NCERT Solutions for Class 9 Maths Chapter 13 Surface Area and Volumes

 

NCERT solutions for class 9 maths chapter 13 Surface Area and Volumes: It is among one of the best topics which has so much use in real-life problems. The concepts of this chapter are used in our daily life for calculations like- how many liters of paint is needed to paint an entire house? How many boxes are needed to fill these chocolates? what should be the area of the sheet I should by to cover the roof of my house? These types of calculations can be done using the concept of surface areas and volumes. CBSE NCERT solutions for class 9 maths chapter 13 Surface Area and Volumes are covering every problem in a very detailed manner. In this chapter, you will study the shapes- cube, cuboid, cylinder, cone, sphere, hemisphere, etc. You have to face the questions related to the curved surface area, total surface area, volumes, and many others.

Solutions of NCERT for class 9 maths chapter 13 Surface Area and Volumes are designed to provide you assistance while solving the practice exercises. The language of the questions sometimes is not direct, you have to identify that what the question is specifically about. There are a total of 9 exercises consisted of a total of 102 questions including the optional exercise. NCERT solutions for class 9 maths chapter 13 Surface Area and Volumes have the exam-oriented solutions to all the questions of practice exercises. NCERT solutions are available for free which you can get by clicking on the given link.

 

NCERT solutions for class 9 maths chapter 13 Surface Area and Volumes Excercise: 13.1

Q1 (i) A plastic box \small 1.5\hspace {1mm}m long,\small 1.25\hspace {1mm}m wide and \small 65\hspace {1mm}cm deep is to be made. It is opened at the top. Ignoring the thickness of the plastic sheet, determine: The area of the sheet required for making the box.

Answer:

Given, dimensions of the plastic box

Length, l = 1.5\ m

Width, b = 1.25\ m

Depth, h = 65\ cm = 0.65\ m

(i) The area of the sheet required for making the box (open at the top)= Lateral surface area of the box. + Area of the base.

  = 2(bh+hl)+lb

\\ = 2(1.25\times0.65+0.65\times1.5)+1.5\times1.25 \\ = 2(0.8125+ 0.975)+1.875= 5.45

The required area of the sheet required for making the box is 5.45\ m^2

 

Q1 (ii) A plastic box \small 1.5\hspace {1mm} m long, \small 1.25\hspace {1mm} m wide and \small 65\hspace {1mm} cm deep is to be made. It is opened at the top. Ignoring the thickness of the plastic sheet, determine: The cost of sheet for it, if a sheet measuring \small 1m^2  costs Rs 20.

Answer:

Given, dimensions of the plastic box

Length, l = 1.5\ m

Width, b = 1.25\ m

Depth, h = 65\ cm = 0.65\ m

We know, area of the sheet required for making the box is 5.45\ m^2

(ii) Cost for  \small 1m^2 of sheet = Rs 20

\therefore Cost for 5.45\ m^2 of sheet = 5.45\times20 = 109

Required cost of the sheet is Rs.\ 109

 

Q2 The length, breadth and height of a room are \small 5\hspace {1mm}m, \small 4\hspace {1mm}m and \small 3\hspace {1mm}m respectively. Find the cost of white washing the walls of the room and the ceiling at the rate of \small Rs \hspace {1mm} 7.50   per \small m^2.

Answer:

Given,

Dimensions of the room = 5\ m\times4\ m\times3\ m

Required area to be whitewashed = Area of the walls + Area of the ceiling

2(lh+bh) + lb

\\ = 2(5\times4+4\times3)+ 5\times4 \\ = 2(32)+20 = 74\ m^2
Cost of white-washing per \small m^2 area = \small Rs \hspace {1mm} 7.50
Cost of white-washing 74\ m^2area = Rs (74 \times 7.50)
= Rs.\ 555

Therefore, the required cost of whitewashing the walls of the room and the ceiling is Rs.\ 555

 

Q3 The floor of a rectangular hall has a perimeter \small 250\hspace {1mm} m. If the cost of painting the four walls at the rate of Rs 10 per \small m^2 is Rs 15000, find the height of the hall. [Hint : Area of the four walls \small = Lateral surface area.]

Answer:

Given,

The perimeter of rectangular hall = \small 250\hspace {1mm} m

Cost of painting the four walls at the rate of Rs 10 per \small m^2 = Rs 15000

Let the height of the wall be h\ m

\therefore Area to be painted = Perimeter\times height

= 250h\ m^2

\therefore Required cost = 250h\times10\ m^2 = 15000\ m^2

\\ \implies 2500h = 15000 \\ \implies h = \frac{150}{25} = 6\ m

Therefore, the height of the hall is 6\ m

 

Q4 The paint in a certain container is sufficient to paint an area equal to \small 9.375\hspace{1mm} m^2.  How many bricks of dimensions \small 22.5\hspace {1mm}cm \times 10\hspace {1mm}cm \times 7.5\hspace {1mm}cm can be painted out of this container?

Answer:

Given, dimensions of the brick = \small 22.5\hspace {1mm}cm \times 10\hspace {1mm}cm \times 7.5\hspace {1mm}cm

We know, Surface area of a cuboid =2(lb+bh+hl)

\therefore The surface area of a single brick = 2(22.5\times10+10\times7.5+7.5\times22.5)

= 2(225+75+166.75) = 937.5\ cm^2 = 0.09375\ m^2

\therefore Number of bricks that can be painted = \frac{Total\ area\ the\ container\ can\ paint}{Surface\ area\ of\ a\ single\ brick}

= \frac{9.375}{0.09375} = 100

Therefore, the required number of bricks that can be painted = 100

Q5 (i) A cubical box has each edge \small 10 \hspace{1mm}cm and another cuboidal box is \small 12.5 \hspace{1mm}cm long, \small 10 \hspace{1mm}cm wide and \small 8 \hspace{1mm}cm high.

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

Answer:

Given,

Edge of the cubical box = \small 10 \hspace{1mm}cm

Dimensions of the cuboid = 12.5\cm \times 10\ cm \times 8\ cm

The lateral surface area of the cubical box = 4\times(10\times10)\ cm^2 = 400\ cm^2

The lateral surface area of the cuboidal box = 2[lh + bh]

\\ = [2(12.5 � 8 + 10 � 8)]\ cm^2 \\ = (2 \times 180)\ cm^2 \\ = 360\ cm^2

Clearly, Lateral surface area of the cubical box is greater than the cuboidal box.

Difference between them = 400-360 = 40\ cm^2

 

Q5 (ii) A cubical box has each edge \small 10 \hspace {1mm}cm and another cuboidal box is \small 12.5 \hspace {1mm}cm long, \small 10 \hspace {1mm}cm wide and \small 8 \hspace {1mm}cm high.

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

Answer:

Given,

Edge of the cubical box = \small 10 \hspace{1mm}cm

Dimensions of the cuboid = 12.5\cm \times 10\ cm \times 8\ cm

(ii) The total surface area of the cubical box = 6\times(10\times10)\ cm^2 = 600\ cm^2

The total surface area of the cuboidal box = 2[lh + bh+lb]

\\ = [2(12.5 � 8 + 10 � 8 + 12.5\times10)]\ cm^2 \\ = 610\ cm^2

Clearly, the total surface area of a cuboidal box is greater than the cubical box.

Difference between them = 610-600 = 10\ cm^2

 

Q6 (i) A small indoor greenhouse (herbarium) is made entirely of glass panes (including base) held together with tape. It is \small 30 \hspace{1mm}cm long, \small 25 \hspace{1mm}cm wide and \small 25 \hspace{1mm}cm high.What is the area of the glass?

Answer:

Given, dimensions of the greenhouse = 30\ cm \times25\ cm \times25\ cm

Area of the glass = 2[lb + lh + bh]
\\ = [2(30 \times 25 + 30 \times 25 + 25 \times 25)] \\ = [2(750 + 750 + 625)] \\ = (2 \times 2125) \\ = 4250\ cm^2
Therefore, the area of glass is 4250\ cm^2

 

Q6 (ii) A small indoor greenhouse (herbarium) is made entirely of glass panes (including base) held together with tape. It is 30 cm long, 25 cm wide and 25 cm high. How much of tape is needed for all the 12 edges?

Answer:

Given, dimensions of the greenhouse = 30\ cm \times25\ cm \times25\ cm

(ii) Tape needed for all the 12 edges = Perimeter =4(l+b+h)

4(30+25+25) = 320\ cm

Therefore, 320\ cm of tape is needed for the edges.

Q7 Shanti Sweets Stall was placing an order for making cardboard boxes for packing their sweets. Two sizes of boxes were required. The bigger of dimensions \small 25\hspace {1mm}cm \times 20\hspace {1mm}cm \times 5 \hspace {1mm}cm, and the smaller of dimensions \small 15\hspace {1mm}cm \times 12\hspace {1mm}cm \times 5 \hspace {1mm}cm. For all the overlaps, \small 5\% of the total surface area is required extra. If the cost of the cardboard is Rs 4 for \small 1000\hspace {1mm}cm^2,  find the cost of cardboard required for supplying 250 boxes of each kind.

Answer:

Given,

Dimensions of the bigger box = \small 25\hspace {1mm}cm \times 20\hspace {1mm}cm \times 5 \hspace {1mm}cm,

Dimensions of smaller box =  \small 15\hspace {1mm}cm \times 12\hspace {1mm}cm \times 5 \hspace {1mm}cm

We know,

Total surface area of a cuboid = 2(lb+bh+hl)

\therefore Total surface area of the bigger box = 2(25\times20+20\times5+5\times25)

= 2(500+100+125) = 1450\ cm^2

\therefore Area of the overlap for the bigger box = 5\%\ of\ 1450\ cm^2 = \frac{5}{100}\times1450 = 72.5\ cm^2

Similarly,

 

Total surface area of the smaller box = 2(15\times12+12\times5+5\times15)

= 2(180+60+75) = 630\ cm^2

\therefore Area of the overlap for the smaller box = 5\%\ of\ 630\ cm^2 = \frac{5}{100}\times630 = 31.5\ cm^2

Since, 250 of each box is required,

\therefore Total area of carboard required = 250[(1450+72.5)+(630 + 31.5)] = 546000\ cm^2

Cost of \small 1000\hspace {1mm}cm^2 of the cardboard = Rs 4 

\therefore Cost of \small 546000\hspace {1mm}cm^2 of the cardboard = \small Rs. (\frac{4}{1000}\times546000) = Rs. 2184

Therefore, the cost of the cardboard sheet required for 250 such boxes of each kind is \small Rs.\ 2184

 

Solutions of NCERT for class 9 maths chapter 13 Surface Area and Volumes Excercise: 13.2

Q1 The curved surface area of a right circular cylinder of height  \small 14\hspace{1mm}cm is  \small 88\hspace{1mm}cm^2. Find the diameter of the base of the cylinder.

