NCERT SOLUTIONS FOR CLASS 9 MATHS CHAPTER 13 SURFACE AREAS AND VOLUMES

The National Council of Educational Research and Training (NCERT) is an autonomous body of the Indian government that formulates the curricula for schools in India that are governed by the Central Board of Secondary Education (CBSE) and certain state boards. Therefore, students who will be taking the Class 10 tests administered by various boards should consult this NCERT Syllabus in order to prepare for those examinations, which in turn will assist those students get a passing score.

When working through the exercises in the NCERT textbook, if you run into any type of difficulty or uncertainty, you may use the swc NCERT Solutions for class 9 as a point of reference. While you are reading the theory form textbook, it is imperative that you always have notes prepared. You should make an effort to understand things from the very beginning so that you may create a solid foundation in the topic. Use the NCERT as your parent book to ensure that you have a strong foundation. After you have finished reading the theoretical section of the textbook, you should go to additional reference books.

NCERT Solutions for Class 9 Maths Exercise 13.1

Question 1. A plastic box 1.5 m long, 1.25 m wide and 65 cm deep is to be made. It is to be open at the 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 cost Rs.20.

Solution:
(i) Given: Length Chapter 13-Surface Areas And Volumes/image001.png= 1.5 m, Breadth Surface Areas And Volumes/image002.png= 1.25 m and Depth Surface Areas And Volumes/image003.png

= 65 cm = 0.65 m

Area of the sheet required for making the box open at the top = Chapter 13-Surface Areas And Volumes/image004.png

Surface Areas And Volumes/image005.png

Chapter 13-Surface Areas And Volumes/image006.png

Surface Areas And Volumes/image007.png

= 3.575 + 1.875 = 5.45 m2

(ii) Since, Cost of 1 m2 sheet = Rs. 20

Surface Areas And Volumes/image008.pngCost of 5.45 m2 sheet = 20 x 5.45 = Rs. 109

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

Solution:
Given: Length Chapter 13-Surface Areas And Volumes/image001.png= 5 m, Breadth Surface Areas And Volumes/image002.png= 4 m and Height Surface Areas And Volumes/image003.png= 3 m

Surface Areas And Volumes/image009.pngArea of the four walls = Lateral surface area = Chapter 13-Surface Areas And Volumes/image010.pngChapter 13-Surface Areas And Volumes/image011.png

= 2 x 3 (4 + 5)

= 2 x 9 x 3 = 54 m2

Area of ceiling = = 5 x 4 = 20 m2

Surface Areas And Volumes/image008.pngTotal area of walls and ceiling of the room = 54 + 20 = 74 m2

Now Cost of white washing for 1 m2 = Rs. 7.50

Surface Areas And Volumes/image008.pngCost of white washing for 74 m= 74 x 7.50 = Rs. 555

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

Solution:
Given: Perimeter of rectangular wall

Surface Areas And Volumes/image013.png= 250 m ……….(i)

Now Area of the four walls of the room

Chapter 13-Surface Areas And Volumes/image014.png

Surface Areas And Volumes/image015.png= 1500 m2 ……….(ii)

Surface Areas And Volumes/image009.pngArea of the four walls = Lateral surface area

Chapter 13-Surface Areas And Volumes/image010.pngChapter 13-Surface Areas And Volumes/image011.pngChapter 13-Surface Areas And Volumes/image011.png= 1500

Surface Areas And Volumes/image016.pngChapter 13-Surface Areas And Volumes/image017.png[using eq. (i) and (ii)

Surface Areas And Volumes/image016.pngSurface Areas And Volumes/image018.png= 6 m

Hence required height of the hall is 6 m.

Question 4. The paint in a certain container is sufficient to paint an area equal to 9.375 m2. How many bricks of dimensions 22.5 cm x 10 cm x 7.5 cm can be painted out of this container?

Solution:
Given: Length of the brick Chapter 13-Surface Areas And Volumes/image001.png

= 22.5 cm, Breadth Surface Areas And Volumes/image002.png

= 10 cm and Height Surface Areas And Volumes/image003.png= 7.5 m

Surface Areas And Volumes/image008.pngSurface area of the brick

Chapter 13-Surface Areas And Volumes/image019.png

= 2 (22.5 x 10 + 10 x 7.5 + 7.5 x 22.5)

= 2 (225 + 75 + 468.75)

= 937.5 cm2

= 0.09375 m2 [1 cm = 0.01 m]

Now No. of bricks to be painted

Surface Areas And Volumes/image020.pngSurface Areas And Volumes/image021.png

= 100

Hence 100 bricks can be painted.

Question 5. A cubical box has each edge 10 cm and a cuboidal box is 10 cm wide, 12.5 cm long 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 how much?

Solution:
(i) Lateral surface area of a cube = 4 (side)2 = 4 x (10)2 = 400 cm2

Lateral surface area of a cuboid = Chapter 13-Surface Areas And Volumes/image022.png

= 2 x 8 (12.5 + 10) = 16 x 22.5 = 360 cm2

Surface Areas And Volumes/image008.pngLateral surface area of cubical box is greater by (400 – 360) = 40 cm2

(ii) Total surface area of a cube = 6 (side)2

= 6 x (10)2 = 600 cm2

Total surface area of cuboid = Chapter 13-Surface Areas And Volumes/image019.png

= 2 (12.5 x 10 + 10 x 8 + 8 x 12.5)

= 2 (125 + 80 + 100)

= 2 x 305 = 610 cm2

Surface Areas And Volumes/image008.pngTotal surface area of cuboid box is greater by (610 – 600) = 10 cm2

Question 6. A small indoor green house (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.

(i) What is the surface area of the glass?

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

Solution:
(i) Given: Length of glass herbarium Chapter 13-Surface Areas And Volumes/image001.png= 30 cm,

Breadth Surface Areas And Volumes/image002.png= 25 cm and Height Surface Areas And Volumes/image003.png= 25 m

Surface Areas And Volumes/image023.png

Total surface area of the glass

Chapter 13-Surface Areas And Volumes/image019.png

= 2 (30 x 25 + 25 x 25 + 25 x 30)

= 2 (750 + 625 + 750)

= 2 x 2125 = 4250 cm2

Hence 4250 cm2 of the glass is required to make a herbarium.

(ii) Tape is used at 12 edges.

Surface Areas And Volumes/image016.pngTape is used at 4 lengths, 4 breadths and 4 heights.

Surface Areas And Volumes/image016.pngTotal length of the tape = Chapter 13-Surface Areas And Volumes/image024.png

= 2 (30 + 25 + 25) = 320 cm

Hence 320 cm of the tape if needed to fix 12 edges of herbarium.

Question 7. 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 25 cm by 20 cm by 5 cm and the smaller of dimensions 15 cm by 12 cm by 5 cm. 5% of the total surface area is required extra, for all the overlaps. If the cost of the card board is Rs. 4 for 1000 cm2, find the cost of cardboard required for supplying 250 boxes of each kind.

Solution:
Given: Length of bigger cardboard box (L) = 25 cm

Breadth (B) = 20 cm and Height (H) = 5cm

Chapter 13-Surface Areas And Volumes/image025.png

Total surface area of bigger cardboard box

= 2 (LB + BH + HL)

= 2 (25 x 20 + 20 x 5 + 5 x 25)

= 2 (500 + 100 + 125)

= 1450 cm2

5% extra surface of total surface area is required for all the overlaps.

Surface Areas And Volumes/image016.png5% of 1450 = Chapter 13-Surface Areas And Volumes/image026.png= 72.5 cm2

Now, total surface area of bigger cardboard box with extra overlaps

= 1450 + 72.5 = 1522.5 cm2

Surface Areas And Volumes/image016.pngTotal surface area with extra overlaps of 250 such boxes

= 250 x 1522.5 = 380625 cm2

Since, Cost of the cardboard for 1000 cm2 = Rs. 4

Surface Areas And Volumes/image008.pngCost of the cardboard for 1cm2 = Rs. Surface Areas And Volumes/image027.png

Surface Areas And Volumes/image008.pngCost of the cardboard for 380625 cm2

= Rs. Chapter 13-Surface Areas And Volumes/image028.png= Rs. 1522.50

Now length of the smaller box Chapter 13-Surface Areas And Volumes/image001.png= 15 cm,

Chapter 13-Surface Areas And Volumes/image029.jpg

Breadth Surface Areas And Volumes/image002.png= 12 cm and Height Surface Areas And Volumes/image003.png= 5 cm

Total surface area of the smaller cardboard box

Chapter 13-Surface Areas And Volumes/image019.png

= 2 (15 x 12 + 12 x 5 + 5 x 15)

= 2 (180 + 60 +75) = 2 x 315 = 630 cm2

5% of extra surface of total surface area is required for all the overlaps.

