 28 cm. The length of the pipe is 35 cm. Find the mass of the pipe, if 1 cm3 of wood has a mass of 0.6 g.

Ans.

The inner diameter of a cylinderical wooden pipe        =24cm      The inner radius of a cylinderical wooden pipe        =12cmThe outer diameter of a cylinderical wooden pipe        =28cm      The outer radius of a cylinderical wooden pipe        =14cm       The length of pipe=35 cm              Mass of 1 cm3 wood=0.6gmVolume of wood in cylinderical wooden pipe=227{(14)2(12)2}×35=227×2×26×35=5720cm3Mass of wood in cylinderical wooden pipe=0.6×5720gm=3432gm=3.432kgThus, the mass of wood in cylinderical wooden pipe is 3.432 kg.

Q.46 A 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?

Ans.

(i)Capacity of cuboidal tin can=5×4×15      =300cm3(ii)Capacity of plastic cylinder=πr2h      =227×(72)2×10cm3      =385cm3In this way we see that capacity of plastic container is more than tin container.More capacity of cylinderical container than tin container      =385300      =85cm3

Q.47 It costs ₹ 2200 to paint the inner curved surface of a cylindrical vessel 10 m deep. If the cost of painting is at the rate of ₹ 20 per m2, find
(i) inner curved surface area of the vessel,
(ii) radius of the base,
(iii) capacity of the vessel.

Ans.

              Depth of cylinder=10 mCost of painting the inner curved surface area of cylinder             =₹ 2200Rate of painting of 1 m2=  ₹ 20 (i)Inner curved surface area of cylinder             =Cost of paintingRate of painting             =220020             =110m2(ii)   2πrh=110m2    2×227×r×10=110r=110×744×10             =1.75mThus, the radius of the base of cylinder is 1.75 m.(iii)      Capacity of vessel=πr2h              =227×1.75×1.75×10              =22×0.25×17.5              =96.25m3              =96.25kl[1m3=1kl]

Q.48 The volume 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?

Ans.

  Height of a closed cylinder=1mThe capacity of a closed cylinder=15.4litresπr2h=15.41000m3[1litre=11000m3]227×r2×1=15.41000m3r2=15.41000×722=7×710000r=7100=0.07mRequired metal sheet for closed cylinder=2πr(h+r)=2×227×0.07(1+0.07)=44×0.01×1.07=0.4708m2Thus, 0.4708m2of metal sheet would be needed to make closed cylinder.

Q.49 A lead pencil consists of a cylinder of wood with a solid cylinder of graphite filled in the interior. The diameter of the pencil is 7 mm and the diameter of the graphite is 1 mm. If the length of the pencil is 14 cm, find the volume of the wood and that of the graphite.

Ans.

 Diameter of the pencil=7 mm Radius(r1) of the pencil=72 mmDiameter of the graphite=1 mmRadius(r2) of the graphite=12 mmThe length of the pencil=14 cm=140mmVolume of the wood=π(r12r22)h=227{(72)2(12)2}×140=22(72+12)(7212)×20=44×4×30=5280mm3=52801000cm3[1cm=10mm]=5.28cm3

Thus, the volume of wood in the pencil is 5.28 cm3.

Volume of graphite=π r 2 2 h = 22 7 × 1 2 × 1 2 ×140 =11×10 =110m m 3 = 110 1000 c m 3 =0.11c m 3 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaakqaaeeqaaiaadAfacaWGVbGaamiBaiaadwhacaWGTbGaamyzaiaabccacaqGVbGaaeOzaiaabccacaqGNbGaaeOCaiaabggacaqGWbGaaeiAaiaabMgacaqG0bGaaeyzaiabg2da9GGaaiab=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@7F03@

Thus, the volume of graphite in the pencil is 0.11 cm3.

Q.50 A patient in a hospital is given soup daily in a cylindrical bowl of diameter 7 cm. If the bowl is filled with soup to a height of 4 cm, how much soup the hospital has to prepare (in litres) daily to serve 250 patients?

Ans.

The diameter of bowl=7cmThe radius of bowl=72cmHeight of soup in the bowl=4cmVolume of soup in bowl=πr2h=227×72×72×4=154cm3Volume of soup required for one patient=154cm3Volume of soup required for 250 patients=250×154cm3=250×1541000litres[1cm3=11000litres]=38.5litres

Thus, 38.5 litres soup the hospital has to prepare daily to serve 250 patients.

Q.51 Find the volume of the right circular cone with

(i) radius 6 cm, height 7 cm

(ii) radius 3.5 cm, height 12 cm

Ans.