Answer:

Given,

The curved surface area of the cylinder = \small 88\hspace{1mm}cm^2

And, the height of the cylinder, h= 14\ cm

We know, Curved surface area of a right circular cylinder = 2\pi rh

 \\ \therefore 2\pi rh = 88 \\ \implies 2.\frac{22}{7}. r. (14) = 88 \\ \\ \implies r = 1

Therefore, the diameter of the cylinder = 1\ cm

 

Q2 It is required to make a closed cylindrical tank of height \small 1\hspace{1mm}m and base diameter \small 140\hspace{1mm}cm from a metal sheet. How many square metres of the sheet are required for the same?

Answer:

Given,

Height of the cylindrical tank = h = 1\ m

Base diameter =  \small d = 140\ cm = 1.4\ m

We know, 

The total surface area of a cylindrical tank = \small 2\pi r h+2\pi r^2 = 2\pi r(r+h)

\small = 2.\frac{22}{7}. \frac{1.4}{2}.(0.7+1) = 2.\frac{22}{7}. (0.7).(1.7)

\small = 44\times0.17

\small = 7.48\ m^2

Therefore, square metres of the sheet is \small 7.48\ m^2

 

Q3 (i) A metal pipe is \small 77\hspace{1mm}cm long. The inner diameter of a cross section is \small 4\hspace{1mm}cm, the outer diameter being \small 4.4\hspace{1mm}cm (see Fig.  \small 13.11). Find its inner curved surface area,

            

 

Answer:

Note: There are two surfaces, inner and outer.

Given,

Height of the cylinder, h = 77\ cm

Outer diameter  = r_1 = 4.4\ cm

Inner diameter  = r_2 = 4\ cm

 Inner curved surface area = 2\pi r_2h

\\ = 2\times\frac{22}{7}\times2\times77 \\ = 968\ cm^2

Therefore, the inner curved surface area of the cylindrical pipe is 968\ cm^2

 

Q3 (ii) A metal pipe is \small 77\hspace{1mm}cm long. The inner diameter of a cross section is \small 4\hspace{1mm}cm, the outerdiameter being  \small 4.4\hspace{1mm}cm (see Fig. \small 13.11). Find its outer surface area.

                

 

Answer:

Note: There are two surfaces, inner and outer.

Given,

Height of the cylinder, h = 77\ cm

Outer diameter  = r_1 = 4.4\ cm

Inner diameter  = r_2 = 4\ cm

Outer curved surface area = 2\pi r_1h

\\ = 2\times\frac{22}{7}\times2.2\times77 \\ = 1064.8\ cm^2

Therefore, the outer curved surface area of the cylindrical pipe is 1064.8\ cm^2

 

Q3 (iii) A metal pipe is \small 77\hspace{1mm}cm long. The inner diameter of a cross section is 4 cm, the outer diameter being \small 4.4 \hspace{1mm}cm (see Fig. \small 13.11). Find its total surface area.

          

 

Answer:

Note: There are two surfaces, inner and outer.

Given,

Height of the cylinder, h = 77\ cm

Outer diameter  = r_1 = 4.4\ cm

Inner diameter  = r_2 = 4\ cm

Outer curved surface area = 2\pi r_1h

Inner curved surface area = 2\pi r_2h

Area of the circular rings on top and bottom = 2\pi(r_2^2-r_1^2)

\therefore The total surface area of the pipe = 2\pi r_1h +2\pi r_2h+ 2\pi(r_2^2-r_1^2)

\\ = [968 + 1064.8 + 2\pi {(2.2)^2 - (2)^2}]\\ \\ = (2032.8 + 2\times \frac{22}{7}\times 0.84) \\ \\ = (2032.8 + 5.28) \\ = 2038.08\ cm^2

Therefore, the total surface area of the cylindrical pipe is 2038.08\ cm^2

 

Q4 The diameter of a roller is \small 84\hspace{1mm}cm and its length is \small 120\hspace{1mm}cm. It takes \small 500 complete revolutions to move once over to level a playground. Find the area of the playground in \small m^2.

Answer:

Given,

The diameter of the cylindrical roller = \small d = 84\hspace{1mm}cm

Length of the cylindrical roller = \small h = 120\hspace{1mm}cm

The curved surface area of the roller = 2\pi r h = \pi dh

\\ = \frac{22}{7}\times84\times120 \\ \\ = 31680\ cm^2

\therefore Area of the playground = Area\ covered\ in\ 1\ rotation \times500

\\ = 31680 \times500 \\ = 15840000 cm^2 \\ = 1584\ m^2

Therefore, the required area of the playground = 1584\ m^2

 

Q5 A cylindrical pillar is  \small 50\hspace{1mm}cm in diameter and 3.5\ m in height. Find the cost of painting the curved surface of the pillar at the rate of \small Rs \hspace{1mm} 12.50  per \small m^2

Answer:

Given,

Radius of the cylindrical pillar, r = \frac{50}{2}\ cm= 25\ cm= 0.25\ m

Height of the cylinder, h  = 3.5\hspace{1mm}m

We know,

Curved surface area of a cylinder = 2\pi r h

\therefore Curved surface area of the pillar = 2\times\frac{22}{7}\times 0.25 \times3.5

= 1\times22\times 0.25\ m^2

= 5.5\ m^2

Now,

Cost of painting \small 1\ m^2 of the pillar =\small Rs \hspace{1mm} 12.50

\therefore Cost of painting the curved surface area of the pillar = \small Rs.\ (12.50\times5.5)

\small = Rs.\ 68.75

Therefore, the cost of painting curved surface area of the pillar is \small Rs.\ 68.75

 

Q6 Curved surface area of a right circular cylinder is \small 4.4\hspace{1mm}m^2. If the radius of the base of the cylinder is \small 0.7\hspace{1mm}m, find its height.

Answer:

Given, a right circular cylinder

Curved surface area of the cylinder = \small 4.4\hspace{1mm}m^2

The radius of the base = r = 0.7\ m

Let the height of the cylinder be  h

We know, 

Curved surface area of a cylinder of radius r and height  h = 2\pi r h

\therefore  2\pi r h = 4.4

\\ \Rightarrow 2\times\frac{22}{7}\times0.7 \times h = 4.4 \\ \\ \Rightarrow 44\times0.1 \times h = 4.4 \\ \Rightarrow h = 1\ m

Therefore, the required height of the cylinder is 1\ m

 

Q7 (i) The inner diameter of a circular well is \small 3.5\hspace{1mm}m. It is \small 10\hspace{1mm}m deep. Find its inner curved surface area.

Answer:

Given,

The inner diameter of the circular well =  d = \small 3.5\hspace{1mm}m

Depth of the well = h = 10\ m

We know,

The curved surface area of a cylinder = 2\pi rh

\therefore The curved surface area of the well = 2\times\frac{22}{7}\times\frac{3.5}{2}\times10

\\ = 44 \times 0.25 \times 10 \\ = 110\ m^2

Therefore, the inner curved surface area of the circular well is 110\ m^2

Q7 (ii) The inner diameter of a circular well is \small 3.5\hspace{1mm}m. It is \small 10\hspace{1mm}m deep. Find the cost of plastering this curved surface at the rate of Rs 40 per \small m^2.

Answer:

Given,

The inner diameter of the circular well =  d = \small 3.5\hspace{1mm}m

Depth of the well = h = 10\ m

\therefore The inner curved surface area of the circular well is 110\ m^2

Now, the cost of plastering the curved surface per \small m^2 = Rs. 40

\therefore Cost of plastering the curved surface of  110\ m^2 = Rs.\ (110 \times 40) = Rs.\ 4400

Therefore, the cost of plastering the well is Rs.\ 4400

 

Q8 In a hot water heating system, there is a cylindrical pipe of length \small 28\hspace{1mm}m and diameter \small 5\hspace{1mm}cm. Find the total radiating surface in the system.

Answer:

Given,

Length of the cylindrical pipe = l = \small 28\hspace{1mm}m

Diameter =  \small d = 5\hspace{1mm}cm = 0.05\ m

The total radiating surface will be the curved surface of this pipe.

We know,

The curved surface area of a cylindrical pipe of radius r and length l = 2\pi r l

\therefore CSA of this pipe = 

= 2\times\frac{22}{7}\times\frac{0.05}{2}\times28

\\ = 22\times0.05\times4 \\ = 4.4\ m^2

Therefore, the total radiating surface of the system is 4.4\ m^2

 

Q9 (i) Find the lateral or curved surface area of a closed cylindrical petrol storage tank that is \small 4.2\hspace{1mm}m in diameter and \small 4.5\hspace{1mm}m high.

Answer:

Given, a closed cylindrical petrol tank.

The diameter of the tank = \small d = 4.2\hspace{1mm}m

Height of the tank = \small h = 4.5\hspace{1mm}m

We know, 

The lateral surface area of a cylinder of radius \small r and height \small h = \small 2\pi r h

\small \therefore The lateral surface area of a cylindrical tank = \small 2\times\frac{22}{7}\times\frac{4.2}{2}\times4.5

\small \\ = (44 \times 0.3 \times 4.5) \\ = 59.4\ m^2

Therefore, the lateral or curved surface area of a closed cylindrical petrol storage tank is \small 59.4\ m^2

 

Q9 (ii) Find:  how much steel was actually used, if \small \frac{1}{12}  of the steel actually used was wasted in making the tank.