Surface Areas And Volumes/image008.png5% of 630 = Surface Areas And Volumes/image030.png= 31.5 cm2

Total surface area with extra overlaps = 630 + 31.5 = 661.5 cm2

Now Total surface area with extra overlaps of 250 such smaller boxes

= 661.5 x 250 = 165375 cm2

Cost of the cardboard for 1000 cm2 = Rs. 4

Cost of the cardboard for 1cm2 = Rs. Surface Areas And Volumes/image027.png

Cost of the cardboard for 165375 cm2 = Rs. Surface Areas And Volumes/image031.png= Rs. 661.50

Surface Areas And Volumes/image008.pngTotal cost of the cardboard required for supplying 250 boxes of each kind

= Total cost of bigger boxes + Total cost of smaller boxes

= Rs. 1522.50 + Rs. 661.50

= Rs. 2184

Question 8. 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 simensions 4 m x 3 m?

Solution:
Given: Length of base Chapter 13-Surface Areas And Volumes/image001.png= 4 m, Breadth Surface Areas And Volumes/image002.png

= 3 m and Height Surface Areas And Volumes/image003.png= 2.5 m

Tarpaulin required to make shelter

= Surface area of 4 walls + Area of roof

Surface Areas And Volumes/image032.png

= 2 (4 + 3) 2.5 + 4 x 3

= 35 + 12 = 47 m2

Hence 47 mof the tarpaulin is required to make the shelter for the car.

NCERT Solutions for Class 9 Maths Exercise 13.2

Assume NCERT Solutions for Class 9 Maths Chapter 13-Surface Areas And Volumesimage001.pngunless stated otherwise

Question 1. The curved surface area of a right circular cylinder of height 14 cm is 88 cm2. Find the diameter of the base of the cylinder.

Solution:
Given: Height of cylinder NCERT Solutions for Class 9 Maths Chapter 13-Surface Areas And Volumesimage002.png= 14 cm, Curved Surface Area = 88 cm2

Let radius of base of right circular cylinder = Chapter 13-Surface Areas And Volumes/image003.pngcm

Chapter 13-Surface Areas And Volumes/image004.jpg

Chapter 13-Surface Areas And Volumes/image005.png= 88

NCERT Solutions for Class 9 Maths Chapter 13-Surface Areas And Volumesimage006.png
Chapter 13-Surface Areas And Volumes/image007.png
NCERT Solutions for Class 9 Maths Chapter 13-Surface Areas And Volumesimage006.png
NCERT Solutions for Class 9 Maths Chapter 13-Surface Areas And Volumesimage008.png

NCERT Solutions for Class 9 Maths Chapter 13-Surface Areas And Volumesimage006.pngChapter 13-Surface Areas And Volumes/image009.png1 cm

Diameter of the base of the cylinder = NCERT Solutions for Class 9 Maths Chapter 13-Surface Areas And Volumesimage010.png= 2 x 1 = 2 cm

Question 2. It is required to make a closed cylindrical tank of height 1 m and base diameter 140 cm from a metal sheet. How many square meters of the sheet are required for the same?

NCERT Solutions for Class 9 Maths Chapter 13-Surface Areas And Volumesimage011.png

Solution:
Given: Diameter = 140 cm

NCERT Solutions for Class 9 Maths Chapter 13-Surface Areas And Volumesimage006.pngRadius (r) = 70 cm = 0.7 m

Height of the cylinder (h) = 1 m

Total Surface Area of the cylinder

= 2πr(r + h)

NCERT Solutions for Class 9 Maths Chapter 13-Surface Areas And Volumesimage014.png

= 2 x 22 x 0.1 x 1.7 = 7.48 m2

Hence 7.48 m2 metal sheet is required to make the close cylindrical tank.

Question 3. A metal pipe is 77 cm long. The inner diameter of a cross section is 4 cm, the outer diameter being 4.4 cm. [See fig.]. Find its :

(i) Inner curved surface area

(ii) Outer curved surface area

(iii) Total surface area

NCERT Solutions for Class 9 Maths Chapter 13-Surface Areas And Volumesimage015.png

Solution:
(i) Length of the pipe = 77 cm, Inner diameter of cross-section = 4 cm

Inner radius of cross-section = 2 cm

Inner curved surface area of pipe = Chapter 13-Surface Areas And Volumes/image005.pngChapter 13-Surface Areas And Volumes/image017.png

= 2 x 22 x 2 x 11 = 968 cm2

(ii) Length of pipe = 77 cm, Outer diameter of pipe = 4.4 cm

NCERT Solutions for Class 9 Maths Chapter 13-Surface Areas And Volumesimage006.pngOuter radius of the pipe = 2.2 cm

Outer surface area of the pipe = Chapter 13-Surface Areas And Volumes/image005.png

Chapter 13-Surface Areas And Volumes/image017.png

= 44 x 2.2 x 11 = 1064.8 cm2

(iii) Now there are two circles of radii 2 cm and 2.2 cm at both the ends of the pipe.

NCERT Solutions for Class 9 Maths Chapter 13-Surface Areas And Volumesimage019.pngArea of two edges of the pipe = 2 (Area of outer circle – area of inner circle)

NCERT Solutions for Class 9 Maths Chapter 13-Surface Areas And Volumesimage020.png

= 5.28 cm2

Total surface area of pipe

= Inner curved surface area + Outer curved surface area + Area of two edges

= 968 + 1064.8 + 5.28 = 2038.08 cm2

Question 4. The diameter of a roller is 84 cm and its length is 120 cm. It takes 500 complete revolutions to move once over to level a playground. Find the area of the playground in m2.

Solution:
Diameter of roller = 84 cm

NCERT Solutions for Class 9 Maths Chapter 13-Surface Areas And Volumesimage025.jpg
Radius of the roller = 42 cm

Length (Height) of the roller = 120 cm

Curved surface area of the roller = Chapter 13-Surface Areas And Volumes/image005.png

Chapter 13-Surface Areas And Volumes/image026.png

= 3.1680 m2

Now area leveled by roller in one revolution = 3.1680 m2

Area leveled by roller in 500 revolutions

= 3.1680 x 500 = 1584.0000 = 1584 m2

Question 5. A cylindrical pillar is 50 cm in diameter and 3.5 m in height. Find the cost of white washing the curved surface of the pillar at the rate of Rs. 12.50 per m2.

Solution:
Diameter of pillar = 50 cm

Radius of pillar = 25 cm = Chapter 13-Surface Areas And Volumes/image028.png

Height of the pillar = 3.5 m

Now, Curved surface area of the pillar

Chapter 13-Surface Areas And Volumes/image005.png

NCERT Solutions for Class 9 Maths Chapter 13-Surface Areas And Volumesimage029.pngNCERT Solutions for Class 9 Maths Chapter 13-Surface Areas And Volumesimage030.pngm2

Cost of white washing 1 m2 = Rs. 12.50

Cost of white washing NCERT Solutions for Class 9 Maths Chapter 13-Surface Areas And Volumesimage030.pngm2

= 12.50 x NCERT Solutions for Class 9 Maths Chapter 13-Surface Areas And Volumesimage030.png= Rs. 68.75

Question 6. Curved surface area of a right circular cylinder is 4.4 m2. If the radius of the base of the cylinder is 0.7 m, find its height.

Solution:
Curved surface area of the cylinder

= 4.4 m2, Radius of cylinder = 0.7 m

Let height of the cylinder = Chapter 13-Surface Areas And Volumes/image031.png

Chapter 13-Surface Areas And Volumes/image032.png

Question 7. The inner diameter of a circular well is 3.5 m. It is 10 m deep. Find :

(i) its inner curved surface area.

(ii) thecosr of plastering this curved surface at the rate of Rs. 40 per m2.

NCERT Solutions for Class 9 Maths Chapter 13-Surface Areas And Volumesimage034.jpg

Solution:
Inner diameter of circular well = 3.5 m

Inner radius of circular well

= 3.5/2= 1.75 m

And Depth of the well = 10 m

(i) Inner surface area of the well = Chapter 13-Surface Areas And Volumes/image005.png

NCERT Solutions for Class 9 Maths Chapter 13-Surface Areas And Volumesimage036.png

(ii) Cost of plastering 1 m2 = Rs. 40

Cost of plastering 100 m2 = 40 x 110 = Rs. 4400

Question 8. Find:

(i) the lateral or curved surface area of a 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?

