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 1m^{2} cost Rs.20.

**Solution:**

(i) Given: Length = 1.5 m, Breadth = 1.25 m and Depth

= 65 cm = 0.65 m

Area of the sheet required for making the box open at the top =

=

=

=

= 3.575 + 1.875 = 5.45 m^{2}

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

Cost of 5.45 m^{2} 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 m ^{2}.**

**Solution:**

Given: Length = 5 m, Breadth = 4 m and Height = 3 m

Area of the four walls = Lateral surface area = =

= 2 x 3 (4 + 5)

= 2 x 9 x 3 = 54 m^{2}

Area of ceiling = = 5 x 4 = 20 m^{2}

Total area of walls and ceiling of the room = 54 + 20 = 74 m^{2}

Now Cost of white washing for 1 m^{2} = Rs. 7.50

Cost of white washing for 74 m^{2 }= 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 m ^{2} is Rs. 15000, find the height of the hall.**

**Solution:**

Given: Perimeter of rectangular wall

= = 250 m ……….(i)

Now Area of the four walls of the room

=

= = 1500 m^{2} ……….(ii)

Area of the four walls = Lateral surface area

= = = 1500

[using eq. (i) and (ii)

= 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 m ^{2}. 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

= 22.5 cm, Breadth

= 10 cm and Height = 7.5 m

Surface area of the brick

=

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

= 2 (225 + 75 + 468.75)

= 937.5 cm^{2}

= 0.09375 m2 [1 cm = 0.01 m]

Now No. of bricks to be painted

= =

= 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 cm^{2}

Lateral surface area of a cuboid =

= 2 x 8 (12.5 + 10) = 16 x 22.5 = 360 cm^{2}

Lateral surface area of cubical box is greater by (400 – 360) = 40 cm^{2}

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

= 6 x (10)^{2} = 600 cm^{2}

Total surface area of cuboid =

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

= 2 (125 + 80 + 100)

= 2 x 305 = 610 cm^{2}

Total surface area of cuboid box is greater by (610 – 600) = 10 cm^{2}

**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 = 30 cm,

Breadth = 25 cm and Height = 25 m

Total surface area of the glass

=

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

= 2 (750 + 625 + 750)

= 2 x 2125 = 4250 cm^{2}

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

(ii) Tape is used at 12 edges.

Tape is used at 4 lengths, 4 breadths and 4 heights.

Total length of the tape =

= 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 cm ^{2}, 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

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 cm^{2}

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

5% of 1450 = = 72.5 cm^{2}

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

= 1450 + 72.5 = 1522.5 cm^{2}

Total surface area with extra overlaps of 250 such boxes

= 250 x 1522.5 = 380625 cm^{2}

Since, Cost of the cardboard for 1000 cm^{2} = Rs. 4

Cost of the cardboard for 1cm^{2} = Rs.

Cost of the cardboard for 380625 cm^{2}

= Rs. = Rs. 1522.50

Now length of the smaller box = 15 cm,

Breadth = 12 cm and Height = 5 cm

Total surface area of the smaller cardboard box

=

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

= 2 (180 + 60 +75) = 2 x 315 = 630 cm^{2}

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

5% of 630 = = 31.5 cm2

Total surface area with extra overlaps = 630 + 31.5 = 661.5 cm^{2}

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

= 661.5 x 250 = 165375 cm^{2}

Cost of the cardboard for 1000 cm^{2} = Rs. 4

Cost of the cardboard for 1cm^{2} = Rs.

Cost of the cardboard for 165375 cm^{2} = Rs. = Rs. 661.50

Total 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 = 4 m, Breadth

= 3 m and Height = 2.5 m

Tarpaulin required to make shelter

= Surface area of 4 walls + Area of roof

=

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

= 35 + 12 = 47 m^{2}

Hence 47 m^{2 }of the tarpaulin is required to make the shelter for the car.