(i)Volume of cone=13πr2h=13×227×6×6×7=264cm3Thus, the volume of right circular cone is 264 cm3.(ii)Volume of cone=13πr2h=13×227×3.5×3.5×12=154cm3Thus, the volume of right circular cone is 154 cm3.

Q.52 Find 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

Ans.

(i) Radius of cone=7 cm, Slant height of cone=25 cmHeight of cone(h)=l2r2=25272=62549=24cmCapacity of vessel=13πr2h=13×227×7×7×24=22×7×8=1232cm3=12321000litres[1cm3=11000litres]=1.232litres(ii) Height of cone=12 cm, Slant height of cone=13 cmRadius of cone(r)=l2h2=132122=169144=5cmCapacity of vessel=13πr2h=13×227×5×5×12=227×25×4=22007cm3=22007×1000litres[1cm3=11000litres]=1135litres

Q.53 A conical pit of top diameter 3.5 m is 12 m deep of height. What is its capacity in kilolitres?

Ans.

Diameter of conical pit=3.5 m  Radius of conical pit=3.52m Depth of conical pit=12 m Volume of conical pit=13πr2h=13×227×3.52×3.52×12=11×3.5=38.5m3=38.5 kilolitres [1m3=1 kilolitre]Thus, the capacity of conical pit is 38.5 kilolitres.

Q.54 The 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 cone.

Ans.

Diameter of a right circular cone=28cm Radius of a right circular cone=14cm(i)Volume of a right circular cone=9856cm313×227×14×14×h=9856cm3h=9856×322×2×14=48cm(ii) Slantheight of the cone(l)=r2+h2=142+482=2500=50cm(iii) Curved surface area of cone=πrl=227×14×50=2200 cm2

Thus, the curved surface area of cone is 2200 cm2.

Q.55 A right triangle ABC with sides 5 cm, 12 cm and 13 cm is revolved about the side 12 cm. Find the volume of the solid so obtained..

Ans.

InΔABC,  let BC=5cm, AB=12 cm and AC=13 cmWhen ΔABC is revolved around AB, a cone is obtained,whose radius is 5 cm, height is 12 cm and slant height is 13 cm.So, Volume of cone=13πr2h    =13π(5)2×12    =100πcm2Thus, the volume of cone is 100π cm2.

Q.56 If the triangle ABC in the Question 7 above is revolved about the side 5 cm, then find the volume of the solid so obtained. Find also the ratio of the volumes of the two solids obtained in Questions 7 and 8.

Ans.

When Δ ABC is revolved around BC, then radius of cone is 12 cm, height is 5 cm and slant height is 13 cm.Volume of cone=13πr2h=13π(12)2×5=13π×12×12×5=240πcm3Ratio of volumes of the two solids obtained on revolving triangle=100π240π=5:12

Q.57 A heap of wheat is in the form of a cone whose diameter is 10.5 m and height is 3 m. Find its volume. The heap is to be covered by canvas to protect it from rain. Find the area of the canvas required.

Ans.

Diameter of conical heap=10.5m  Radius of conical heap=10.52m Height of conical heap=3m  Volume of conical heap=13πr2h=13×227×10.52×10.52×3=227×1051520×10520=1110×15×214=86.625cm3Slant height of conical heap=32+(10.52)2=9+110.254=36+110.254=146.254=12.12=6.05mCurved surface area of conical heap=πrl=227×10.52×6.05=99.825m2Thus, the area of canvas required is 99.825 m2.

Q.58 Find the volume of a sphere whose radius is

(i) 7 cm (ii) 0.63 m.

Ans.

 Volume of sphere=43πr3(i)When radius=7 cm Volume of sphere=43π(7)3=43×227×7×7×7=1437.33cm3(ii)   When radius=0.63 cm Volume of sphere=43π(0.63)3=43×227×0.63×0.63×0.63=4×22×0.03×0.63×0.63=1.047816cm3=1.05cm3(approx.)

Q.59 Find the amount of water displaced by a solid spherical ball of diameter (i) 28 cm (ii) 0.21 m, when the solid spherical ball is placed in a large tank containing water.

Ans.

 Volume of sphere=43πr3Water displaced by a spherical ball is equal to the volume of the sphere.(i) When diameterof sphere=28 cm  Radius of sphere=14cm Volume of sphere=43π(14)3=43×227×14×14×14=11498.64cm3(ii)When diameterof sphere=0.21 mRadius of sphere=0.212m Volume of sphere=43π(0.212)3=43×227×0.212×0.212×0.212=11×0.01×0.21×0.21=0.004851m3

Q.60 The diameter of a metallic ball is 4.2 cm. What is the mass of the ball, if the density of the metal is 8.9 g per cm3?