Solution:

The length (height) of the cylindrical pipe = 28 m

Diameter = 5 cm

NCERT Solutions for Class 9 Maths Chapter 13-Surface Areas And Volumesimage006.pngRadius = 5/2 cm

NCERT Solutions for Class 9 Maths Chapter 13-Surface Areas And Volumesimage019.pngCurved surface area of the pipe = Chapter 13-Surface Areas And Volumes/image005.png

NCERT Solutions for Class 9 Maths Chapter 13-Surface Areas And Volumesimage039.png= 44000 cm2 = NCERT Solutions for Class 9 Maths Chapter 13-Surface Areas And Volumesimage040.png= 4.4 m2

Solution:
(i) Diameter of cylindrical petrol tank = 4.2 m

NCERT Solutions for Class 9 Maths Chapter 13-Surface Areas And Volumesimage019.pngRadius of the cylindrical petrol tank

= 4.2/2= 2.1 m

And Height of the tank = 4.5 m

NCERT Solutions for Class 9 Maths Chapter 13-Surface Areas And Volumesimage019.pngCurved surface area of the cylindrical tank = Chapter 13-Surface Areas And Volumes/image005.png

Chapter 13-Surface Areas And Volumes/image042.png= 59.4 m2

(ii) Let the actual area of steel used be NCERT Solutions for Class 9 Maths Chapter 13-Surface Areas And Volumesimage043.pngmeters

Since 1/12 of the actual steel used was wasted, the area of steel which has gone into the tank.

NCERT Solutions for Class 9 Maths Chapter 13-Surface Areas And Volumesimage045.png

NCERT Solutions for Class 9 Maths Chapter 13-Surface Areas And Volumesimage045.png= 59.4

NCERT Solutions for Class 9 Maths Chapter 13-Surface Areas And Volumesimage046.png= 64.8 m2

Hence steel actually used is 64.8 m2.

Question 9. In the adjoining figure, you see the frame of a lampshade. It is to be covered with a decorative 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. [See fig.]

NCERT Solutions for Class 9 Maths Chapter 13-Surface Areas And Volumesimage047.jpg

Solution:
Height of each of the folding at the top and bottom (Chapter 13-Surface Areas And Volumes/image031.png)= 2.5 cm

Height of the frame (H) = 30 cm

Diameter = 20 cm

NCERT Solutions for Class 9 Maths Chapter 13-Surface Areas And Volumesimage006.pngRadius = 10 cm

Now cloth required for covering the lampshade

= CSA of top part + CSA of middle part + CSA of bottom part

Chapter 13-Surface Areas And Volumes/image048.png

= 2200 cm2

Question 10. The students of a Vidyalayawere asked to participate in a competition for making and decorating penholders in the shape of a cylinder with a base, using cardboard. Each penholder was to be of radius 3 cm and height 10.5 cm. The Vidyalaya was to supply the competitors with cardboard. If there were 35 competitors, how much cardboard was required to be bought for the competition?

Solution:
Radius of a cylindrical pen holder (r)= 3 cm

Height of the cylindrical pen holder (Chapter 13-Surface Areas And Volumes/image031.png)

= 10.5 cm

Cardboard required for pen holder = CSA of pen holder + Area of circular base

Chapter 13-Surface Areas And Volumes/image053.png

Chapter 13-Surface Areas And Volumes/image052.png= 226.28 cm2

Since Cardboard required for making 1 pen holder = 226.28 cm2

NCERT Solutions for Class 9 Maths Chapter 13-Surface Areas And Volumesimage019.pngCardboard required for making 35 pen holders = 226.28 x 35 = 7919.8 cm2

= 7920 cm(approx.)

NCERT Solutions for Class 9 Maths Exercise 13.3

Assume NCERT Solutions for Class 9 Maths Chapter 13-Surface Areas And Volumesimage001.pngunless ^stated otherwise.

Question 1. Diameter of the base of a cone is 10.5 cm and its slant height is 10 cm. Find its curved surface area and its total surface area.

Solution:
Diameter = 10.5 cm

Chapter 13-Surface Areas And Volumes/image002.jpg

NCERT Solutions for Class 9 Maths Chapter 13-Surface Areas And Volumes/image003.pngRadius (r)= 10.5/2= 21/4cm

Slant height of cone (l)= 10 cm

Curved surface area of cone= πr(l + r)

Chapter 13-Surface Areas And Volumes/image009.png= 165 cm2

Total surface area of cone = πr(l + r)

NCERT Solutions for Class 9 Maths Chapter 13-Surface Areas And Volumes/image011.png

Chapter 13-Surface Areas And Volumes/image012.png= 251.625 cm2

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

Solution:
Slant height of cone (l)= 21 m

Diameter of cone = 24 m

NCERT Solutions for Class 9 Maths Chapter 13-Surface Areas And Volumes/image013.jpg

NCERT Solutions for Class 9 Maths Chapter 13-Surface Areas And Volumes/image003.pngRadius of cone (r)= 24/2= 12 m

Total surface area of cone = πr(l + r)

Chapter 13-Surface Areas And Volumes/image015.png

Chapter 13-Surface Areas And Volumes/image016.png= 1244.57 m2

Question 3. Curved surface area of a cone is 308 cm2 and its slant height is 14 cm. Find (i) radius of the base and  (ii) total surface area of the cone.

Solution:
(i) Slant height of cone (l)= 14 cm

Curved surface area of cone = 308 cm2

NCERT Solutions for Class 9 Maths Chapter 13-Surface Areas And Volumes/image003.pngπrl= 308

Chapter 13-Surface Areas And Volumes/image017.png

(ii) Total surface area of the cone

= Curved surface area + Area of circular base

Chapter 13-Surface Areas And Volumes/image020.png

= 462 cm2

Question 4. 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 a m2 canvas is Rs. 70.

Solution:
Height of the conical tent(Chapter 13-Surface Areas And Volumes/image031.png)= 10 m

Radius of the conical tent (r)= 24 m

(i) Slant height of the tent Chapter 13-Surface Areas And Volumes/image023.png

NCERT Solutions for Class 9 Maths Chapter 13-Surface Areas And Volumes/image024.png

(ii) Canvas required to make the tent

= Curved surface area of the tent= πrl

Chapter 13-Surface Areas And Volumes/image027.pngNCERT Solutions for Class 9 Maths Chapter 13-Surface Areas And Volumes/image028.pngm2

Cost of 1 mcanvas = Rs. 70

Cost of NCERT Solutions for Class 9 Maths Chapter 13-Surface Areas And Volumes/image028.pngm2 canvas

= 70 x NCERT Solutions for Class 9 Maths Chapter 13-Surface Areas And Volumes/image028.png= Rs. 137280

Question 5. 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)

Solution:
Height of the conical tent (Chapter 13-Surface Areas And Volumes/image031.png)= 8 m and Radius of the conical tent (r)= 6 m

Slant height of the tent Chapter 13-Surface Areas And Volumes/image023.png

Surface area

Area of tarpaulin = Curved surface area of tent = πrl= 3.14 x 6 x 10 = 188.4 m2

Width of tarpaulin = 3 m

Let Length of tarpaulin = L

Area of tarpaulin = Length x Breadth

= L x 3 = 3L

Now According to question,

3L = 188.4

NCERT Solutions for Class 9 Maths Chapter 13-Surface Areas And Volumes/image003.pngL = Chapter 13-Surface Areas And Volumes/image035.png= 62.8 m

The extra length of the material required for stitching margins and cutting is 20 cm = 0.2 m.

So the total length of tarpaulin bought is (62.8 + 0.2) m = 63 m

Question 6. The slant height and base diameter of a conical tomb are 25 m and 14 m respectively. Find the cost of whitewashing its curved surface at the rate of Rs. 210 per 100 m2.

Solution:
Slant height of conical tomb (l)

= 25 m, Diameter of tomb = 14 m

Radius of the tomb (r) = 14/2 = 7 m

Curved surface are of tomb = πrl

Chapter 13-Surface Areas And Volumes/image037.png= 550 m2

Cost of white washing 100 m2

= Rs. 210

So, Cost of white washing 1 mNCERT Solutions for Class 9 Maths Chapter 13-Surface Areas And Volumes/image038.png

So, Cost of white washing 550 m2

Chapter 13-Surface Areas And Volumes/image039.png= Rs. 1155

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

Solution:
Radius of cap (r)= 7 cm,

Height of cap (Chapter 13-Surface Areas And Volumes/image031.png) = 24 cm

Slant height of the cone Chapter 13-Surface Areas And Volumes/image023.png

Chapter 13-Surface Areas And Volumes/image041.png

= 25 cm

Area of sheet required to make a cap

= CSA of cone = πrl

Chapter 13-Surface Areas And Volumes/image037.png= 550 cm2

Area of sheet required to make 10 caps = 10 x 550 = 5500 cm2

Question 8. 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 m2, what will be the cost of painting all these cones? 