### NCERT Solutions for Class 9 Maths Exercise 13.2

Assume unless stated otherwise

**Question 1. The curved surface area of a right circular cylinder of height 14 cm is 88 cm ^{2}. Find the diameter of the base of the cylinder.**

**Solution:**

Given: Height of cylinder = 14 cm, Curved Surface Area = 88 cm^{2}

Let radius of base of right circular cylinder = cm

= 88

1 cm

Diameter of the base of the cylinder = = 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?**

**Solution:**

Given: Diameter = 140 cm

Radius (r) = 70 cm = 0.7 m

Height of the cylinder (h) = 1 m

Total Surface Area of the cylinder

= 2πr(r + h)

=

= 2 x 22 x 0.1 x 1.7 = 7.48 m^{2}

Hence 7.48 m^{2} 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

**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 = =

= 2 x 22 x 2 x 11 = 968 cm^{2}

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

Outer radius of the pipe = 2.2 cm

Outer surface area of the pipe =

=

= 44 x 2.2 x 11 = 1064.8 cm^{2}

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

Area of two edges of the pipe = 2 (Area of outer circle – area of inner circle)

= 5.28 cm^{2}

Total surface area of pipe

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

= 968 + 1064.8 + 5.28 = 2038.08 cm^{2}

**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 m ^{2}.**

**Solution:**

Diameter of roller = 84 cm

Radius of the roller = 42 cm

Length (Height) of the roller = 120 cm

Curved surface area of the roller =

=

= 3.1680 m^{2}

Now area leveled by roller in one revolution = 3.1680 m^{2}

Area leveled by roller in 500 revolutions

= 3.1680 x 500 = 1584.0000 = 1584 m^{2}

**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 =

Height of the pillar = 3.5 m

Now, Curved surface area of the pillar

=

= = m^{2}

Cost of white washing 1 m2 = Rs. 12.50

Cost of white washing m^{2}

= 12.50 x = Rs. 68.75

**Question 6. Curved surface area of a right circular cylinder is 4.4 m ^{2}. 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 =

**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 m^{2}.

**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 =

(ii) Cost of plastering 1 m^{2} = Rs. 40

Cost of plastering 100 m^{2} = 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

Radius = 5/2 cm

Curved surface area of the pipe =

= = 44000 cm2 = = 4.4 m^{2}

**Solution:**

(i) Diameter of cylindrical petrol tank = 4.2 m

Radius of the cylindrical petrol tank

= 4.2/2= 2.1 m

And Height of the tank = 4.5 m

Curved surface area of the cylindrical tank =

= = 59.4 m^{2}

(ii) Let the actual area of steel used be meters

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

= 59.4

= 64.8 m^{2}

Hence steel actually used is 64.8 m^{2}.

**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.]**

**Solution:**

Height of each of the folding at the top and bottom ()= 2.5 cm

Height of the frame (H) = 30 cm

Diameter = 20 cm

Radius = 10 cm

Now cloth required for covering the lampshade

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

= 2200 cm^{2}

**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 ()

= 10.5 cm

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

=

= = 226.28 cm^{2}

Since Cardboard required for making 1 pen holder = 226.28 cm^{2}

Cardboard required for making 35 pen holders = 226.28 x 35 = 7919.8 cm^{2}

= 7920 cm^{2 }(approx.)

### NCERT Solutions for Class 9 Maths Exercise 13.3

Assume unless ^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

Radius (r)= 10.5/2= 21/4cm

Slant height of cone (l)= 10 cm

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

= = 165 cm^{2}

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

=

= = 251.625 cm^{2}

**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

Radius of cone (r)= 24/2= 12 m

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

=

= = 1244.57 m^{2}

**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 cm^{2}

πrl= 308

(ii) Total surface area of the cone

= Curved surface area + Area of circular base

= 462 cm^{2}

**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 m^{2} canvas is Rs. 70.

**Solution:**

Height of the conical tent()= 10 m

Radius of the conical tent (r)= 24 m

(i) Slant height of the tent

(ii) Canvas required to make the tent

= Curved surface area of the tent= πrl

= = m^{2}

Cost of 1 m^{2 }canvas = Rs. 70

Cost of m^{2} canvas

= 70 x = 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 ()= 8 m and Radius of the conical tent (r)= 6 m

Slant height of the tent

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

L = = 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 m ^{2}.**

**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

= = 550 m^{2}

Cost of white washing 100 m^{2}

= Rs. 210

So, Cost of white washing 1 m^{2 }=

So, Cost of white washing 550 m^{2}

= = 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 () = 24 cm

Slant height of the cone

= 25 cm

Area of sheet required to make a cap

= CSA of cone = πrl

= = 550 cm^{2}

Area of sheet required to make 10 caps = 10 x 550 = 5500 cm^{2}

**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 m ^{2}, 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

Height of cone ()= 1 m

Slant height of cone

= m

Curved surface area of cone =

= 3.14 x 0.2 x

= 0.64056 m^{2}

Cost of painting 1 m^{2} of a cone

= Rs. 12

Cost of painting 0.64056 m^{2} 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 unless 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 = 4πr^{2}