Answer:

Given, a closed cylindrical petrol tank.

The diameter of the tank = \small d = 4.2\hspace{1mm}m

Height of the tank = \small h = 4.5\hspace{1mm}m

Now, Total surface area of the tank = 2\pi r (r + h)

\\ = 2 \times \frac{22}{7} \times 2.1 \times (2.1 + 4.5) \\ = (44 \times 0.3 \times 6.6) \\ = 87.12\ m^2

Now, let x\ m^2 of steel sheet be actually used in making the tank

Since \small \frac{1}{12} of steel was wasted, the left \small \frac{11}{12} of the total steel sheet was used to made the tank.
\small \therefore The total surface area of the tank = \small \frac{11}{12}\times (Total\ area\ of\ steel\ sheet )
\small \\ \Rightarrow 87.12= \frac{11}{12}x \\ \Rightarrow x = \frac{87.12\times12}{11} \\ \Rightarrow x = 95.04\ m^2

Therefore, \small 95.04\ m^2 of steel was actually used in making the tank.

Q10 In Fig. \small 13.12, you see the frame of a lampshade. It is to be covered with a decorative cloth. The frame has a base diameter of \small 20\hspace{1mm}cm and height of \small 30\hspace{1mm}cm. A margin of \small 2.5\hspace{1mm}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.

                

 

Answer:

Given, a cylindrical lampshade

The diameter of the base = d = 20\ cm

Height of the cylinder  = 30\ cm

The total height of lampshade= (30+ 2.5 + 2.5) = 35\ cm 

We know, 

Curved surface area of a cylinder of radius r and height  h = 2\pi r h

Now, Cloth required for covering the lampshade = Curved surface area of the cylinder

\\ = 2\times\frac{22}{7} \times10 \times 35 \\ \\ = 22\times10\times10 = 2200\ cm^2

Therefore, 2200\ cm^2 cloth will be required for covering the lampshade.

Q11 The 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 \small 3\hspace{1mm}cm and height  \small 10.5\hspace{1mm}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?

Answer:

Given, a cylinder with a base.

The radius of the cylinder = r = 3\ cm

Height of the cylinder = h = 10.5\ cm

We know, 

The lateral surface area of a cylinder of radius r and height  h = 2\pi r h

\therefore  Area of the cylindrical penholder = Lateral areal + Base area

= 2\pi r h + \pi r^2 = \pi r (2h+r)

= \frac{22}{7}\times3\times[2(10.5)+3]

\\ = \frac{22}{7}\times3\times[24] \\ \\ = \frac{1584}{7}\ cm^2
Area of 35 penholders  = \frac{1584}{7}\times35\ cm^2

= 7920\ cm^2

Therefore, the area of carboard required is 7920\ cm^2

 

NCERT solutions for class 9 maths chapter 13 Surface Area and Volumes Excercise: 13.3

Q1 Diameter of the base of a cone is \small 10.5 \hspace{1mm}cm and its slant height is \small 10 \hspace{1mm}cm. Find its curved surface area.

Answer:

Given, 

Base diameter of the cone = d=10.5\ cm

Slant height = l=10\ cm

We know, Curved surface area of a cone = \pi r l

\therefore  Required curved surface area of the cone=

\\ = \frac{22}{7}\times \frac{10.5}{2}\times10 \\ \\ = 165\ cm^2

 

Q2 Find the total surface area of a cone, if its slant height is \small 21\hspace{1mm}m and diameter of its base is \small 24\hspace{1mm}m.

Answer:

Given, 

Base diameter of the cone = d=24\ m

Slant height = l=21\ cm

We know, Total surface area of a cone = Curved surface area + Base area

= \pi r l + \pi r^2 = \pi r (l + r)

\therefore  Required total surface area of the cone=

\\ = \frac{22}{7}\times\frac{24}{2}\times(21+12) \\ \\ =\frac{22}{7}\times\frac{24}{2}\times33 \\ = 1244.57 \ m^2

 

Q3 (i) Curved surface area of a cone is \small 308\hspace{1mm}cm^2 and its slant height is 14 cm. Find radius of the base.

Answer:

Given,

The curved surface area of a cone =  \small 308\hspace{1mm}cm^2

Slant height = l = 14\ cm

(i) Let the radius of cone be r\ cm

We know, the curved surface area of a cone= \pi rl

\therefore \\ \pi rl = 308 \\ \\ \Rightarrow \frac{22}{7}\times r\times14 = 308 \\ \Rightarrow r = \frac{308}{44} = 7

Therefore, the radius of the cone is 7\ cm

 

Q3 (ii) Curved surface area of a cone is \small 308\hspace{1mm}cm^2 and its slant height is \small 14\hspace{1mm}cm. Find total surface area of the cone.

Answer:

Given,

The curved surface area of a cone =  \small 308\hspace{1mm}cm^2

Slant height = l = 14\ cm

The radius of the cone is r = 7\ cm

(ii) We know, Total surface area of a cone = Curved surface area + Base area

= \pi r l + \pi r^2

\\ = 308+\frac{22}{7}\times 7^2 \\ = 308+154 = 462\ cm^2

Therefore, the total surface area of the cone is 462\ cm^2

Q4 (i) A conical tent is 10 m high and the radius of its base is 24 m. Find slant height of the tent.

Answer:

Given, 

Base radius of the conical tent  = r=24\ m

Height of the conical tent = h=10\ m

\therefore Slant height = l=\sqrt{h^2+r^2}

\\ =\sqrt{10^2+24^2} \\ = \sqrt{676} \\ = 26\ m

Therefore, the slant height of the conical tent is 26\ m

 

Q4 (ii) A conical tent is 10 m high and the radius of its base is 24 m. Find cost of the canvas required to make the tent, if the cost of \small 1\hspace{1mm}m^2 canvas is Rs 70.

Answer:

Given, 

Base radius of the conical tent  = r=24\ m

Height of the conical tent = h=10\ m

\therefore Slant height = l=\sqrt{h^2+r^2} = 26\ m

We know, Curved surface area of a cone = \pi r l

\therefore  Curved surface area of the tent 

\\ = \frac{22}{7}\times24\times26 \\ \\ =\frac{13728}{7}\ m^2

Cost of 1\ m^2 of canvas = Rs.\ 70

\therefore Cost of \frac{13728}{7}\ m^2 of canvas =

Rs.\ (\frac{13728}{7}\times70) = Rs.\ 137280

Therefore, required cost of canvas to make tent is Rs.\ 137280

Q5 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 \small \pi =3.14).

Answer:

Given,

Base radius of the conical tent = r =6\ m

Height of the tent = h =8\ m

We know,

Curved surface area of a cone = \pi rl = \pi r\sqrt{h^2 + r^2}

\therefore Area of tarpaulin required = Curved surface area of the tent

\\ =3.14\times6\times \sqrt{8^2+ 6^2} \\ = 3.14\times6\times 10 \\ = 188.4\ m^2

Now, let the length of the tarpaulin sheet be x\ m

Since 20\ cm is wasted, effective length = x - 20 cm = (x - 0.2)\ m

Breadth of tarpaulin = 3\ m

 \\ \therefore [(x - 0.2) \times 3] = 188.4 \\ \Rightarrow x - 0.2 = 62.8 \\ \Rightarrow x = 63\ m

Therefore, the length of the required tarpaulin sheet will be 63 m.

Q6 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 \small 100\hspace{1mm}m^2.

Answer:

Given, a conical tomb

The base diameter of the cone = d =14\ m

Slant height = l = 25\ m

We know, Curved surface area of a cone = \pi r l

\\ = \frac{22}{7}\times\frac{14}{2}\times25 \\ \\ = 22\times25 \\ = 550\ m^2

Now, Cost of whitewashing per \small 100\hspace{1mm}m^2 =\small Rs.\ 210

\therefore Cost of whitewashing per \small 550\hspace{1mm}m^2 = \small \\ Rs. (\frac{210}{100}\times550 )

\small \\ = Rs.\ (21\times55 ) = Rs.\ 1155

Therefore, the cost of white-washing its curved surface of the tomb is  \small Rs.\ 1155.

Q7 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.

Answer:

Given, a right circular cone cap (which means no base)

Base radius of the cone = r=7\ cm

Height = h = 24\ cm

\therefore l = \sqrt{h^2+r^2}

We know, Curved surface area of a right circular cone = \pi r l

\therefore The curved surface area of a cap =

\\ = \frac{22}{7}\times7\times\sqrt{24^2+7^2} \\ \\ = 22\times\sqrt{625} \\ = 22\times25\ \\ = 550\ cm^2

\therefore The curved surface area of 10 caps = 550\times10 = 5500\ cm^2

Therefore, the area of the sheet required for 10 caps = 5500\ cm^2

 

Q8 A bus stop is barricaded from the remaining part of the road, by using 50 hollow cones made of recycled cardboard. Each cone has a base diameter of 40 cm and height 1 m. If the outer side of each of the cones is to be painted and the cost of painting is Rs 12 per \small m^2, what will be the cost of painting all these cones? (Use \small \pi =3.14 and take \small \sqrt{1.04}=1.02)

Answer:

Given, hollow cone.

The base diameter of the cone = d = 40\ cm = 0.4\ m

Height of the cone = h = 1\ m

\therefore Slant height = l = \sqrt{h^2+r^2}= \sqrt{1^2+0.2^2}

We know, Curved surface area of a cone =\pi r l = \pi r\sqrt{h^2+r^2}

\therefore The curved surface area of 1 cone = 3.14\times0.2\times\sqrt{1.04} = 3.14\times0.2\times1.02

= 0.64056\ m^2

\therefore The curved surface area of 50 cones= (50\times0.64056)\ m^2

= 32.028\ m^2

Now, the cost of painting \small 1\ m^2 area = \small Rs.\ 12

\therefore  Cost of the painting 32.028\ m^2 area = Rs.\ (32.028\times12)

= Rs.\ 384.336

Therefore, the cost of painting 50 such hollow cones is Rs.\ 384.34\ (approx)

 

Solutions of NCERT for class 9 maths chapter 13 Surface Area and Volumes Excercise: 13.4

Q1 (i) Find the surface area of a sphere of radius: \small 10.5\hspace{1mm}cm.