Solution:
Diameter of cone = 40 cm

Radius of cone (r)=40/2 = 20 cm

= 20/100 m = 0.2 m

Chapter 13-Surface Areas And Volumes/image047.jpg

Height of cone (Chapter 13-Surface Areas And Volumes/image031.png)= 1 m

Slant height of cone Chapter 13-Surface Areas And Volumes/image023.png

Chapter 13-Surface Areas And Volumes/image048.png m

Curved surface area of cone = Chapter 13-Surface Areas And Volumes/image049.png

= 3.14 x 0.2 x Chapter 13-Surface Areas And Volumes/image049.png

= 0.64056 m2

Cost of painting 1 m2 of a cone

= Rs. 12

Cost of painting 0.64056 m2 of a cone

= 12 x 0.64056= Rs. 7.68672

Cost of painting of 50 such cones

= 50 x 7.68672 = Rs. 384.34 (approx.)

NCERT Solutions for Class 9 Maths Exercise 13.4

Assume NCERT Solutions for Class 9 Maths Chapter 13-Surface Areas And Volumesimage001.pngunless stated otherwise.

Question 1. Find the surface area of a sphere of radius:

(i) 10.5 cm

(ii) 5.6 cm

(iii) 14 cm

Solution:
(i) Radius of sphere = 105 cm

Surface area of sphere = 4πr2

Chapter 13-Surface Areas And Volumes/image003.png= 1386 cm2

(ii) Radius of sphere = 5.6 m

Surface area of sphere = 4πr2

Chapter 13-Surface Areas And Volumes/image004.png= 3.94.84 m2

(iii) Radius of sphere = 14 cm

Surface area of sphere = 4πr2

Chapter 13-Surface Areas And Volumes/image005.png= 2464 cm2

Question 2. Find the surface area of a sphere of diameter:

(i) 14 cm

(ii) 21 cm

(iii) 3.5 cm

Solution:
(i) Diameter of sphere = 14 cm,

Therefore Radius of sphere = 14/2 = 7 cm

Surface area of sphere = 4πr2

Chapter 13-Surface Areas And Volumes/image007.png= 616 cm2

(ii) Diameter of sphere = 21 cm

Radius of sphere = 21/2 cm

Surface area of sphere = 4πr2

Chapter 13-Surface Areas And Volumes/image010.png= 1386 cm2

(iii) Diameter of sphere = 3.5 cm

Radius of sphere = 3.5/2 = 1.75 cm

Surface area of sphere = 4πr2

NCERT Solutions for Class 9 Maths Chapter 13-Surface Areas And Volumes/image012.png= 38.5 cm2

Question 3. Find the total surface area of a hemisphere of radius 10 cm.

(Use π =3.14)

Solution:
Radius of hemisphere (r) = 10 cm

Total surface area of hemisphere = 3πr2

= 3 x 3.14 x 10 x 10 = 942 cm2

Hence total surface area of hemisphere is

942 cm2

Question 4. The radius of a spherical balloonincreases from 7 cm to 14 cm as air isbeing pumped into it. Find the ratio ofsurface areas of the balloon in the two cases.

Solution:
I case: Radius of balloon (r) = 7 cm

Surface area of balloon = 4πr2

=4π x 7 x 7 cm2……….(i)

NCERT Solutions for Class 9 Maths Chapter 13-Surface Areas And Volumes/image018.jpg
II case: Radius of balloon (R) = 14 cm

Surface area of balloon = 4πr2

= 4π x 14 x 14 cm2 …….(ii)

Now, Ratio [from eq. (i) and (ii)],

NCERT Solutions for Class 9 Maths Chapter 13-Surface Areas And Volumes/image021.png

Hence, required ratio = 1 : 4

Question 5. A hemispherical bowl made of brass has inner diameter 105 cm. Find the cost of tin-plating it on the inside at the rate of Rs. 16 per 100 cm2.

Chapter 13-Surface Areas And Volumes/image024.jpg

Solution:
Inner diameter of bowl

= 10.5 cm

Inner radius of bowl (r) = 10.5/2

= 5.25 cm

Now, Inner surface area of bowl

= 2πr2

NCERT Solutions for Class 9 Maths Chapter 13-Surface Areas And Volumes/image027.png

NCERT Solutions for Class 9 Maths Chapter 13-Surface Areas And Volumes/image030.pngCost of tin-plating per 100 cm2

= Rs. 16

NCERT Solutions for Class 9 Maths Chapter 13-Surface Areas And Volumes/image008.pngCost of tin-plating per 1 cm2 = Chapter 13-Surface Areas And Volumes/image031.png

NCERT Solutions for Class 9 Maths Chapter 13-Surface Areas And Volumes/image008.pngCost of tin-plating per Chapter 13-Surface Areas And Volumes/image029.pngcm2

Chapter 13-Surface Areas And Volumes/image032.png= Rs. 27.72

Question 6. Find the radius of a sphere whose surface area is 154 cm2.

Solution:
Surface area of sphere = 154 cm2

NCERT Solutions for Class 9 Maths Chapter 13-Surface Areas And Volumes/image003.png4πr2= 154

Chapter 13-Surface Areas And Volumes/image035.png

Question 7. The diameter of the moon is approximately one fourth the diameter of the earth. Find the ratio of their surface areas.

Solution:
Let diameter of Earth = NCERT Solutions for Class 9 Maths Chapter 13-Surface Areas And Volumes/image038.png

NCERT Solutions for Class 9 Maths Chapter 13-Surface Areas And Volumes/image008.pngRadius of Earth (r) = NCERT Solutions for Class 9 Maths Chapter 13-Surface Areas And Volumes/image039.png

NCERT Solutions for Class 9 Maths Chapter 13-Surface Areas And Volumes/image008.pngSurface area of Earth = 4πr2

Chapter 13-Surface Areas And Volumes/image040.png

Chapter 13-Surface Areas And Volumes/image042.jpg

Now, Diameter of Moon = Chapter 13-Surface Areas And Volumes/image043.pngof diameter of Earth = Chapter 13-Surface Areas And Volumes/image044.png

NCERT Solutions for Class 9 Maths Chapter 13-Surface Areas And Volumes/image008.pngRadius of Moon (r) = NCERT Solutions for Class 9 Maths Chapter 13-Surface Areas And Volumes/image045.png

Surface area of Moon = 4πr2

Chapter 13-Surface Areas And Volumes/image046.png

Now, Ratio = Chapter 13-Surface Areas And Volumes/image048.png

NCERT Solutions for Class 9 Maths Chapter 13-Surface Areas And Volumes/image049.png

NCERT Solutions for Class 9 Maths Chapter 13-Surface Areas And Volumes/image008.pngRequired ratio = 1 : 16

Question 8. A hemispherical bowl is made of steel, 0.25 cm thick. The inner radius of the bowl is 5 cm. Find the outer curved surface area of the bowl.

Solution:
Inner radius of bowl (r) = 5 cm

Thickness of steel (t) = 0.25 cm

Outer radius of bowl (R) = r + t

= 5 +0.25 = 5.25 cm

Outer curved surface area of bowl

NCERT Solutions for Class 9 Maths Chapter 13-Surface Areas And Volumes/image054.png

= 173.25 cm2

Question 9. 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).

NCERT Solutions for Class 9 Maths Chapter 13-Surface Areas And Volumes/image056.jpg

Solution:
(i) Radius of sphere = r

Surface area of sphere

2π(radius)2 = 2πr2

The cylinder just encloses the sphere in it.

The height of cylinder will be equal to diameter of sphere.

And The radius of cylinder will be equal to radius of sphere.

(ii) Curved surface area of cylinder

= 2πrh2πr x πr surface area106

= 4πr2

NCERT Solutions for Class 9 Maths Chapter 13-Surface Areas And Volumes/image061.png

So, Required ratio = 1 :1

NCERT Solutions for Class 9 Maths Exercise 13.5

Question 1. A matchbox 4 cm x 2.5 cm x 1.5 cm. What will be the volume a packet containing 12 such boxes?

Chapter 13-Surface Areas And Volumes/image001.jpg
Solution:
Given: Length (l)= 4 cm, Breadth (b) = 2.5 cm, Height (h) = 1.5 cm

Volume of a matchbox = l x b x h

= 4 x 2.5 x 1.5

= 15 cm3

Volume of a packet containing 12

such matchboxes = 12 x 15 = 180 cm3

Question 2. A cubical 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)

Chapter 13-Surface Areas And Volumes/image008.jpg

Solution:
Volume of water in cuboidal tank = 6 m x 5 m x 4.5 m

= 135 m3= 135 x 1000 liters

= 135000 liters

Hence tank can hold 135000 liters of water.