Ans.

Diameter of a metallic ball=4.2cm Radius of a metallic ball=2.1cm Volume​ of metallic ball=43πr3=43×227×2.1×2.1×2.1=4×22×0.1×2.1×2.1=38.808cm3 Density of metal=8.9gpercm3  Mass of metallic ball=Volume×Density=38.808×8.9=345.39 grams.

Q.61 The diameter of the moon is approximately one-fourth of the diameter of the earth..What fraction of the volume of the earth is the volume of the moon?

Ans.

Radius of earth(Re)=Diameter of earth2=De2Radius of moon(Rm)=14×De2=De8Volume of moonVolume of earth=43π(Rm)343π(Re)3=43π(De8)343π(De2)3=18×18×1812×12×12=164 Volume of moon=164×Volume of earth

Q.62 How many litres of milk can a hemispherical bowl of diameter 10.5 cm hold?

Ans.

Diameter of hemispherical bowl=10.5cm  Radius of hemispherical bowl=10.52cm Volume of hemispherical bowl=23πr3=23×227×10.52×10.52×10.52=11×520×10510×10510=11×14×212×212=303.1875cm3=303.18751000litres(approx.)[1cm3=11000litres]=0.3031875litres  Capacity of hemispherical bowl=0.303litres.

Q.63 A hemispherical tank is made up of an iron sheet 1 cm thick. If the inner radius is 1 m, then find the volume of the iron used to make the tank.

Ans.

The inner radius of hemispherical tank(r2)=1m Thickness of hemispherical tank=1cm=0.01m  Outer radius of spherical tank(r1)=1+0.01=1.01m   Volume of iron used to make the tank=23π(r13r23)=23×227{(1.01)3(1)3}=4421(1.0303011)=4421(0.030301)=0.06348m3Thus, the volume of iron used to make the tank is 0.06348m3.

Q.64 Find the volume of a sphere whose surface area is 154 cm2.

Ans.

Surface area of a sphere=154cm24×227×r2=154cm2r2=154×74×22=7×74r=7×74=72cm Volume of a sphere=43×227×72×72×72=11×493=5393=17923cm3Thus, volume of sphere is 17923cm3.

Q.65 A dome of a building is in the form of a hemisphere. From inside, it was white-washed at the cost of ₹ 498.96. If the cost of white-washing is ₹ 2.00 per square metre, find the
(i) inside surface area of the dome,
(ii) volume of the air inside the dome.

Ans.

(i)Cost of white-washing the hemisphere= 498.96Rate​ of white-washing the hemisphere= 2.00/m2Therefore,  Surface area of the hemisphere=498.962      =249.48m2Thus, inside surface area of the dome is 249.48m2.(ii)        Surface area of the hemisphere=249.48m2       2πr2=249.48m2     2×227×r2=249.48m2    r=249.48×72×22   r=6.3m                   Volume of air inside the tomb=43πr3      =43×227×6.3×6.3×6.3      =88×0.3×6.3×6.3      =1047.816m3Thus, the volume of air inside the tomb is 1047.816m3.

Q.66 Twenty seven solid iron spheres, each of radius r and surface area S are melted to form a sphere with surface area S′. Find the
(i) radius r ′ of the new sphere,
(ii) ratio of S and S′.

Ans.

(i) Volume of new sphere=27×volume of solid iron sphere       43π(r)3=27×43πr3  r=27r3  r=3rThus, the radius(r) of new sphere is 3r.(ii)             Ratio of S and S’=4πr24πr2      =r2(3r)2[r=3r]      =19Thus, ration of S and S’ is 1:9.

Q.67 A capsule of medicine is in the shape of a sphere of diameter 3.5 mm. How much medicine material (in mm3) is needed to fill this capsule?

Ans.

Diameter of spherical capsule=3.5 mmRadius of spherical capsule=3.52 mm Volume of spherical capsule=43πr3=43×227×3.52×3.52×3.52=113×0.5×3.5×3.5=22.458mm3=22.46mm3(Approx.)Thus,22.46mm3 medicine materials are required to fill the capsule.

Q.68 A wooden bookshelf has external dimensions as follows: Height = 110 cm, Depth = 25 cm, Breadth = 85 cm (see Figure). 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.

Ans.