Answer:

We know,

The surface area of a sphere of radius r = 4\pi r^2

\therefore Required surface area = 4\times\frac{22}{7} \times(10.5)^2

\\ =88\times1.5\times10.5 \\ = 1386\ cm^2

 

Q1 (ii) Find the surface area of a sphere of radius: \small 5.6\hspace{1mm}cm

Answer:

We know,

The surface area of a sphere of radius r = 4\pi r^2

\therefore Required surface area = 4\times\frac{22}{7} \times(5.6)^2

\\ =88\times0.8\times5.6 \\ = 394.24\ cm^2

 

Q1 (iii) Find the surface area of a sphere of radius: \small 14\hspace{1mm}cm

Answer:

We know,

The surface area of a sphere of radius r = 4\pi r^2

\therefore Required surface area = 4\times\frac{22}{7} \times(14)^2

\\ = 88\times2\times14 \\ = 2464 \ cm^2

 

Q2 (i) Find the surface area of a sphere of diameter: 14 cm

Answer:

Given,

The diameter of the sphere = 14\ cm

We know,

The surface area of a sphere of radius r = 4\pi r^2

\therefore Required surface area = 4\times\frac{22}{7} \times\left (\frac{14}{2} \right )^2

= 4\times\frac{22}{7} \times\frac{14}{2}\times\frac{14}{2}

\\ = 22\times2\times14 \\ = 616\ cm^2

 

Q2 (ii) Find the surface area of a sphere of diameter: 21 cm

Answer:

Given,

The diameter of the sphere = 21\ cm

We know,

The surface area of a sphere of radius r = 4\pi r^2

\therefore Required surface area = 4\times\frac{22}{7} \times\left (\frac{21}{2} \right )^2

= 4\times\frac{22}{7} \times\frac{21}{2}\times\frac{21}{2}

\\ = 22\times3\times21 \\ = 1386\ cm^2

 

Q2 (iii) Find the surface area of a sphere of diameter: \small 3.5\hspace{1mm}m

Answer:

Given,

The diameter of the sphere = 3.5\ m

We know,

The surface area of a sphere of radius r = 4\pi r^2

\therefore Required surface area = 4\times\frac{22}{7} \times\left (\frac{3.5}{2} \right )^2

= 4\times\frac{22}{7} \times\frac{3.5}{2}\times\frac{3.5}{2}

\\ = 22\times0.5\times3.5 \\ = 38.5\ m^2

 

Q3 Find the total surface area of a hemisphere of radius 10 cm. (Use \small \pi =3.14)

Answer:

We know,

The total surface area of a hemisphere = Curved surface area of hemisphere + Area of the circular end

= 2\pi r^2 + \pi r^2 = 3\pi r^2

\therefore The required total surface area of the hemisphere = 3\times3.14\times(10)^2

\\ = 942\ cm^2

 

Q4 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.

Answer:

Given,

r_1 = 7\ cm

r_2 = 14\ cm

We know,

The surface area of a sphere of radius r = 4\pi r^2

\therefore The ratio of surface areas of the ball in the two cases = \frac{Initial}{Final} = \frac{4\pi r_1^2}{4\pi r_2^2}

= \frac{r_1^2}{r_2^2}

\\ = \left (\frac{7}{14} \right )^2 \\ \\ = \left (\frac{1}{2} \right )^2 \\ \\ = \frac{1}{4}

Therefore, the required ratio is 1:4

 

Q5 A hemispherical bowl made of brass has inner diameter \small 10.5\hspace{1mm}cm. Find the cost of tin-plating it on the inside at the rate of  Rs 16 per \small 100\hspace{1mm}cm^2

Answer:

Given,

The inner radius of the hemispherical bowl = r = \frac{10.5}{2}\ cm

We know,

The curved surface area of a hemisphere = 2\pi r^2

\therefore The surface area of the hemispherical bowl = 2\times\frac{22}{7}\times\left (\frac{10.5}{2} \right )^2

=11\times1.5\times10.5

= 173.25 \ cm^2

Now,

Cost of tin-plating \small 100\hspace{1mm}cm^2 =  Rs 16

\therefore Cost of tin-plating \small 33\hspace{1mm}cm^2 =  \small \\ Rs. \left (\frac{16}{100}\times173.25 \right )

\small = Rs. 27.72

Therefore, the cost of tin-plating it on the inside is \small Rs. 27.72

Q6 Find the radius of a sphere whose surface area is \small 154\hspace{1mm}cm^2.

Answer:

Given,

The surface area of the sphere = \small 154\hspace{1mm}cm^2

We know,

The surface area of a sphere of radius r = 4\pi r^2

\therefore 4\pi r^2 = 154

\\ \Rightarrow 4\times\frac{22}{7}\times r^2 = 154

\\ \Rightarrow r^2 = \frac{154\times7}{4\times22} = \frac{7\times7}{4}

\\ \Rightarrow r = \frac{7}{2}

\\ \Rightarrow r = 3.5\ cm

Therefore, the radius of the sphere is 3.5\ cm

Q7 The diameter of the moon is approximately one fourth of the diameter of the earth. Find the ratio of their surface areas.

Answer:

Let diameter of Moon be d_m and diameter of Earth be d_e

We know,

The surface area of a sphere of radius r = 4\pi r^2

\therefore The ratio of their surface areas = \frac{Surface\ area\ of\ moon}{Surface\ area\ of\ Earth}

= \frac{4\pi \left (\frac{d_m}{2} \right )^2}{4\pi \left (\frac{d_e}{2} \right )^2}

= \frac{d_m^2}{d_e^2}

=\left ( \frac{\frac{1}{4}d_e}{d_e} \right )^2

= \frac{1}{16}

Therefore, the ratio of the surface areas of the moon and earth is = 1:16 

Q8 A hemispherical bowl is made of steel, \small 0.25\hspace{1mm}cm thick. The inner radius of the bowl is 5 cm. Find the outer curved surface area of the bowl.

Answer:

Given,

The inner radius of the bowl = r_1 = 5\ cm

The thickness of the bowl = \small 0.25\hspace{1mm}cm

\therefore Outer radius of the bowl = (Inner radius + thickness) = 

r_2 = 5+0.25 = 5.25\ cm

We know, Curved surface area of a hemisphere of radius r2\pi r^2

\therefore The outer curved surface area of the bowl = 2\pi r_2^2

= 2\times\frac{22}{7}\times (5.25)^2

= 2\times\frac{22}{7}\times5.25\times5.25 = 173.25\ cm^2

Therefore, the outer curved surface area of the bowl is 173.25\ cm^2

Q9 (i) A right circular cylinder just encloses a sphere of radius \small r (see Fig. \small 13.22). Find surface area of the sphere,

            

Answer:

Given,

The radius of the sphere = r

\therefore Surface area of the sphere = 4\pi r^2

 

Q9 (ii) A right circular cylinder just encloses a sphere of radius \small r (see Fig. \small 13.22). Find curved surface area of the cylinder,

            

Answer:

Given,

The radius of the sphere = r

\therefore The surface area of the sphere = 4\pi r^2

According to the question, the cylinder encloses the sphere.

Hence, the diameter of the sphere is the diameter of the cylinder. 

Also, the height of the cylinder is equal to the diameter of the sphere.

 We know, the curved surface area of a cylinder = 2\pi rh

= 2\pi r(2r) = 4\pi r^2

Therefore, the curved surface area of the cylinder is 4\pi r^2

 

Q9 (iii) A right circular cylinder just encloses a sphere of radius \small r (see Fig. \small 13.22). Find ratio of the areas obtained in (i) and (ii).

            

Answer:

The surface area of the sphere = 4\pi r^2

And,  Surface area of the cylinder = 4\pi r^2

So, the ratio of the areas = \frac{4\pi r^2}{4\pi r^2} = 1

NCERT solutions for class 9 maths chapter 13 Surface Area and Volumes Excercise: 13.5

Q1 A matchbox measures \small 4\hspace{1mm}cm\times 2.5\hspace{1mm}cm\times 1.5\hspace{1mm}cm. What will be the volume of a packet containing 12 such boxes?

Answer:

Given,

Dimensions of a matchbox =  \small 4\hspace{1mm}cm\times 2.5\hspace{1mm}cm\times 1.5\hspace{1mm}cm

We know,

The volume of a cuboid = l\times b \times h

\small \therefore The volume of a matchbox= \small (4 \times2.5\times 1.5)\ cm^3 = 15\ cm^3

\small \therefore Volume of 12 such matchboxes = \small (15\times12)\ cm^3 = 180\ cm^3

Therefore, the volume of a packet containing 12 matchboxes is \small 180\ cm^3

Q2 A cuboidal water tank is 6 m long, 5 m wide and \small 4.5\hspace{1mm}m deep. How many litres of water can it hold? (\small 1\hspace{1mm}m^3=1000\hspace{1mm}l)

Answer:

Given,

Dimensions of the cuboidal water tank = 6\ m \times5\ m\times 4.5\ m

We know,

Volume of a cuboid = l\times b \times h

\small \therefore Volume of the water tank = \small (6 \times5\times 4.5)\ m^3 = 135\ m^3

We know,

\small 1\hspace{1mm}m^3=1000\hspace{1mm}l

\small \therefore Volume of water the tank can hold = \small (135\times1000) = 135000\ litres

Q3 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?