= = 1386 cm^{2}

(ii) Radius of sphere = 5.6 m

Surface area of sphere = 4πr^{2}

= = 3.94.84 m^{2}

(iii) Radius of sphere = 14 cm

Surface area of sphere = 4πr^{2}

= = 2464 cm^{2}

**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 = 4πr^{2}

= = 616 cm^{2}

(ii) Diameter of sphere = 21 cm

Radius of sphere = 21/2 cm

Surface area of sphere = 4πr^{2}

= = 1386 cm^{2}

(iii) Diameter of sphere = 3.5 cm

Radius of sphere = 3.5/2 = 1.75 cm

Surface area of sphere = 4πr^{2}

= = 38.5 cm^{2}

**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 = 3πr^{2}

= 3 x 3.14 x 10 x 10 = 942 cm^{2}

Hence total surface area of hemisphere is

942 cm^{2}

**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 = 4πr^{2}

=4π x 7 x 7 cm^{2}……….(i)

II case: Radius of balloon (R) = 14 cm

Surface area of balloon = 4πr^{2}

= 4π^{ }x 14 x 14 cm^{2} …….(ii)

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

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 cm ^{2}.**

**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

= 2πr^{2}

=

Cost of tin-plating per 100 cm^{2}

= Rs. 16

Cost of tin-plating per 1 cm^{2} =

Cost of tin-plating per cm^{2}

= = Rs. 27.72

**Question 6. Find the radius of a sphere whose surface area is 154 cm ^{2}.**

**Solution:**

Surface area of sphere = 154 cm^{2}

4πr^{2}= 154

**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 =

Radius of Earth (r) =

Surface area of Earth = 4πr^{2}

=

Now, Diameter of Moon = of diameter of Earth =

Radius of Moon (r) =

Surface area of Moon = 4πr^{2}

=

Now, Ratio =

=

Required 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

= 173.25 cm^{2}

**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).**

**Solution:(i) Radius of sphere = r**

**Surface area of sphere**

**= **2π(radius)^{2} = 2πr^{2}

**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**

= 2πrh**= **2πr x πr

= 4πr^{2}

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?**

**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 cm^{3}

Volume of a packet containing 12

such matchboxes = 12 x 15 = 180 cm^{3}

**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 m ^{3} = 1000 L)**

**Solution:**

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

= 135 m^{3}= 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 = m

Volume of liquid in cuboidal vessel

= 380 m^{3}

= 380 cm^{3}

=> 10 m x 8 m x = 380

=> = = 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 m ^{3}.**

**Solution:**

Volume of cuboidal pit = 8 m x 6 m x 3 m= 144 m^{3}

Cost of digging 1 m^{3} cuboidal pit

= Rs. 30

Cost of digging 144 m^{3 }cuboidal 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 m ^{3} = 1000 L)**