Question 3. A cuboidal vessel is 10 m long and 8 m wide. How high must it be to hold 380 cubic meters of a liquid?

Solution:
Let height of cuboidal vessel = Chapter 13-Surface Areas And Volumes/image031.pngm

Volume of liquid in cuboidal vessel

= 380 m3

NCERT Solutions for Class 9 Maths Chapter 13-Surface Areas And Volumes/image005.png= 380 cm3

=> 10 m x 8 m x Chapter 13-Surface Areas And Volumes/image031.png= 380

=> Chapter 13-Surface Areas And Volumes/image031.pngNCERT Solutions for Class 9 Maths Chapter 13-Surface Areas And Volumes/image011.png= 4.75 m

Hence cuboidal vessel is 4.75 m high.

Question 4. 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.

Chapter 13-Surface Areas And Volumes/image012.jpg

Solution:
Volume of cuboidal pit = 8 m x 6 m x 3 m= 144 m3

Cost of digging 1 m3 cuboidal pit

= Rs. 30

Cost of digging 144 mcuboidal pit

= 30 x 144 = Rs. 4320

Question 5. 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.

(1 m3 = 1000 L)

Solution:
Capacity of cuboidal tank

= 50000 liters

NCERT Solutions for Class 9 Maths Chapter 13-Surface Areas And Volumes/image005.png= 50000 liters

NCERT Solutions for Class 9 Maths Chapter 13-Surface Areas And Volumes/image010.png2.5 m x b x 10 m = Chapter 13-Surface Areas And Volumes/image015.pngm3

NCERT Solutions for Class 9 Maths Chapter 13-Surface Areas And Volumes/image016.png

Hence breadth of cuboidal tank is 2 m.

Question 6. A village having a population of 4000 requires 150 litres of water per head per day. It has a tank measuring 20 m by 15 m by 6 m. For how many days will the water of this tank last?

Solution:
Capacity of cuboidal tank

 NCERT Solutions for Class 9 Maths Chapter 13-Surface Areas And Volumes/image005.png= 20 m x 15 m x 6 m

= 1800 m3 = 1800 x 1000 liters

NCERT Solutions for Class 9 Maths Chapter 13-Surface Areas And Volumes/image016.png

= 1800000 liters

Water required by her head per day

= 150 liters

Water required by 4000 persons per day = 150 x 4000 = 600000 liters

Number of days the water will last

Chapter 13-Surface Areas And Volumes/image018.png

NCERT Solutions for Class 9 Maths Chapter 13-Surface Areas And Volumes/image019.png= 3

Hence water of the given tank will last for 3 days.

Question 7. A godown measures 40 m x 25 m x 15 m. Find the maximum number of wooden crates each measuring 1.5 m x 1.25 m x 0.5 m that can be stored in the godown.

Solution:
Capacity of cuboidal godown

= 40 m x 25 m x 15 m = 15000 m3

Capacity of wooden crate = 1.5 m x 1.25 m x 0.5 m = 0.9375 m3

Maximum number of crates that can be stored in the godown

Chapter 13-Surface Areas And Volumes/image020.png

NCERT Solutions for Class 9 Maths Chapter 13-Surface Areas And Volumes/image021.png= 16000

Hence maximum 16000 crates can be stored in the godown.

Question 8. Find the minimum number of bricks each measuring 22.5 cm x 11.5 cm x 7.5 cm required to construct a wall 10 m long, 6 m high and 1.5 m thick.

NCERT Solutions for Class 9 Maths Chapter 13-Surface Areas And Volumes/image022.jpg

Solution:
Volume of one cuboidal brick

= l x b x h

= 22.5 cm x 11. 5 cm x 7.5 cm3

= 1940.625 cm3

= 0.001940625 m3

Volume of cuboidal wal

= 10 m x 6 m x 1.5 m= 90 m3

Minimum number of bricks required

Chapter 13-Surface Areas And Volumes/image023.png

= 46377 [Since bricks cannot be in fraction]

Question 9. 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.

Solution:
Volume of solid cube = (side)3

= (12)= 1728 cm3

According to question, Volume of each new cube= 1/8 (Volume of original cube)

NCERT Solutions for Class 9 Maths Chapter 13-Surface Areas And Volumes/image030.png= 216 cm3

Side of new cube = NCERT Solutions for Class 9 Maths Chapter 13-Surface Areas And Volumes/image031.png= 6 cm

Now, Surface area of original solid cube = 6 (side)2

= 6 x 12 x 12 = 864 cm2

Now, Surface area of original solid cube = 6 (side)2= 6 x 6 x 6 = 216 cm2

Now according to the question,

Chapter 13-Surface Areas And Volumes/image032.png

Hence required ration between surface area of original cube to that of new cube = 4 : 1.

Question 10. A river 3 m deep and 40 m wide is flowing at the rate of 2 km per hour. How much water eill fall into the sea in a minute?

NCERT Solutions for Class 9 Maths Chapter 13-Surface Areas And Volumes/image035.jpg

Solution:
Since water flows at the rate of 2 km per hour, the water from 2 km of the river flows into the sea in one hour.

Therefore, the volume of water flowing into the sea in one hour

= Volume of a cuboid

NCERT Solutions for Class 9 Maths Chapter 13-Surface Areas And Volumes/image005.png= 2000 m x 40 m x 3 m

= 240000 m3 [1 km = 1000 m]

Now, Volume of water flowing into sea in 1 hour (in 60 minutes)

= 240000 m3

Volume of water flowing into sea in 1 minute = Chapter 13-Surface Areas And Volumes/image036.png= 4000 m3

Question 11. Find the length of a wooden plank of width 2.5 m, thickness 0.025 m and volume 0.25 m3.

Chapter 13-Surface Areas And Volumes/image037.jpg

Solution:
Given: Volume of wooden plank

= 0.25 m3

=> lx 2.5 x 0.025 = 0.25

Chapter 13-Surface Areas And Volumes/image038.png

l = 4 m

Hence required length of wooden plank is 4 m.

NCERT Solutions for Class 9 Maths Exercise 13.6

Assume unless stated otherwise.

Question 1. 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? (1 m3 = 1000 L)

Solution:
Height of vessel = h= 25 cm

Chapter 13-Surface Areas And Volumes/image004.jpg

Circumference of base of vessel

= 132 cm

=>2πr = 132

Chapter 13-Surface Areas And Volumes/image007.png

= 21 cm

Now, Volume of cylindrical vessel

Chapter 13-Surface Areas And Volumes/image009.pngChapter 13-Surface Areas And Volumes/image010.png= 34650 cm3

Chapter 13-Surface Areas And Volumes/image011.pngliters [1000 cm= 1 liter]

= 34.65 liters

Question 2. The inner diameter of a cylindrical wooden pipe is 24 cm and its out diameter is 28 cm. The length of the pipe is 35 cm. Find the mass of the pipe, if 1cm3 of wood has a mass of 0.5 g.

Solution:
Inner diameter of pipe = 28 cm

Inner radius of pipe (r)

= 24/2 = 12 cm

And Outer diameter of pipe = 28 cm

Outer radius of pipe (R)

= 28/2 = 14 m

Length of pipe (l)= 35 cm

Chapter 13-Surface Areas And Volumes/image017.jpg

Volume of wood = Volume of outer cylinder – Volume of inner cylinder

Chapter 13-Surface Areas And Volumes/image019.png

= 110 [ 196 – 144 ] = 110 x 52

= 5720 cm3

Chapter 13-Surface Areas And Volumes/image012.pngWeight of 1 cm3 of wood = 0.6 g

Weight of 5720 cm3 of wood

= 0.6 x 5720 = 3432 g = 3.432 kg

Question 3. 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 height of 15 cm (ii) a plastic cylinder with circular base of diameter 7 cm and height 10 cm. Which container has greater capacity and how much?

Solution:
I case: Length of tin NCERT Solutions for Class 9 Maths Chapter 13-Surface Areas And Volumes/image005.png= 5 cm, Width of tin (b)= 4 cm

and Height of tin (h) = 15 cm

Chapter 13-Surface Areas And Volumes/image023.jpg
Then, Capacity of tin = NCERT Solutions for Class 9 Maths Chapter 13-Surface Areas And Volumes/image005.png= 5 x 4 x 15 = 300 cm3

II case: Diameter of base of cylinder = 7 cm

Chapter 13-Surface Areas And Volumes/image025.png

Radius of base of cylinder (r) = 7/2 cm

Height of cylinder Chapter 13-Surface Areas And Volumes/image027.png= 10 cm

Capacity of cylinder = Chapter 13-Surface Areas And Volumes/image028.png

Chapter 13-Surface Areas And Volumes/image029.png= 385 cm3

From the cases I and II, we observed that

cylindrical container has greater capacity

by (385 – 300) = 85 cm3.