Answer:

Let the height of the vessel be h\ m

Dimensions of the cuboidal water tank = 10\ m \times8\ m\times h\ m

We know,

The volume of a cuboid = l\times b \times h

\small \therefore The volume of the water tank = \small (10 \times8\times h)\ m^3 = 80h\ m^3

According to question,

\small \\ \Rightarrow h= \frac{380}{80} = 4.75

Therefore, the required height of the vessel is \small 4.75\ m

Q4 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 \small m^3.

Answer:

Given,

Dimensions of the cuboidal pit = 8\ m\times 6\ m\times 3\ m

We know , Volume of a cuboid = l\times b\times h

\therefore Volume of the cuboidal pit = (8\times 6\times 3\) = 144\ m^3

Now, Cost of digging \small 1\ m^3 = Rs. 30

\therefore Cost of digging 144\ m^3 = Rs.(144\times30)

= Rs.4320

Q5 The capacity of a cuboidal tank is 50000 litres of water. Find the breadth of the tank, if its length and depth are respectively \small 2.5\hspace{1mm}m and 10 m.

Answer:

Given,

Length of the tank = \small l = 2.5\hspace{1mm}m

Depth of the tank = h = 10\ m

Let the breadth of the tank be \small b\ m

We know, Volume of a cuboid = l\times b\times h

\therefore The volume of the cuboidal tank = (2.5\times b\times 10) = 25b\ m^3

We know, 1\ m^3 = 1000\ litre

\therefore 25b\ m^3 = 50000\ litre = 50\ m^3

\\ \Rightarrow 25b = 50 \\ \Rightarrow b = 2\ m

Therefore, the breadth of the tank is 2\ m

Q6 A village, having a population of 4000, requires 150 litres of water per head per day. It has a tank measuring \small 20\hspace{1mm}m\times 15\hspace{1mm}m\times 6\hspace{1mm}m. For how many days will the water of this tank last?

Answer:

Given,

Dimensions of the tank = \small 20\hspace{1mm}m\times 15\hspace{1mm}m\times 6\hspace{1mm}m

4000 people requiring 150 litres of water per head per day

The total volume of water required = 4000\times150 = 600000\ litres

Let the number of days the water will last be n

We know,

The volume of a cuboid = l\times b \times h

\small \therefore The volume of the water tank = \small (20 \times15\times 6)\ m^3 = 1800\ m^3 = 1800000\ litres

According to question,

\small n\times600000 = 1800000

\small \\ \Rightarrow n= 3

Therefore, the water in the tank will last for 3 days.

Q7 A godown measures \small 40\hspace{1mm}m\times 25\hspace{1mm}m\times 15\hspace{1mm}m. Find the maximum number of wooden crates each measuring 1.5\hspace{1mm}m\times 1.25\hspace{1mm}m\times 0.5\hspace{1mm}m that can be stored in the godown.

Answer:

Given,

Dimensions of the godown = \small 40\hspace{1mm}m\times 25\hspace{1mm}m\times 15\hspace{1mm}m

Dimension of each wooden crate = \small 1.5\hspace{1mm}m\times 1.25\hspace{1mm}m\times 0.5\hspace{1mm}m

We know , Volume of a cuboid = l\times b\times h

\therefore Volume of the godown = (40\times 25\times 15)\ m^3

\therefore Volume of the each crate= (1.5\times 1.25\times 0.5)\ m^3

Let number of wooden crates be n

\therefore Volume of n wooden crates = Volume of the godown

n\times(1.5\times 2.5\times 0.5)\ m^3 = (40\times 25\times 15)\ m^3

\\ \Rightarrow n = \frac{40\times 25\times 15}{1.5\times 1.25\times 0.5} \\ \\ \Rightarrow n = 80\times 20\times 10 = 16000

Q8 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.

Answer:

This is an important question.

Given,

Side of a solid cube = l = 12\ cm

We know, the volume of a cube of side l = l^3

\therefore The volume of the given cube =12^3\ cm^3

Now, the cube is cut into 8 equal cubes of side a (let)

\therefore The total volume of these 8 cubes = Volume of the bigger cube

\Rightarrow (8\times a^3)\ cm^3 = 12\ cm^3

\\ \Rightarrow a^3 = \frac{12}{8} \\ \Rightarrow a^3 = \left (\frac{12}{2} \right ) \\ \Rightarrow a = 6\ cm

Therefore, the side of the new cube is 6\ cm

Now, we know,

The surface area of a cube of side l = 6l^2

\therefore The ratio between their surface areas = \frac{Surface\ area\ of\ bigger\ cube}{Surface\ area\ of\ a\ smaller\ cube}

\\ = \frac{6l^2}{6a^2} \\ = \left (\frac{l}{a} \right )^2 \\ = \left (\frac{12}{6} \right )^2 \\ = 4

Therefore, the ratio of their surface areas is 4:1

Q9 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?

Answer:

Given,
Rate of water flow = 2\ km\ per\ hour
= \frac{2\times1000}{60}\ m/min
= \frac{100}{3}\ m/min
Depth of river, h = 3\ m
Width of the river, b= 40\ m
The volume of water flowing in 1 min = Rate\ of\ flow \times Cross\ sectional\ area

= \frac{100}{3}\times40\times3

= 4000\ m^3

Therefore, water falling into the sea in a minute = 4000\ m^3

Solutions of NCERT for class 9 maths chapter 13 Surface Area and Volumes Excercise: 13.6

Q1 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? (\small 1000\hspace{1mm}cm^3=1l)

Answer:

Given, 

Circumference of the base of a cylindrical vessel = 132\ cm

Height = h = 25\ cm

Let the radius of the base of the cylinder be r\ cm

Now, Circumference of the circular base of the cylinder = 2\pi r = 132\ cm

\\ \Rightarrow 2\times\frac{22}{7}\times r = 12\times11 \\ \\ \Rightarrow r = \frac{12\times7}{4} = 21\ cm

\therefore The volume of the cylinder = \pi r^2 h

\\ = \frac{22}{7}\times (21)^2 \times25 \\ = 22\times3\times21\times25 \\ = 34650\ cm^3

Also, \small 1000\hspace{1mm}cm^3=1l

\therefore 34650\ cm^3 = \frac{34650}{1000} = 34.65\ litres

Therefore, the cylindrical vessel can hold 34.65\ litres of water.

Q2 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 \small 1\hspace{1mm}cm^3 of wood has a mass of  \small 0.6\hspace{1mm}g.

Answer:

Given, A hollow cylinder made of wood.

The inner diameter of the cylindrical wooden pipe = r_1 = 24\ cm
Outer diameter =  r_2 = 28\ cm

Length of the pipe = h= 35\ cm

\therefore The volume of the wooden cylinder = \pi r^2 h = \pi (r_2^2 - r_1^2)h

\\ = \frac{22}{7}\times (14^2-12^2) \times35 \\ = 22\times (14-12)(14+12) \times5 \\ = 22\times (2)(26) \times5 \\ = 5720\ cm^3

Mass of \small 1\hspace{1mm}cm^3 wood = \small 0.6\hspace{1mm}g
Mass of 5720\ cm^3 wood = 5720\times0.6\ g
\\ = 3432\ g \\ = 3.432\ kg

Therefore, the mass of the wooden hollow cylindrical pipe is 3.432\ kg

 

Q3 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?

Answer:

Given, 

(i) The dimension of the rectangular base of the tin can = 5\ cm\times4\ cm

Height of the can = h = 15\ cm

\therefore The volume of the tin can = Rectangular\ area\times height

(5\times4\times15)\ cm^3 = 300\ cm^3

(ii) The radius of the circular base of the plastic cylinder = r = \frac{7}{2}= 3.5\ cm

Height of the cylinder = h = 10\ cm

\therefore The volume of the plastic cylinder = \pi r^2 h

\\ = \frac{22}{7}\times (3.5)^2 \times10 \\ = 22\times0.5\times3.5\times10 \\ = 11\times35 \\ = 385\ cm^3

Clearly, the plastic cylinder has more capacity than the rectangular tin can.

The difference in capacity = (385-300)\ cm^3 = 85\ cm^3

 

Q4 (i) If the lateral surface of a cylinder is \small 94.2\hspace{1mm}cm^2 and its height is 5 cm, then find radius of its base (Use \small \pi =3.14)

Answer:

Given,

The lateral surface area of the cylinder = \small 94.2\hspace{1mm}cm^2

Height of the cylinder = h = 5\ cm

(i) Let the radius of the base be r\ cm

We know,

The lateral surface area of a cylinder = 2\pi r h

\\ \therefore 2\pi r h = 94.2 \\ \Rightarrow 2\times3.14\times r\times5 = 94.2 \\ \\ \Rightarrow r = \frac{94.2}{31.4} = 3\ cm

Therefore, the radius of the base is 3\ cm

Q4 (ii) If the lateral surface of a cylinder is \small 94.2\hspace{1mm}cm^2 and its height is 5 cm, then find it's volume. (Use \small \pi =3.14)

Answer:

Given,

The lateral surface area of the cylinder = \small 94.2\hspace{1mm}cm^2

Height of the cylinder = h = 5\ cm

The radius of the base is 3\ cm

(ii) We know,

The volume of a cylinder = \pi r^2 h

\\ = 3.14\times3^2\times5 \\ = 3.14 \times 9 \times 5= 3.14 \times 45 \\ = 141.3\ cm^3

Therefore, the volume of the cylinder is 141.3\ cm^3.

Q5 (i) It costs Rs 2200 to paint the inner curved surface of a cylindrical vessel 10 m deep. If the cost of painting is at the rate of Rs 20 per \small m^2, find inner curved surface area of the vessel.

Answer:

(i) Given,

Rs 20 is the cost of painting 1\small m^2 area of the inner curved surface of the cylinder. 

\therefore Rs 2200 is the cost of painting \frac{1}{20}\times2200 = 110\ m^2area of the inner curved surface of the cylinder. 

\therefore The inner curved surface area of the cylindrical vessel = 110\ m^2

Q5 (ii) It costs Rs 2200 to paint the inner curved surface of a cylindrical vessel 10 m deep. If the cost of painting is at the rate of Rs 20 per \small m^2, find radius of the base.