**Solution:**

Capacity of cuboidal tank

= 50000 liters

= 50000 liters

2.5 m x b x 10 m = m^{3}

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

= 20 m x 15 m x 6 m

= 1800 m3 = 1800 x 1000 liters

= 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

=

= = 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 m^{3}

Capacity of wooden crate = 1.5 m x 1.25 m x 0.5 m = 0.9375 m^{3}

Maximum number of crates that can be stored in the godown

=

= = 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.**

**Solution:**

Volume of one cuboidal brick

= l x b x h

= 22.5 cm x 11. 5 cm x 7.5 cm^{3}

= 1940.625 cm^{3}

= 0.001940625 m^{3}

Volume of cuboidal wal

= 10 m x 6 m x 1.5 m= 90 m^{3}

Minimum number of bricks required

= 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)^{3 }= 1728 cm^{3}

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

= = 216 cm^{3}

Side of new cube = = 6 cm

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

= 6 x 12 x 12 = 864 cm^{2}

Now, Surface area of original solid cube = 6 (side)^{2}= 6 x 6 x 6 = 216 cm^{2}

Now according to the question,

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?**

**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

= = 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 m^{3}

Volume of water flowing into sea in 1 minute = = 4000 m^{3}

**Question 11. Find the length of a wooden plank of width 2.5 m, thickness 0.025 m and volume 0.25 m ^{3}.**

**Solution:**

Given: Volume of wooden plank

= 0.25 m^{3}

=> lx 2.5 x 0.025 = 0.25

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 m ^{3} = 1000 L)**

**Solution:**

Height of vessel = h= 25 cm

Circumference of base of vessel

= 132 cm

=>2πr = 132

= 21 cm

Now, Volume of cylindrical vessel

= = = 34650 cm^{3}

= liters [1000 cm^{3 }= 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 1cm ^{3} 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

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

= 110 [ 196 – 144 ] = 110 x 52

= 5720 cm^{3}

Weight of 1 cm3 of wood = 0.6 g

Weight of 5720 cm^{3} 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 = 5 cm, Width of tin (b)= 4 cm

and Height of tin (h) = 15 cm

Then, Capacity of tin = = 5 x 4 x 15 = 300 cm3

II case: Diameter of base of cylinder = 7 cm

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

Height of cylinder = 10 cm

Capacity of cylinder =

= = 385 cm^{3}

From the cases I and II, we observed that

cylindrical container has greater capacity

by (385 – 300) = 85 cm^{3}.

**Question 4. If the lateral surface of a cylinder is 94.2 cm ^{2} and its height is 5 cm, then (i) radius of its base (ii) volume of the cylinder.**

**Solution:**

Height of the cylinder = 5 cm

Lateral surface area of the cylinder

= 94.2 cm^{2}

=> 2πrh = 94.2

=> 2 x 3.14 x r x 5 = 94.2

= 3 cm

Volume of cylinder = πr^{2}h

= 3.14 x 3 x 3 x 5 = 141.3 cm^{3}

**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 m ^{2}, 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 =

= 110 m^{2}

(ii) Depth of the vessel ()= 10 m

Now, Inner surface area of vessel

= 110 m^{2}

= 110

= 1.75 m

(iii) Since r = 1.75 m and

= 10 m

Capacity of vessel

= Volume of cylinder = πr^{2}h

= = 96.25 m^{3}

= 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 ()= 1 m

Capacity of vessel = 15.4 liters

= kilo liters

= 0.0154 m3 [1 m3 = 1 kl]

=> πr^{2}h = 0.0154

=> r^{2} = 0.0007 x 7 = 0.0048

=> r = 0.07 m

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

= 0.4708 m^{2}

**Question 7. A bag of grain contains 2.8 m ^{3} 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 ()= 5 m

Volume of a drum = πr^{2}h

=

= 22 x 0.6 x 4.2 x 5 = 277.2 m^{3}

Now, Number of bags

=

= = 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

Height of graphite ()= 14 cm

Volume of graphite = πr^{2}h

= = 0.11 cm3

Diameter of pencil = 7 mm

Radius of pencil (R) = 3.5 mm

= 0.35 cm

Volume of pencil=

= = 5.39 cm^{3}

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

= 5.39 – 0.11 = 5.28 cm^{3}

**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

Radius of circular base of cylindrical

bowl (r)= 7/2 cm

Height of the bowl ()= 4 cm

Now, Volume of cylindrical bowl = πr^{2}h

= = 22 x 7 = 154 cm3

Quantity of soup filled in a bowl

= 154 cm^{3}

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

= 38500 cm^{3}

= liter [1 liter = 1000 cm^{3}]

= 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**

**Solution:(i) **Given: r = 6 cm, = 7 cm

Volume of cone =

=

= 264 cm^{3}

(ii) Given: r= 3.5 cm, = 12 cm

Volume of cone =

=

= 154 cm^{3}

**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**

Solution:

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

= 24 cm

Capacity of conical vessel =

= = 1232 cm^{3}

= 1.232 liters [1000 cm^{3}= 1liter]

(ii) Given: = 12 cm, l = 13 cm

= 5 cm

Capacity of conical vessel =

= = cm^{3}

= liters

[1000 cm^{3}= 1liter]

**= **liter

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

**Solution:**

Height of the cone = 15 cm

Volume of cone = 1570 cm^{3}

= 1570

= 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 cm ^{3}, find the diameter of the base.**