Question 4. If the lateral surface of a cylinder is 94.2 cm2 and its height is 5 cm, then (i) radius of its base (ii) volume of the cylinder.

Solution:
Height of the cylinder Chapter 13-Surface Areas And Volumes/image031.png= 5 cm

Lateral surface area of the cylinder

= 94.2 cm2

=> 2πrh = 94.2

=> 2 x 3.14 x r x 5 = 94.2

Chapter 13-Surface Areas And Volumes/image032.png= 3 cm

Volume of cylinder = πr2h

= 3.14 x 3 x 3 x 5 = 141.3 cm3

Question 5. 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 m2, find:

(i) inner curved surface area of the vessel.

(ii) radius of the base.

(iii) capacity of the vessel.

Solution:
v Total cost to paint inner curved surface area of the vessel = Rs. 2200

Rate = Rs. 20 per square meter

(i) Inner curved surface area of vessel = Chapter 13-Surface Areas And Volumes/image033.png

= 110 m2

Chapter 13-Surface Areas And Volumes/image035.png
(ii) Depth of the vessel (Chapter 13-Surface Areas And Volumes/image031.png)= 10 m

Now, Inner surface area of vessel

= 110 m2

Chapter 13-Surface Areas And Volumes/image030.png= 110

Chapter 13-Surface Areas And Volumes/image037.png

= 1.75 m

(iii) Since r = 1.75 m and

Chapter 13-Surface Areas And Volumes/image031.png= 10 m

NCERT Solutions for Class 9 Maths Chapter 13-Surface Areas And Volumes/image008.pngCapacity of vessel

= Volume of cylinder = πr2h

Chapter 13-Surface Areas And Volumes/image039.png= 96.25 m3

= 96.25 kl [1 m3 = 1 kl]

Question 6. The capacity of a closed cylindrical vessel of height 1 m is 15.4 litres. How many square meters of metal sheet would be needed to make it?

Solution:
Height of the vessel (Chapter 13-Surface Areas And Volumes/image031.png)= 1 m

Chapter 13-Surface Areas And Volumes/image040.jpg

Capacity of vessel = 15.4 liters

Chapter 13-Surface Areas And Volumes/image041.pngkilo liters

= 0.0154 m3 [1 m3 = 1 kl]

=> πr2h = 0.0154

Chapter 13-Surface Areas And Volumes/image042.png

=> r2 = 0.0007 x 7 = 0.0048

=> r = 0.07 m

Now, Area of metal sheet required = TSA of cylindrical vessel

Chapter 13-Surface Areas And Volumes/image045.png
Chapter 13-Surface Areas And Volumes/image046.png

= 0.4708 m2

Question 7. A bag of grain contains 2.8 m3 of grain. How many bags are needed to fill a drum of radius 4.2 m and height 5 m?

Solution:
Radius of drum (r) = 4.2 m and Height of drum (Chapter 13-Surface Areas And Volumes/image031.png)= 5 m

Chapter 13-Surface Areas And Volumes/image049.png

Volume of a drum = πr2h

Chapter 13-Surface Areas And Volumes/image050.png

= 22 x 0.6 x 4.2 x 5 = 277.2 m3

Now, Number of bags

Chapter 13-Surface Areas And Volumes/image051.png

Chapter 13-Surface Areas And Volumes/image052.png= 99

Hence 99 bags are needed to fill the drum.

Question 8. 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 diameter of graphite is 1 mm. If the length of the pencil is 14 cm, find the columns of the wood and that of the graphite.

Solution:
Diameter of graphite = 1 mm

Radius of drum = 0.5 mm

= 0.05 cm

Chapter 13-Surface Areas And Volumes/image053.jpg

Height of graphite (Chapter 13-Surface Areas And Volumes/image031.png)= 14 cm

Volume of graphite = πr2h

Chapter 13-Surface Areas And Volumes/image054.png= 0.11 cm3

Diameter of pencil = 7 mm

Radius of pencil (R) = 3.5 mm

= 0.35 cm

Volume of pencil= Chapter 13-Surface Areas And Volumes/image055.png

Chapter 13-Surface Areas And Volumes/image056.png= 5.39 cm3

Now, Volume of wood = Volume of pencil – Volume of graphite

= 5.39 – 0.11 = 5.28 cm3

Question 9. A patient in a hospital is given soup daily in a cylindrical bowl of diameter 7 cm. If the bowl is filled with soup to a height of 4 cm, how much soup the hospital has to prepare daily to serve 250 patients?

Solution:
Diameter of circular base of cylindrical bowl = 7 cm

Chapter 13-Surface Areas And Volumes/image057.jpg

Radius of circular base of cylindrical

bowl (r)= 7/2 cm

Height of the bowl (Chapter 13-Surface Areas And Volumes/image031.png)= 4 cm

Now, Volume of cylindrical bowl = πr2h

Chapter 13-Surface Areas And Volumes/image058.png= 22 x 7 = 154 cm3

Quantity of soup filled in a bowl

= 154 cm3

Therefore, total quantity of soup to be prepared by the hospital = 250 x 154

= 38500 cm3

Chapter 13-Surface Areas And Volumes/image059.pngliter [1 liter = 1000 cm3]

= 38.5 liters

NCERT Solutions for Class 9 Maths Exercise 13.7

Assume unless stated otherwise.

Question 1. Find the volume of the right circular cone with:

(i) Radius 6 cm, Height 7 cm

(ii) Radius 3.5 cm, Height 12 cm

Chapter 13-Surface Areas And Volumes/image002.jpg

Solution:
(i) 
Given: r = 6 cm, Chapter 13-Surface Areas And Volumes/image031.png= 7 cm

Volume of cone = Chapter 13-Surface Areas And Volumes/image005.png

Chapter 13-Surface Areas And Volumes/image006.png

= 264 cm3

(ii) Given: r= 3.5 cm, Chapter 13-Surface Areas And Volumes/image031.png= 12 cm

Volume of cone = Chapter 13-Surface Areas And Volumes/image005.png

Chapter 13-Surface Areas And Volumes/image008.png

= 154 cm3

Question 2. Find the capacity of a conical vessel with:

(i) Radius 7 cm, Slant height 25 cm

(ii) Height 12 cm, Slant height 13 cm

Chapter 13-Surface Areas And Volumes/image009.jpg

Solution:
(i) Given: r = 7 cm, l = 25 cm

Chapter 13-Surface Areas And Volumes/image011.png= 24 cm

Capacity of conical vessel = Chapter 13-Surface Areas And Volumes/image005.png

Chapter 13-Surface Areas And Volumes/image015.png= 1232 cm3

= 1.232 liters [1000 cm3= 1liter]

Chapter 13-Surface Areas And Volumes/image017.jpg

(ii) Given: Chapter 13-Surface Areas And Volumes/image031.png= 12 cm, l = 13 cm

Chapter 13-Surface Areas And Volumes/image018.png

= 5 cm

Capacity of conical vessel = Chapter 13-Surface Areas And Volumes/image005.png

Chapter 13-Surface Areas And Volumes/image022.pngChapter 13-Surface Areas And Volumes/image023.pngcm3

Chapter 13-Surface Areas And Volumes/image024.pngliters

[1000 cm3= 1liter]

liter

Question 3. The height of a cone is 15 cm. If its volume is 1570 cm3, find the radius of the base. (Use π = 3.14)

Solution:
Height of the cone Chapter 13-Surface Areas And Volumes/image031.png= 15 cm

Chapter 13-Surface Areas And Volumes/image028.jpg

Volume of cone = 1570 cm3

Chapter 13-Surface Areas And Volumes/image005.png= 1570

Chapter 13-Surface Areas And Volumes/image030.png

Chapter 13-Surface Areas And Volumes/image032.png= 10 cm

Hence required radius of the base is 10 cm.

Question 4. If the volume of a right circular cone of height 9 cm is 48π cm3, find the diameter of the base.

Solution:
Height of the cone (Chapter 13-Surface Areas And Volumes/image031.png)= 9 cm

Chapter 13-Surface Areas And Volumes/image034.jpg

Volume of cone = 48π cm3

Chapter 13-Surface Areas And Volumes/image035.png

Chapter 13-Surface Areas And Volumes/image029.pngr = 4 cm

NCERT Solutions for Class 9 Maths Chapter 13-Surface Areas And Volumes/image008.pngDiameter of base = 2r = 2 x 4 = 8 cm

Question 5. A conical pit of top diameter 3.5 m is 12 m deep. What is its capacity in kiloliters?