Answer:

The inner curved surface area of the cylindrical vessel = 110\ m^2

Height of the cylinder = h = 10\ m

Let the radius of the circular base be r \ m

\therefore The inner curved surface area of the cylindrical vessel = 2\pi r h

\\ \Rightarrow 2\times\frac{22}{7}\times r \times10 = 110 \\ \Rightarrow 44\times r = 11\times7 \\ \Rightarrow 4\times r = 7 \\ \Rightarrow r = \frac{7}{4} = 1.75\ m

Therefore, the radius of the base of the vessel is 1.75 \ m

Q5 (iii) It costs Rs 2200 to paint the inner curved surface of a cylindrical vessel 10 m deep. If the cost of painting is at the rate of Rs 20 per \small m^2, find capacity of the vessel.

Answer:

(iii) Height of the cylinder = h = 10\ m

Radius of the base of the vessel =  r =1.75 \ m

\therefore Volume of the cylindrical vessel = \pi r^2 h

\\ = \frac{22}{7}\times (1.75)^2 \times10 \\ = 22\times2.5\times1.75\times10 \\ = 55\times17.5 \\ = 96.25 \ m^3

Therefore, the capacity of the cylindrical vessel is 96.25 \ m^3

Q6 The capacity of a closed cylindrical vessel of height 1 m is \small 15.4 litres. How many square metres of metal sheet would be needed to make it?

Answer:

(Using capacity(volume), we will find the radius and then find the surface area)

The capacity of the vessel = Volume of the vessel =\small 15.4 litres

Height of the cylindrical vessel = h = 1\ m

Let the radius of the circular base be r \ m

\therefore The volume of the cylindrical vessel = \pi r^2 h

\\ \Rightarrow \frac{22}{7}\times r^2 \times1 = 15.4\ litres= 0.0154\ m^3 \\ \Rightarrow r^2 = \frac{0.0014\times11\times7}{22} \\ \Rightarrow r^2 = \frac{7\times7}{10000} \\ \Rightarrow r = \frac{7}{100}= 0.07\ m

Therefore, the total surface area of the vessel = 2\pi r h+ 2\pi r^2 = 2\pi r(r+h)

\\ = 2\times\frac{22}{7}\times0.07\times(0.07+1) \\ = 0.44 \times 1.07 \\ = 0.4708\ m^2

Therefore, square metres of metal sheet needed to make it is 0.4708\ m^2 

Q7 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.

Answer:

Given,

Length of the cylindrical pencil = h = 14\ cm

The radius of the graphite (Inner solid cylinder)  = r_1 = \frac{1}{2}\ mm = 0.05\ cm

Radius of the pencil (Inner solid graphite cylinder + Hollow wooden cylinder) =

r_2 = \frac{7}{2}\ mm = 0.35\ cm

We know, Volume of a cylinder= \pi r^2 h

\therefore The volume of graphite = \pi r_1^2 h

\\ = \frac{22}{7}\times 0.05^2 \times14 \\ = 44\times0.0025 \\ = 0.11\ cm^3

And, Volume of wood = \pi (r_2^2- r_1^2) h

\\ = \frac{22}{7}\times (0.35^2-0.05^2) \times14 \\ = 44\times(0.35-0.05)(0.35+0.05) \\ = 44\times(0.30)(0.40) \\ = 5.28 \ cm^3

Therefore, the volume of wood is 5.28 \ cm^3 and the volume of graphite is 0.11 \ cm^3

Q8 A patient in a hospital is given soup daily in a cylindrical bowl of diameter 7 cm. If the bowl is filled with soup to a height of 4 cm, how much soup the hospital has to prepare daily to serve 250 patients?

Answer:

Given,

Height = h = 4\ cm

The radius of the cylindrical bowl  = r_1 = \frac{7}{2}\ cm = 3.5\ cm

\therefore The volume of soup in a bowl for a single person = \pi r^2 h

\\ = \frac{22}{7}\times (3.5)^2 \times4 \\ = 88\times0.5\times3.5 \\ = 154\ cm^3
\therefore The volume of soup given for 250 patients = (250 \times 154)\ cm^3
\\ = 38500\ cm^3 \\ = 38.5\ litres

Therefore, the amount of soup the hospital has to prepare daily to serve 250 patients is 38.5\ litres

NCERT solutions for class 9 maths chapter 13 Surface Area and Volumes Excercise: 13.7

Q1 (i) Find the volume of the right circular cone with radius 6 cm, height 7 cm

Answer:

Given,

Radius = r =6\ cm

Height = h =7\ cm

We know,

Volume of a right circular cone = \frac{1}{3}\pi r^2 h

\therefore Required volume = \frac{1}{3}\times\frac{22}{7}\times6^2\times7

\\ = 22\times2\times6 \\ = 264\ cm^3

Q1 (ii) Find the volume of the right circular cone with: radius \small 3.5 cm, height 12 cm

Answer:

Given,

Radius = r =3.5\ cm

Height = h =12\ cm

We know,

Volume of a right circular cone = \frac{1}{3}\pi r^2 h

\therefore Required volume = \frac{1}{3}\times\frac{22}{7}\times3.5^2\times12

\\ = 22\times0.5\times3.5\times4 \\ = 11\times14 \\ = 154\ cm^3

Q2 (i) Find the capacity in litres of a conical vessel with radius 7 cm, slant height 25 cm

Answer:

Given,

Radius = r =7\ cm

Slant height = l = \sqrt{r^2 + h^2} = 25\ cm

Height = h =\sqrt{l^2-r^2} = \sqrt{25^2-7^2}

= \sqrt{(25-7)(25+7)} = \sqrt{(18)(32)}

= 24\ cm

We know,
Volume of a right circular cone = \frac{1}{3}\pi r^2 h

\therefore Volume of the vessel= \frac{1}{3}\times\frac{22}{7}\times7^2\times24

\\ = 22\times7\times8\\ = 154\times8 \\ = 1232\ cm^3

\therefore Required capacity of the vessel = 

= \frac{1232}{1000} = 1.232\ litres

Q2 (ii) Find the capacity in litres of a conical vessel with height 12 cm, slant height 13 cm 

Answer:

Given,

Height = h =12\ cm

Slant height = l = \sqrt{r^2 + h^2} = 13\ cm

Radius = r =\sqrt{l^2-h^2} = \sqrt{13^2-12^2}

= \sqrt{(13-12)(13+12)} = \sqrt{(1)(25)}

= 5\ cm

We know,
Volume of a right circular cone = \frac{1}{3}\pi r^2 h

\therefore Volume of the vessel= \frac{1}{3}\times\frac{22}{7}\times5^2\times12

\\ = \frac{22}{7}\times25\times4\\ = \frac{2200}{7}\ cm^3

\therefore Required capacity of the vessel = 

= \frac{2200}{7\times1000} = \frac{11}{35}\ litres

Q3 The height of a cone is 15 cm. If its volume is 1570 \small cm^3, find the radius of the base. (Use \small \pi =3.14)

Answer:

Given,

Height of the cone = h =15\ cm

Let the radius of the base of the cone be r\ cm

We know,
The volume of a right circular cone = \frac{1}{3}\pi r^2 h

\therefore \frac{1}{3}\times3.14\times r^2\times15 = 1570

\\ \Rightarrow 3.14\times r^2\times5 = 1570 \\ \Rightarrow r^2 = \frac{1570}{15.7} \\ \Rightarrow r^2 = 100 \\ \Rightarrow r = 10\ cm

Q4 If the volume of a right circular cone of height 9 cm is \small 48\pi \hspace{1mm}cm^3, find the diameter of its base.

Answer:

Given,

Height of the cone = h =9\ cm

Let the radius of the base of the cone be r\ cm

We know,
The volume of a right circular cone = \frac{1}{3}\pi r^2 h

\therefore \frac{1}{3}\times\pi\times r^2\times9 = 48\pi

\\ \Rightarrow 3r^2 = 48 \\ \Rightarrow r^2 = 16\\ \Rightarrow r = 4\ cm

Therefore the diameter of the right circular cone is 8\ cm

Q5 A conical pit of top diameter \small 3.5 m is 12 m deep. What is its capacity in kilolitres?

Answer:

Given,

Depth of the conical pit  = h =12\ m

The top radius of the conical pit = r = \frac{3.5}{2}\ m

We know,
The volume of a right circular cone = \frac{1}{3}\pi r^2 h

\therefore The volume of the conical pit =

  = \frac{1}{3}\times\frac{22}{7}\times \left (\frac{3.5}{2} \right )^2\times12

\\ = \frac{1}{3}\times\frac{22}{7}\times \frac{3.5\times 3.5}{4}\times12 \\ \\ = 22\times 0.5\times 3.5 \\ = 38.5\ m^3

Now, 1\ m^3 = 1\ kilolitre

\therefore The capacity of the pit = 38.5\ kilolitre

Q6 (i) The volume of a right circular cone is \small 9856\hspace{1mm}cm^3. If the diameter of the base is 28 cm, find height of the cone

Answer:

Given, a right circular cone.

The radius of the base of the cone = r = \frac{28}{2} = 14\ cm

The volume of the cone = \small 9856\hspace{1mm}cm^3

(i) Let the height of the cone be h\ m

We know,
The volume of a right circular cone = \frac{1}{3}\pi r^2 h

\therefore   \frac{1}{3}\times\frac{22}{7}\times(14)^2\times h = 9856

\\ \Rightarrow \frac{1}{3}\times\frac{22}{7}\times14\times14\times h = 9856 \\ \Rightarrow \frac{1}{3}\times22\times2\times14\times h = 9856 \\ \Rightarrow h = \frac{9856\times3}{22\times2\times14} \\ \\ \Rightarrow h =48\ cm

Therefore, the height of the cone is 48\ cm

Q6 (ii) The volume of a right circular cone is \small 9856\hspace{1mm}cm^3. If the diameter of the base is 28 cm, find slant height of the cone

Answer:

Given, a right circular cone.