**Solution:**

Height of the cone ()= 9 cm

Volume of cone = 48π cm^{3}

r = 4 cm

Diameter 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 = 12 m

Capacity of pit =

= m^{3} = 35.8 m^{3}

= 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

Radius of cone = 14 cm

Volume of cone = 9856 cm^{3}

= 9856

= 48 cm

(ii) Slant height of cone =

= 50 cm

(iii) Curved surface area of cone = πrl = = 2200 cm^{2}

**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 () = 12 cm

And radius (r) = 5 cm

Therefore, Volume of cone =

=

= 100π cm^{3}

**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 ()= 5 cm

And radius ( r )= 12 cm

Therefore, Volume of cone =

=

= 240π cm^{3}

Now,

=

Required 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

2r = 14 cm

r = 7 cm

And also Height of the cone = Edge of cube = 14 cm

Now, Volume of the largest cone

=

= 718.66 cm^{3}

**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

Height (h) of heap = 3 m

Volume of heap

=

Therefore, the volume of the heap of wheat is 86.625 m^{3}.

Area of canvas required = CSA of cone

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 =

=

= cm^{3}

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

Volume of sphere =

= 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

=

=

= cm^{3}

(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

=

= 0.004851 m^{3}

**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 cm ^{3}?**

**Solution:**

Diameter of metallic ball = 4.2 cm

Radius of metallic ball (r)= 4.2/2

= 2.1 cm

Volume of metallic ball =

= 38.808 cm^{3}

Density of metal = 8.9 g per cm^{3}

Mass of 1 cm^{3} = 8.9 g

Mass of 38.808 cm^{3} = 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 =

[Earth is considered to be a sphere]

= ……..(i)

According to question,

Diameter of moon = x Diameter of earth=

Radius of moon (R) =

Now, Volume of Moon =

[Moon is considered to be a sphere]

[From eq. (i)]

So, Volume of moon is the 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 )

= = 5.25 cm

Volume of milk in hemispherical bowl

= 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

Thickness of sheet = 1 cm

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

Volume of iron of hemisphere

= 63487.81 cm^{3}

= 0.06348 m^{3}

**Question 7. Find the volume of a sphere whose surface area is 154 cm ^{2}.**

**Solution:**

Surface area of sphere = 154 cm^{2}

4πr^{2} = 154

Now, Volume of sphere

**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

= = 249.48 cm^{2}

Inside surface area of the dome

= 249.48 cm^{2}

2πr^{2 }= 249.48

= 5.67 x 7

r = 6.3

So, Volume of the dome

=

= 523.9 cm^{3}

**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

27 x =

**Question 10. A capsule of medicine is in the shape of a sphere of diameter 3.5 mm. How much medicine (in mm ^{3}) is needed to fill this capsule?**

**Solution:**

Diameter of spherical capsule = 3.5 mm

Radius of spherical capsule ( r )

= mm

Medicine needed to fill the capsule

= Volume of sphere =

= 22.46 mm^{3} (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 cm ^{2}, find the total expenses required for polishing and painting the surface of the bookshelf.**

**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 cm^{2}

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

= Rs. 0.20 per cm^{2} = 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 cm^{2}

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 cm ^{2}.**

**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

= 1378.928 cm^{2}

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

= 11031.424 cm^{2}

Cost of silver paint over 1 cm^{2}

= 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

= = 66 cm2

Curved surface area of 8 such cylindrical supports = 66 x 8 = 528 cm^{2}

Cost of black paint over 1 cm^{2} of cylindrical support = Rs. 0.50

Cost of black paint over 528 cm^{2} 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 =

According to the question,

Decreased diameter = 25% of D

=

Diameter of new sphere =

Radius of new sphere =

Now, curved surface area of new sphere =

Change in curved surface area

Percent change in the curved surface area

**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

Radius of conical cap ( r )= 5 cm

Slant height of cone

= 13 cm

Curved surface area of a cap

= πrl = 3.14 x 5 x 13 = 204.1 cm^{2}

Curved surface area of 15 caps

= 15 x 204.1 = 3061.5 cm^{2}

Area of a sheet of paper used for making caps = 25 x 40 = 1000 cm^{2}

82% of sheet is used after cutting

= 82% of 1000 cm^{2}

= 820 cm^{2}

Number of sheet = = 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.