Outer surface area=2(110+85)×25+85×110   =2×195×25+9350   =19100cm2Area​ vertical edges=110×5+110×5   =1100cm2Area horizontal edges   =75×5+75×5+75×5+75×5   =75(5+5+5+5)   =75×20   =1500cm2Total external area=19100cm2+1100cm2+1500cm2   =21700cm2Internal length of each rack of bookshelf=8555=75cmInternal breadth of each rack of bookshelf=110203=30cmInternal depth of each rack of bookshelf=255=20​ cmSurface area of a bookshelf=2(75+30)×20+75×30=2×105×20+2250 =4200+2250=6450cm2Surface area of three bookshelves=3×6450cm2=19350cm2Rate of polishing external surface of bookshelf=  20100=0.20 per m2Rate of painting internal surface of bookshelf=  10100=0.10 per m2Total cost of polishing and painting the bookshelf=0.20×21700+0.10×19350=4340+1935=6275Thus, total expenses required for polishing and painting the surface of the bookshelf is ₹ 6275.

Q.69 The front compound wall of a house is decorated by wooden spheres of diameter 21 cm, placed on small supports as shown in Figure. Eight such spheres are used for this purpose, and are to be painted silver. Each support is a cylinder of radius 1.5 cm and height 7 cm and is to be painted black. Find the cost of paint required if silver paint costs 25 paise per cm2 and black paint costs 5 paise per cm2.

Ans.

Diameter of wooden sphere=21cm Radius of wooden sphere=212cm Radius of each cylinder=1.5cm  Height of each cylinder=7cm Surface area of sphere=4πr2=4×227×212×212=22×3×21=1386cm2 Area of top of cylinder=πr2=227×1.5×1.5=7.07cm2Surface area of sphere to be painted=13867.07=1378.93cm2Curved surface area of cylinder=2×227×1.5×7 =66cm2Cost of silver painting on sphere=25100=Rs.0.25Cost of black painting on cylinder=Rs.5100=Rs.0.05Cost of painting of one sphere and cylinder =Rs.0.25×1378.93+Rs.0.05×66=Rs.(344.7325+3.3)=Rs.348.0325Cost of painting of eight combinatin of sphere and cylinder=8×Rs.348.0325=Rs.2784.26

Q.70 The diameter of a sphere is decreased by 25%. By what per cent does its curved surface area decrease?

Ans.

Let diameter of sphere be 2r. Curved surface area of sphere=4π r 2 Decreased radius of sphere=r25% of r =r 25 100 ×r MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@B453@ =rr4=3r4New curved surface area of sphere=4π(3r4)2=94πr2Percentage of decreased curved surface area=4πr294πr24πr2×100%=74πr24πr2×100%=716×100%=43.75%Thus, the 43.75% curved surface area of sphere is decreased.

Q.71 Curved surface area of a cone is 308 cm2 and its slant height is 14 cm. Find

(i) radius of the base
(ii) total surface area of the cone.

Ans.

(i) Curved surface area of cone = 308 cm2
πrl = 308 cm2
(22/7) x r x 14 = 308 cm2
r = (308/14) x (7/ 22)
= 7 cm
Thus, the radius of the base of cone is 7 cm.

(ii) Total surface area of the cone = πr(l + r)
= (22/7) x 7(14 + 7)
= 22 x 21

= 462 cm2
Thus, total surface area of the cone is 462 cm2.

Q.72 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 π = 3.14).

Ans.

  Height of conical tent=8m  Base radius of conical tent=6m Slant height of conical tent=82+62=100=10msurface area of conical tent=πrl=3.14×6×10 =188.4m2  Width of tarpaulin=3mRequired length of tarpaulin=188.43=62.8mWastage in cutting of tarpaulin= 20 cm=20100m=0.2mRequired tarpaulin for tent=62.8+0.2=63mThus, the length of the required tarpaulin will be 63 m.

Q.73 If 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 π = 3.14)

Ans.

(i) The height of cylinder=5 cmThe lateral surface area of cylinder =94.2cm22πrh=94.2cm22×3.14×r×5=94.2 cm2r=94.22×3.14×5=3 cmThus, the radius of cylinder is 3 cm.(ii) Volume of cylinder=πr2h=3.14×3×3×5=141.3cm3Thus, the volume of cylinder is 141.2 cm3.

Q.74 The height of a cone is 15 cm. If its volume is 1570 cm3, find the radius of the base. (Use π = 3.14)

Ans.

  Height of the cone=15cmVolume of the cone=1570cm313πr2h=1570cm313×3.14×r2×15=1570cm3r2=15703.14×5r=10Thus, the radius of the base is 10 cm.

Q.75 If the volume of a right circular cone of height 9 cm is 48 π cm3, find the diameter of its base.

Ans.

  Height of the cone=9cmVolume of the cone=48πcm313πr2h=48πcm313×r2×9=48cm3r2=483=16r=4  diameter=2×4=8cmThus, the diameter of the base is 8 cm.

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