The volume of the cone = \small 9856\hspace{1mm}cm^3

The radius of the base of the cone = r = \frac{28}{2} = 14\ cm

And the height of the cone =  h = 48\ cm

(ii) We know, Slant height, l = \sqrt{r^2+h^2}

\\ \Rightarrow l = \sqrt{14^2+48^2} \\ \Rightarrow l = \sqrt{196+2304} = \sqrt{2500} \\ \Rightarrow l = 50\ cm

Therefore, the slant height of the cone is 50\ cm.

Q7 (iii) The volume of a right circular cone is \small 9856\hspace{1mm}cm^3. If the diameter of the base is 28 cm, find curved surface area of the cone

Answer:

Given, a right circular cone.

The radius of the base of the cone = r = \frac{28}{2} = 14\ cm

And Slant height of the cone =  l = 50\ cm

(iii) We know,

The curved surface area of a cone = \pi r l

\therefore Required curved surface area= \frac{22}{7}\times14\times50

\\ = 22\times2\times50 \\ = 2200\ cm^2

Q7 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.

Answer:

When a right-angled triangle is revolved about the perpendicular side, a cone is formed whose,

Height of the cone = Length of the axis= h = 12\ cm

Base radius of the cone = r = 5\ cm

And, Slant height of the cone = l = 13\ cm

We know,

 The volume of a cone = \frac{1}{3}\pi r^2 h

The required volume of the cone formed = \frac{1}{3}\times\pi\times5^2\times12

\\ = \pi\times25\times4 \\ = 100\pi\ cm^3

Therefore, the volume of the solid cone obtained is 100\pi\ cm^3

Q8 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.

Answer:

When a right-angled triangle is revolved about the perpendicular side, a cone is formed whose,

Height of the cone = Length of the axis= h = 5\ cm

Base radius of the cone = r = 12\ cm

And, Slant height of the cone = l = 13\ cm

We know,

 The volume of a cone = \frac{1}{3}\pi r^2 h

The required volume of the cone formed = \frac{1}{3}\times\frac{22}{7}\times12^2\times5

\\ = \pi\times4\times60 \\ = 240\pi\ cm^3

Now, Ratio of the volumes of the two solids = \\ = \frac{100\pi}{240\pi}

\\ = \frac{5}{12}

Therefore, the required ratio is 5:12

Q9 A heap of wheat is in the form of a cone whose diameter is \small 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.

Answer:

Given,

Height of the conical heap =  h = 3\ m

Base radius of the cone = r = \frac{10.5}{2}\ m

We know,

 The volume of a cone = \frac{1}{3}\pi r^2 h

The required volume of the cone formed = \frac{1}{3}\times\frac{22}{7}\times\left (\frac{10.5}{2} \right )^2\times3

\\ = 22\times\frac{1.5\times10.5}{4} \\ = 86.625\ m^3

Now,

The slant height of the cone = l = \sqrt{r^2+h^2}

\\ \Rightarrow l = \sqrt{3^2+5.25^2} = \sqrt{9+27.5625} \approx 6.05

We know, the curved surface area of a cone = \pi r l

The required area of the canvas to cover the heap  = \frac{22}{7}\times\frac{10.5}{2}\times6.05

= 99.825\ m^2

Solutions of NCERT for class 9 maths chapter 13 Surface Area and Volumes Excercise: 13.8

Q1 (i) Find the volume of a sphere whose radius is 7 cm

Answer:

Given,

The radius of the sphere = r = 7\ cm

We know, Volume of a sphere = \frac{4}{3}\pi r^3

The required volume of the sphere = \frac{4}{3}\times\frac{22}{7}\times (7)^3

\\ = \frac{4}{3}\times22\times 7\times 7

\\ = \frac{4312}{3}

\\ = 1437\frac{1}{3}\ cm^3

Q1 (ii) Find the volume of a sphere whose radius is \small 0.63\hspace{1mm}m

Answer:

Given,

The radius of the sphere = r = 0.63\ m

We know, Volume of a sphere = \frac{4}{3}\pi r^3

The required volume of the sphere = \frac{4}{3}\times\frac{22}{7}\times (0.63)^3

\\ = 4\times22\times 0.03\times 0.63\times 0.63

\\ = 1.048\ m^3

\\ = 1.05\ m^3\ \ \ (approx.)

Q2 (i) Find the amount of water displaced by a solid spherical ball of diameter 28 cm

Answer:

The solid spherical ball will displace water equal to its volume. 

Given,

The radius of the sphere = r = \frac{28}{2}\ cm = 14\ cm

We know, Volume of a sphere = \frac{4}{3}\pi r^3

\therefore The required volume of the sphere = \frac{4}{3}\times\frac{22}{7}\times (14)^3

\\ = \frac{4}{3}\times22\times 2\times 14\times 14

\\ = \frac{34469}{3} \\ = 11489\frac{2}{3}\ cm^3

Therefore, the amount of water displaced will be 11489\frac{2}{3}\ cm^3

Q2 (ii) Find the amount of water displaced by a solid spherical ball of diameter \small 0.21\hspace{1mm}m

Answer:

The solid spherical ball will displace water equal to its volume. 

Given,

The radius of the sphere = r = \frac{0.21}{2}\ m

We know, Volume of a sphere = \frac{4}{3}\pi r^3

\therefore The required volume of the sphere = \frac{4}{3}\times\frac{22}{7}\times \left(\frac{0.21}{2} \right )^3

\\ = 4\times22\times \frac{0.01\times 0.21\times 0.21}{8}

\\ = 11\times 0.01\times 0.21\times 0.21

\\ =0.004851\ m^3

Therefore, amount of water displaced will be 0.004851\ m^3

Q3 The diameter of a metallic ball is \small 4.2\hspace{1mm}cm. What is the mass of the ball, if the density of the metal is \small 8.9\hspace{1mm}g per \small cm^3?

Answer:

Given,

The radius of the metallic sphere = r = \frac{4.2}{2}\ cm = 2.1\ cm

We know, Volume of a sphere = \frac{4}{3}\pi r^3

\therefore The required volume of the sphere = \frac{4}{3}\times\frac{22}{7}\times 2.1^3

\\ = 4\times22\times 0.1\times 2.1\times 2.1

\\ =38.808\ cm^3

Now, the density of the metal is \small 8.9\hspace{1mm}g per \small cm^3,which means,

Mass of \small 1\ cm^3 of the metallic sphere = \small 8.9\hspace{1mm}g

Mass of 38.808\ cm^3 of the metallic sphere = \small (8.9\times38.808)\ g

\small \approx 345.39\ g

Q4 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?

Answer:

Given,

Let d_e be the diameters of Earth

\therefore The diameter of the Moon = d_m = \frac{1}{4}d_e

We know, Volume of a sphere =

 \frac{4}{3}\pi r^3 =\frac{4}{3}\pi \left (\frac{d}{2} \right )^3 = \frac{1}{6}\pi d^3

\therefore The ratio of the volumes = \frac{Volume\ of\ the\ Earth}{Volume\ of\ the\ Moon}

\\ = \frac{\frac{1}{6}\pi d_e^3}{\frac{1}{6}\pi d_m^3} \\ = \frac{ d_e^3}{(\frac{d_e}{4})^3} \\ = 64: 1

Therefore, the required ratio of the volume of the moon to the  volume of the earth is 1: 64

Q5 How many litres of milk can a hemispherical bowl of diameter \small 10.5\hspace{1mm}cm hold?

Answer:

The radius of the hemispherical bowl = r = \frac{10.5}{2}\ cm

We know, Volume of a hemisphere = \frac{2}{3}\pi r^3

The volume of the given hemispherical bowl = \frac{2}{3}\times\frac{22}{7}\times \left (\frac{10.5}{2} \right )^3

= \frac{2}{3\times8}\times22\times1.5\times10.5\times10.5

= 303.1875\ cm^3

The capacity of the hemispherical bowl = = \frac{303.1875}{1000} \approx 0.303\ litres\ \ \ (approx.)

Q6 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.

Answer:

Given,

Inner radius of the hemispherical tank = r_1 = 1\ m

Thickness of the tank = 1\ cm = 0.01\ m

\therefore Outer radius = Internal radius + thickness = r_2 = (1+0.01)\ m = 1.01\ m

We know, Volume of a hemisphere = \frac{2}{3}\pi r^3

\therefore Volume of the iron used = Outer volume - Inner volume 

= \frac{2}{3}\pi r_2^3 - \frac{2}{3}\pi r_1^3

= \frac{2}{3}\times\frac{22}{7}\times (1.01^3 - 1^3)

= \frac{44}{21}\times0.030301

= 0.06348\ m^3\ \ (approx)

Q7 Find the volume of a sphere whose surface area is \small 154\hspace{1mm}cm^2.

Answer:

Given,

The surface area of the sphere = \small 154\hspace{1mm}cm^2

We know, Surface area of a sphere = 4\pi r^2

\therefore 4\pi r^2 = 154

\\ \Rightarrow 4\times\frac{22}{7}\times r^2 = 14\times11 \\ \Rightarrow r^2 = \frac{7\times7}{4} \\ \Rightarrow r = \frac{7}{2} \\ \Rightarrow r = 3.5\ cm

\therefore  The volume of the sphere = \frac{4}{3}\pi r^3

= \frac{4}{3}\times\frac{22}{7}\times (3.5)^3

= 179\frac{2}{3}\ cm^3

Q8 (i) A dome of a building is in the form of a hemisphere. From inside, it was white-washed at the cost of \small Rs\hspace{1mm}4989.60. If the cost of white-washing is Rs 20 per square metre, find the inside surface area of the dome

Answer:

Given,

\small Rs\hspace{1mm}20 is the cost of white-washing 1\ m^2 of the inside area

\small Rs\hspace{1mm}4989.60 is the cost of white-washing \frac{1}{20}\times4989.60\ m^2 = 249.48\ m^2 of inside area

(i) Therefore, the surface area of the inside of the dome is 249.48\ m^2

Q8 (ii) A dome of a building is in the form of a hemisphere. From inside, it was white-washed at the cost of \small Rs\hspace{1mm}4989.60. If the cost of white-washing is Rs 20 per square metre, find the volume of the air inside the dome.