Solution:
Diameter of pit = 3.5 m

Radius of pit (r)= 3.5/2 = 1.75 m

Depth of pit Chapter 13-Surface Areas And Volumes/image031.png= 12 m

Capacity of pit = Chapter 13-Surface Areas And Volumes/image005.png

Chapter 13-Surface Areas And Volumes/image043.png

Chapter 13-Surface Areas And Volumes/image047.pngm3 = 35.8 m3

= 38.5 kl [1 m3 = 1 kl]

Question 6. The volume of a right circular cone is 9856 cm3. If the diameter of the base if 28 cm, find:

(i) Height of the cone

(ii) Slant height of the cone

(iii) Curved surface area of the cone.

Solution:
(i) 
Diameter of cone = 28 cm

Chapter 13-Surface Areas And Volumes/image048.jpg

Chapter 13-Surface Areas And Volumes/image039.pngRadius of cone = 14 cm

Volume of cone = 9856 cm3

Chapter 13-Surface Areas And Volumes/image029.pngChapter 13-Surface Areas And Volumes/image005.png= 9856

Chapter 13-Surface Areas And Volumes/image029.png
Chapter 13-Surface Areas And Volumes/image049.png

Chapter 13-Surface Areas And Volumes/image029.pngChapter 13-Surface Areas And Volumes/image050.png= 48 cm

(ii) Slant height of cone = Chapter 13-Surface Areas And Volumes/image052.png

Chapter 13-Surface Areas And Volumes/image053.png

= 50 cm

(iii) Curved surface area of cone = πrl = Chapter 13-Surface Areas And Volumes/image057.png= 2200 cm2

Question 7. 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. (Use π = 3.14 )

Solution:
When right angled triangle ABC is revolved about side 12 cm, then the solid formed is a cone.

In that cone, Height (Chapter 13-Surface Areas And Volumes/image031.png) = 12 cm

And radius (r) = 5 cm

Chapter 13-Surface Areas And Volumes/image059.jpg

Therefore, Volume of cone = Chapter 13-Surface Areas And Volumes/image005.png

Chapter 13-Surface Areas And Volumes/image060.png

= 100π cm3

Question 8. If the triangle ABC in 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 volume of the two solids obtained.

Solution:
When right angled triangle ABC is revolved about side 5 cm, then the solid formed is a cone.

In that cone, Height (Chapter 13-Surface Areas And Volumes/image031.png)= 5 cm

And radius ( r )= 12 cm

Chapter 13-Surface Areas And Volumes/image062.jpg

Therefore, Volume of cone = Chapter 13-Surface Areas And Volumes/image005.png

Chapter 13-Surface Areas And Volumes/image063.png

= 240π cm3

Now, Chapter 13-Surface Areas And Volumes/image065.png

Chapter 13-Surface Areas And Volumes/image066.png

Chapter 13-Surface Areas And Volumes/image039.pngRequired ratio = 5 : 12

Question 9. Find the volume of the largest right circular cone that can be fitted in a cube whose edge is 14 cm.

Solution:
Since, diameter of the largest right circular cone that can be fitted in a cube = Edge of cube

Chapter 13-Surface Areas And Volumes/image068.jpg

Chapter 13-Surface Areas And Volumes/image029.png2r = 14 cm

Chapter 13-Surface Areas And Volumes/image029.pngr = 7 cm

And also Height of the cone Chapter 13-Surface Areas And Volumes/image031.png= Edge of cube = 14 cm

Now, Volume of the largest cone

Chapter 13-Surface Areas And Volumes/image005.png

Chapter 13-Surface Areas And Volumes/image070.png

= 718.66 cm3

Question 10. 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.

Solution:
Radius (r) of heap

Chapter 13-Surface Areas And Volumes/m147226c1.gif

Height (h) of heap = 3 m

Volume of heap

Chapter 13-Surface Areas And Volumes/image005.png

https://cbse.meritnation.com/img/curr/1/9/7/110/3043/Chapter%2013_html_m7bce1e31.gif

Therefore, the volume of the heap of wheat is 86.625 m3.

Area of canvas required = CSA of cone

Chapter 13-Surface Areas And Volumes/6103ffa3.gif

Therefore, 99.825 m2canvas will be required to protect the heap from rain.

NCERT Solutions for Class 9 Maths Exercise 13.8

Assume unless stated otherwise.

Question 1. Find the volume of a sphere whose radius is (i) 7 cm and (ii) 0.63 cm.

Solution:
(i) Radius of sphere ( r )= 7 cm

Volume of sphere = NCERT Solutions for Class 9 Maths Chapter 13-Surface Areas And Volumes/image003.png

Chapter 13-Surface Areas And Volumes/image004.png

NCERT Solutions for Class 9 Maths Chapter 13-Surface Areas And Volumes/image006.pngcm3

(ii) Radius of sphere ( r )= 0.63 m

Volume of sphere = NCERT Solutions for Class 9 Maths Chapter 13-Surface Areas And Volumes/image003.png

Chapter 13-Surface Areas And Volumes/image007.png

= 1.047816 m3 = 1.05 m3 (approx.)

Question 2. Find the amount of water displaced by a solid spherical ball of diameter:

(i) 28 cm

(ii) 0.21 m

Solution:
(i) Diameter of spherical ball

= 28 cm

Radius of spherical ball ( r) = 28/2

= 14 cm

According to question, Volume of water replaced = Volume of spherical ball

NCERT Solutions for Class 9 Maths Chapter 13-Surface Areas And Volumes/image003.png

Chapter 13-Surface Areas And Volumes/image011.png

NCERT Solutions for Class 9 Maths Chapter 13-Surface Areas And Volumes/image013.pngcm3

(ii) Diameter of spherical ball = 0.21 m

Radius of spherical ball (r)= 0.21/2m

According to question,

Volume of water replaced = Volume of spherical ball

NCERT Solutions for Class 9 Maths Chapter 13-Surface Areas And Volumes/image003.png

Chapter 13-Surface Areas And Volumes/image015.png

= 0.004851 m3

Question 3. The diameter of a metallic ball is 4.2 cm. What is the mass of the ball, if the metal weighs 8.9 g per cm3?

Solution:
Diameter of metallic ball = 4.2 cm

Radius of metallic ball (r)= 4.2/2

= 2.1 cm

Volume of metallic ball = NCERT Solutions for Class 9 Maths Chapter 13-Surface Areas And Volumes/image003.png

NCERT Solutions for Class 9 Maths Chapter 13-Surface Areas And Volumes/image019.png

= 38.808 cm3

Density of metal = 8.9 g per cm3

Mass of 1 cm3 = 8.9 g

Mass of 38.808 cm3 = 8.9 x 38.808

= 345.3912 g = 345.39 g (approx).

Question 4. The diameter of the moon is approximately one-fourth the diameter of the earth. What fraction is the volume of the moon of the volume of the earth?

Solution:
Let diameter of earth be x

Radius of earth (r) = x/2

Now, Volume of earth = NCERT Solutions for Class 9 Maths Chapter 13-Surface Areas And Volumes/image003.png

[Earth is considered to be a sphere]

Chapter 13-Surface Areas And Volumes/image024.png ……..(i)

According to question,

Diameter of moon = NCERT Solutions for Class 9 Maths Chapter 13-Surface Areas And Volumes/image026.pngx Diameter of earth= Chapter 13-Surface Areas And Volumes/image027.png

Radius of moon (R) = NCERT Solutions for Class 9 Maths Chapter 13-Surface Areas And Volumes/image028.png

Now, Volume of Moon = Chapter 13-Surface Areas And Volumes/image029.png

[Moon is considered to be a sphere]

Chapter 13-Surface Areas And Volumes/image030.png

[From eq. (i)]

So, Volume of moon is Chapter 13-Surface Areas And Volumes/image034.pngthe volume of earth.

Question 5. How many litres of milk can a hemispherical bowl of diameter 10.5 hold?

Solution:
Diameter of hemispherical bowl

= 10.5 cm

Radius of hemispherical bowl ( r )

NCERT Solutions for Class 9 Maths Chapter 13-Surface Areas And Volumes/image035.png= 5.25 cm

Chapter 13-Surface Areas And Volumes/image036.jpg

Volume of milk in hemispherical bowl

NCERT Solutions for Class 9 Maths Chapter 13-Surface Areas And Volumes/image037.png

= 0.303187 liters

= 0.303 liters (approx.)

Question 6. 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.

Solution:
Inner radius of hemispherical tank ( r ) = 1 m = 100 cm

Chapter 13-Surface Areas And Volumes/image043.jpg

Thickness of sheet = 1 cm

Outer radius of hemispherical tank (R) = 100 + 1 = 101 cm

Volume of iron of hemisphere

Chapter 13-Surface Areas And Volumes/image044.png

= 63487.81 cm3

= 0.06348 m3

Question 7. Find the volume of a sphere whose surface area is 154 cm2.

Solution:
Surface area of sphere = 154 cm2

Chapter 13-Surface Areas And Volumes/image029.png 4πr2 = 154

NCERT Solutions for Class 9 Maths Chapter 13-Surface Areas And Volumes/image049.png

Now, Volume of sphere

NCERT Solutions for Class 9 Maths Chapter 13-Surface Areas And Volumes/image003.png

Question 8. A dome of a building is in the form of a hemisphere. From inside, it was white-washed at the cost of Rs. 498.96. If the cost of white-washing is at the rate of Rs. 2.00 per square meter, find:

(i) the inner surface area of the dome.

(ii) the volume of the air inside the dome.

Solution:
Cost of white washing from inside = Rs. 498.96

Rate of white washing = Rs. 2

Area white washed

Chapter 13-Surface Areas And Volumes/image057.png= 249.48 cm2

Inside surface area of the dome

= 249.48 cm2

Chapter 13-Surface Areas And Volumes/image029.png 2πr= 249.48

Chapter 13-Surface Areas And Volumes/image059.png

= 5.67 x 7

Chapter 13-Surface Areas And Volumes/image029.png r = 6.3

So, Volume of the dome

NCERT Solutions for Class 9 Maths Chapter 13-Surface Areas And Volumes/image037.png

= 523.9 cm3

Question 9. Twenty seven solid iron spheres, each of radius 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’.

Solution:
(i) Let radius of sphere be r and radius of new sphere be R.

27 x Volume of sphere = Volume of new sphere

NCERT Solutions for Class 9 Maths Chapter 13-Surface Areas And Volumes/image047.png27 x NCERT Solutions for Class 9 Maths Chapter 13-Surface Areas And Volumes/image003.pngChapter 13-Surface Areas And Volumes/image029.png

Chapter 13-Surface Areas And Volumes/image064.png

Question 10. A capsule of medicine is in the shape of a sphere of diameter 3.5 mm. How much medicine (in mm3) is needed to fill this capsule?

Solution:
Diameter of spherical capsule = 3.5 mm

NCERT Solutions for Class 9 Maths Chapter 13-Surface Areas And Volumes/image072.jpg

Radius of spherical capsule ( r )

Chapter 13-Surface Areas And Volumes/image073.png mm

Medicine needed to fill the capsule

= Volume of sphere = NCERT Solutions for Class 9 Maths Chapter 13-Surface Areas And Volumes/image003.png

Chapter 13-Surface Areas And Volumes/image076.png

= 22.46 mm3 (Approx.)

NCERT Solutions for Class 9 Maths Exercise 13.9

Question 1. A wooden bookshelf has external dimensions as follows: Height = 110 cm, Depth = 25 cm, Breadth = 85 cm [See fig.]. The thickness of the planks 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.

Chapter 13-Surface Areas And Volumes/image001.jpg

Solution:
External faces to be polished

= Area of six faces of cuboidal bookshelf – 3 (Area of open portion ABCD)

= 2 (110 x 25 + 25 x 85 + 85 x 110) – 3 (75 x 30)

[AB = 85 – 5 – 5 = 75 cm and AD

= [⅓ x 110 – 5 – 5 – 5 – 5 = 30 cm]   

= 2 (2750 + 2125 + 9350) – 3 x 2250

= 2 x 14225 – 6750

= 28450 – 6750

= 21700 cm2

Now, cost of painting outer faces of wodden bookshelf at the rate of 20 paise.

= Rs. 0.20 per cm2 = Rs. 0.20 x 21700 = Rs. 4340

Here, three equal five sides inner faces.

Therefore total surface area

= 3 [ 2 (30 + 75) 20 + 30 x 75] [Depth = 25 – 5 = 20 cm]

= 3 [ 2 x 105 x 20 + 2250] = 3

[ 4200 + 2250]

= 3 x 6450 = 19350 cm2

Now, cost of painting inner faces at the rate of 10 paise i.e. Rs. 0.10 per cm2.

= Rs. 0.10 x 19350 = Rs. 1935

Total expenses required = Rs. 4340 + Rs. 1935 = Rs. 6275

Question 2. 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 he cost of paint required if silver paint costs 25 paise per cm2 and black paint costs 5 paise per cm2.

NCERT Solutions for Class 9 Maths Chapter 13-Surface Areas And Volumes/image004.jpg

Solution:
Diameter of a wooden sphere

= 21 cm.

Radius of wooden sphere ( R )

= 21/2 cm

And Radius of the cylinder ( r )

= 1.5 cm

Surface area of silver painted part

= Surface area of sphere – Upper part of cylinder for support

Chapter 13-Surface Areas And Volumes/image009.png

= 1378.928 cm2

Surface area of such type of 8 spherical part = 8 x 1378.928

= 11031.424 cm2

Cost of silver paint over 1 cm2

= Rs. 0.25

Cost of silver paint over 11031.928 cm2 = 0.25 x 11031.928

= Rs. 2757.85

Now, curved surface area of a cylindrical support = 2πrh

Chapter 13-Surface Areas And Volumes/image016.png= 66 cm2

Curved surface area of 8 such cylindrical supports = 66 x 8 = 528 cm2

Cost of black paint over 1 cm2 of cylindrical support = Rs. 0.50

Cost of black paint over 528 cm2 of cylindrical support = 0.50 x 528

= Rs. 26.40

Total cost of paint required

= Rs. 2757.85 + Rs. 26.4

= Rs. 2784.25

Question 3. If diameter of a sphere is decreased by 25% then what percent does its curved surface area decrease?

Solution:
Diameter of original sphere

= D = 2R

R = D/2

Curved surface area of original sphere = NCERT Solutions for Class 9 Maths Chapter 13-Surface Areas And Volumes/image019.png

According to the question,

Decreased diameter = 25% of D

Chapter 13-Surface Areas And Volumes/image022.png

Diameter of new sphere = NCERT Solutions for Class 9 Maths Chapter 13-Surface Areas And Volumes/image024.png

Radius of new sphere = Chapter 13-Surface Areas And Volumes/image026.png

Now, curved surface area of new sphere =Chapter 13-Surface Areas And Volumes/image027.png

Change in curved surface area

Chapter 13-Surface Areas And Volumes/image021.png

Percent change in the curved surface area

NCERT Solutions for Class 9 Maths Chapter 13-Surface Areas And Volumes/image032.png

Question 4. Sameera wants to celebrate the fifth birthday of her daughter with a party. She bought thick paper to make the conical party caps. Each cap is to have a base diameter of 10 cm and height 12 cm. A sheet of the paper is 25 cm by 40 cm and approximately 82% of the sheet can be effectively used for making the caps after cutting. What is the minimum number of sheets of paper that Sameera would need to buy, if there are to be 15 children at the party?

Solution:
Diameter of base of conical cap

= 10 cm

NCERT Solutions for Class 9 Maths Chapter 13-Surface Areas And Volumes/image036.jpg

Radius of conical cap ( r )= 5 cm

Slant height of cone  NCERT Solutions for Class 9 Maths Chapter 13-Surface Areas And Volumes/image038.png

NCERT Solutions for Class 9 Maths Chapter 13-Surface Areas And Volumes/image039.png

= 13 cm

Curved surface area of a cap

= πrl = 3.14 x 5 x 13 = 204.1 cm2

Curved surface area of 15 caps

= 15 x 204.1 = 3061.5 cm2

Area of a sheet of paper used for making caps = 25 x 40 = 1000 cm2

82% of sheet is used after cutting

= 82% of 1000 cm2

= 820 cm2

Number of sheet = Chapter 13-Surface Areas And Volumes/image044.png= 3.73

Hence 4 sheets area needed .

Conclusions for NCERT SOLUTIONS FOR CLASS 9 MATHS CHAPTER 13 SURFACE AREAS AND VOLUMES

SWC academic staff has developed NCERT answers for this chapter of the ninth grade mathematics curriculum. We have solutions prepared for all of the exercises of this chapter. The answers, broken down into steps, to all of the questions included in the NCERT textbook’s chapter are provided here. Read this chapter on theory. Be certain that you have read the theory section of this chapter of the NCERT textbook and that you have learnt the formulas for the chapter that you are studying. 

swc google search e1651044504923

SSAT