Answer:

Let the radius of the hemisphere be r\ m

Inside the surface area of the dome = 249.48\ m^2

We know, Surface area of a hemisphere = 2\pi r^2

\\ \therefore 2\pi r^2 = 249.48 \\ \Rightarrow r^2 = \frac{249.48\times7}{2\times22} \\ \Rightarrow r = 6.3\ m

\therefore  The volume of the hemisphere = \frac{2}{3}\pi r^3

= \frac{2}{3}\times\frac{22}{7}\times (6.3)^3

= 523.908\ m^3

Q9 (i) Twenty seven solid iron spheres, each of radius r and surface area S are melted to form a sphere with surface area \small S'. Find the radius \small r' of the new sphere

Answer:

Given,

The radius of a small sphere = r

The radius of the bigger sphere = r'

\thereforeThe volume of each small sphere= \frac{4}{3}\pi r^3

And, Volume of the big sphere of radius r'\frac{4}{3}\pi r'^3

According to question,

27\times\frac{4}{3}\pi r^3=\frac{4}{3}\pi r'^3

\\ \Rightarrow r'^3 = 27\times r^3 \\ \Rightarrow r' = 3\times r

\therefore r' = 3r

Q9 (ii) Twenty seven solid iron spheres, each of radius r and surface area S are melted to form a sphere with surface area \small S'. Find the ratio of S and \small S'.

Answer:

Given,

The radius of a small sphere = r

The surface area of a small sphere = S

The radius of the bigger sphere = r'

The surface area of the bigger sphere = S'

And, r' = 3r

We know, the surface area of a sphere =4\pi r^2

\therefore The ratio of their surface areas = \frac{4\pi r'^2}{4\pi r^2}

\\ = \frac{ (3r)^2}{ r^2} \\ = 9

Therefore, the required ratio is 1:9

Q10 A capsule of medicine is in the shape of a sphere of diameter \small 3.5\hspace{1mm}mm. How much medicine (in \small mm^3) is needed to fill this capsule?

Answer:

Given,

The radius of the spherical capsule =r =\frac{3.5}{2}

\thereforeThe volume of the capsule = \frac{4}{3}\pi r^3

= \frac{4}{3}\times\frac{22}{7}\times(\frac{3.5}{2})^3

= \frac{4}{3}\times22\times\frac{0.5\times3.5\times3.5}{8}

= 22.458\ mm^3 \approx 22.46\ mm^3\ \ (approx)

Therefore, 22.46\ mm^3\ \ (approx) of medicine is needed to fill the capsule.

NCERT solutions for class 9 maths chapter 13 Surface Area and Volumes Excercise: 13.9

Q1 A wooden bookshelf has external dimensions as follows: Height \small =110\hspace{1mm}cm, Depth \small =25\hspace{1mm}cm, Breadth \small =85\hspace{1mm}cm (see Fig. \small 13.31). The thickness of the plank is \small 5\hspace{1mm}cm everywhere. The external faces are to be polished and the inner faces are to be painted. If the rate of polishing is \small 20\hspace{1mm} paise per \small cm^2 and the rate of painting is \small 10\hspace{1mm} paise per \small cm^2, find the total expenses required for polishing and painting the surface of the bookshelf.

            

Answer:

External dimension od the bookshelf = 85\ cm\times25\ cm\times110\ cm

(Note: There is no front face)

The external surface area of the shelf = lh + 2 (lb + bh)
\\ = [85 \times 110 + 2 (85 \times 25 + 25 \times 110)] \\ = (9350 + 9750) \\ = 19100\ cm^2

We know, each stripe on the front surface is also to be polished. which is 5 cm stretch.

Area of front face = [85 \times 110 - 75 \times 100 + 2 (75 \times5)]

\\ = 1850 + 750 \\ = 2600\ cm^2

Area to be polished = (19100 + 2600) = 21700\ cm^2

Cost of polishing 1\ cm^2 area = Rs\ 0.20

Cost of polishing 21700\ cm^2 area =  Rs.\ (21700 \times 0.20) = Rs.\ 4340

Now,

Dimension of inner part = 75\ cm\times15\ cm\times100\ cm

Area to be painted in 3 rows = 3\times[2 (l + h) b + lh]

\\ =3\times [2 (75 + 30) \times 20 + 75 \times 30] \\ = 3\times[(4200 + 2250)] \\ = 3\times6450 \\ = 19350\ cm^2

Cost of painting 1\ cm^2 area = Rs\ 0.10

Cost of painting 19350\ cm^2 area = Rs.\ (19350 \times 0.10)= Rs.\ 1935

Total expense required for polishing and painting = Rs.\ (4340 + 1935)

= Rs.\ 6275

Q2 The front compound wall of a house is decorated by wooden spheres of diameter 21 cm, placed on small supports as shown in Fig \small 13.32. Eight such spheres are used for this purpose, and are to be painted silver. Each support is a cylinder of radius \small 1.5\hspace{1mm}cm and height 7 cm and is to be painted black. Find the cost of paint required if silver paint costs 25 paise per \small cm^2 and black paint costs 5 paise per \small cm^2.
 

            

Answer:

Given,

The radius of the wooden spheres = r_1 = \frac{21}{2}\ cm

\therefore The surface area of a single sphere = 4\pi r_1^2

\\ = 4\times\frac{22}{7}\times\frac{21}{2}\times\frac{21}{2} \\ = 22\times3\times21

= 1386\ cm^2

Again, the Radius of the cylinder support = \small r_2 = 1.5\hspace{1mm}cm

Height of the support = h = 7\ cm

\therefore The base area of the cylinder = \pi r_2^2 = \frac{22}{7}\times1.5\times1.5 = 7.07\ cm^2

Now, Cost of painting 1\ cm^2silver = 25\ paise = Rs.\ 0.25

\therefore Cost of painting 1 wooden sphere = Cost of painting (1386-7.07)\ cm^2silver

Rs. (0.25\times1378.93) = Rs.\ 344.7325

Now, Curved surface area of the cylindrical support = 2\pi r_2 h

\\ = 2\times\frac{22}{7}\times1.5\times7 \\ = 22\times3 \\ = 66\ cm^2

Now, Cost of painting 1\ cm^2 black = 5\ paise = Rs.\ 0.05

\therefore Cost of painting 1 such stand = Cost of painting 66\ cm^2silver = Rs. (0.05\times66) = Rs.\ 3.3

\thereforeThe total cost of painting 1 sphere and its support = Rs.\ (344.7325+3.3) = Rs.\ 348.0325

Therefore, total cost of painting 8 such spheres and their supports = Rs.\ (8\times348.0325) = Rs.\ 2784.26

Q3 The diameter of a sphere is decreased by \small 25\%. By what per cent does its curved surface area decrease?

Answer:

Let the radius of the sphere be r

Diameter of the sphere = 2r

According to question,

Diameter is decreased by \small 25\%

So, the new diameter = \frac{3}{4}\times2r = \frac{3r}{2}

So, the new radius = r' = \frac{3r}{4}

\thereforeNew surface area = 4\pi r'^2 = 4\pi (\frac{3r}{4})^2

\thereforeDecrease in surface area = 4\pi r^2 - 4\pi (\frac{3r}{4})^2

= 4\pi r^2[1 -\frac{9}{16} ]

= 4\pi r^2[\frac{7}{16} ]

\therefore Percentage decrease in the surface area = \frac{Difference\ in\ areas}{Original\ surface\ area}

= \frac{4\pi r^2[\frac{7}{16}]}{4\pi r^2}\times100 \%

= \frac{7}{16}\times100 \%

= 43.75 \%

NCERT solutions for class 9 maths chapter wise

Chapter No.

Chapter Name

Chapter 1

NCERT solutions for class 9 maths chapter 1 Number Systems

Chapter 2

CBSE NCERT solutions for class 9 maths chapter 2 Polynomials

Chapter 3

Solutions of NCERT class 9 maths chapter 3 Coordinate Geometry

Chapter 4

NCERT solutions for class 9 maths chapter 4 Linear Equations In Two Variables

Chapter 5

CBSE NCERT solutions for class 9 maths chapter 5 Introduction to Euclid's Geometry

Chapter 6

Solutions of NCERT class 9 maths chapter 6 Lines And Angles

Chapter 7

NCERT solutions for class 9 maths chapter 7 Triangles

Chapter 8

CBSE NCERT solutions for class 9 maths chapter 8 Quadrilaterals

Chapter 9

Solutions of NCERT class 9 maths chapter 9 Areas of Parallelograms and Triangles

Chapter 10

NCERT solutions for class 9 maths chapter 10 Circles

Chapter 11

CBSE NCERT solutions for class 9 maths chapter 11 Constructions

Chapter 12

Solutions of NCERT class 9 maths chapter 12 Heron’s Formula

Chapter 13

NCERT solutions for class 9 maths chapter 13 Surface Area and Volumes

Chapter 14

CBSE NCERT solutions for class 9 maths chapter 14 Statistics

Chapter 15

Solutions of NCERT class 9 maths chapter 15 Probability

NCERT solutions for class 9 subject wise

CBSE NCERT Solutions for Class 9 Maths

Solutions of NCERT Solutions for Class 9 Science

How to use NCERT solutions for class 9 maths chapter 13 Surface Area and Volumes?

  • Revise the formulas and concepts regarding the figures circle, rectangle, square, etc.

  • Learn about the other figures introduced in this chapter.

  • Memorize the formulas for every shape.

  • Learn the formula application by going through some examples.

  • Jump on to the practice exercises to get command over the topic.

  • You can use NCERT solutions for class 9 maths chapter 13 Surface Area and Volumes during the practice.

Keep working hard and happy learning!

 

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