CBSE Class 9

These NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.6 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Exercise 13.6

Question 1.
The circumference of the base of a cylinderical vessel is 132 cm and its height is 25 cm. How many liters of water can it hold? (1000 cm³ = 1L)
Solution:
We have given that

Circumference of base of cylinder = 132 cm
2πr = 132
⇒ r = \(\frac{132 \times 7}{22 \times 2}\) = 21 cm
and height of cylinderical vessel = 25 cm
Volume of cylindrical vessel = πr²h
= \(\frac {22}{7}\) × 21 × 21 × 25
= 34650 cm³
Now, we know that 1000 cm³ = 1L
34650 cm³ = 34.65 L
Therefore, the cylindrical vessel can hold 34.65 liters of water.

Question 2.
The inner diameter of a cylindrical wooden pipe is 24 cm and its outer diameter is 28 cm. The length of the pipe is 35 cm. Find the mass of the pipe, if 1 cm³ is wood has a mass of 0.6 g.
Solution:

Volume of required wood = volume of outer cylinder – volume of inner cylinder.
= π(14)² × 35 – π(12)² × 35
= \(\frac {22}{7}\) × 35((14)² – (12)²)
= \(\frac {22}{7}\) × 35 × (196 – 144)
= \(\frac {22}{7}\) × 35 × 52
= 5720 cm³
Now, we have given that, mass of 1 cm³ of wood = 0.6
Therefore, mass of 5720 cm³ of wood = 0.6 × 5720 g.
= 3432g
= 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 wide 4 cm having a height of 15 cm and (ii) a plastic cylinder with a circular base of diameter 7 cm and height 10 cm. Which container has greater capacity and by how much.
Solution:
In the first case,
The tin can with a rectangular base of length 5 cm and wide 4 cm having a height of 15 cm.
∴ The volume of this can = l × b × h
= 15 × 4 × 5
= 300 cm³ …….(i)

In the second case,
The circular plastic cylinder has a base diameter is 7 cm and the height of the cylinder is 10 cm.

∴ Volume of plastic cylinder = πr²h
= \(\frac {22}{7}\) × 3.5 × 3.5 × 10
= 385 cm³ ……(ii)
From (i) and (ii), it is clear that
The volume of the circular cylinder has a greater capacity of 85 cm³.

Question 4.
If the lateral surface of a cylinder is 94.2 cm³ and its height is 5 cm, then find
(i) radius of its base
(ii) volume of the cylinder.
Solution:
(i) We have given that lateral surface of cylinder = 94.2 cm³ and height is 5 cm.

Now, we know that, lateral surface area of cylinder = 2πrh
94.2 = 2 × 3.14 × r × 5
r = \(\frac{94.2}{2 \times 3.14 \times 5}\) = 3 cm
(ii) We know that
Volume of cylinder = 2πrh
= 3.14 × 3 × 3 × 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:
(i) We have given that

Rate of painting = Rs. 20 per m²
and total cost of curved surface area of cylindrical vessel = Rs. 2200
It means that inner curved surface area of cylindrical vessel = \(\frac{2200}{20}\) = 110 m²

(ii) we know that curved surface ara of cylinder = 2πrh
⇒ 110 = 2 × \(\frac {22}{7}\) × r × 10
⇒ r = \(\frac{110 \times 7}{2 \times 22 \times 10}\) = 1.75 m

(iii) We know that volume of cylinder = πr²h
= \(\frac {22}{7}\) × 1.75 × 1.75 × 10
= 96.25 m³
Therefore, Capacity of the vessel = 96.25 m³ = 96.25 kl
(1 m³ = 1000 l =1 kl)

Question 6.
The capacity of a closed cylindrical vessel of height 1 m is 15.4 liters. How many square meters of the metal sheet would be needed to make it?
Solution:
We have given that
Capacity of closed cylindrical vessel = 15.41 = \(\frac{15.4}{1000}\) m³

and height of the cylindrical vessel = 1 m.
We know that,
volume of cylinder = πr²h
⇒ \(\frac{15.4}{1000}\) = \(\frac {22}{7}\) × r² × 1
⇒ r² = 0.0049
⇒ r = 0.07m
Now, Total surface area of cylindrical vessel = 2πr(h + r)
= 2 × \(\frac {22}{7}\) × 0.07 (1 + 0.07)
= 0.4708 m²
Therefore, a metal sheet would be needed to make the required cylinder is 0.4708 m².

Question 7.
A lead pencil consists of a cylinder of wood will a solid cylinder of graphite filled in the interior. The diameter of the pencil is 7 mm and the diameter of the graphite is 1 mm. If the length of the pencil is 14 cm, find the volume of the wood and that of the graphite.
Solution:
We have given that
Diameter of pencil = 7 mm
Radius of pencil = 3.5 mm
and height of pencil = 14 cm = 140 mm

Volume of pencil = πr²h
= \(\frac {22}{7}\) × 3.5 × 3.5 × 140
= 5390 mm³
Agian, we have
Diamter of lead = 1 mm
Radius of lead = 0.5 mm
and height of lead = 14 cm = 140 mm
Volume of lead = πr²h
= \(\frac {22}{7}\) × 0.5 × 0.5 × 140
= 110 mm³
Volume of graphite = 110 mm³ = 0.11 cm³
and volume of wood = 5390 – 110
= 5280 mm³
= 5.28 cm³

Question 8.
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:

We have given that
Diameter of pencil = 7 mm
Radius of cylindrical bowl = 3.5 cm
and height of soup to serve = 4 cm.
Volume of soup to serve one patient = πr²h
= \(\frac {22}{7}\) × 3.5 × 3.5 × 4
= 154 cm³
Therefore, volume of soup to serve 250 patients are = 250 × 154
= 38500 cm³.
= 38.5 litre.
Hence, Hospital has to prepare daily 38.5 litres of soup to serve 250 patients.

These NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.5 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Exercise 13.5

Question 1.
A matchbox measure 4 cm × 2.5 cm × 1.5 cm. What will be the volume of packets containing 12 such boxes?
Solution:

We have given the dimension of match box = 4 cm × 2.5 cm × 1.5 cm
Volume of matchbox = l × b × h
= 4 × 2.5 × 1.5
= 15 cm³
Volume of 12 such matchbox = 12 × 15 = 180 cm³

Question 2.
A cuboidal water tank 6 m long, 5 m, wide, and 4.5 m deep. How many liters of water can it hold? (1 m³ = 10001)
Solution:
We have given that
length of cuboidal tank = 6m

Breadth of cubodical tank = 5 m
and height of cubodical tank = 4.5 m
Volume of cubodical tank = l × b × h
= 6 × 5 × 4.5
= 135 m³
Again, we have given that
1 m² = 10001
135m³ = 135 × 10001 = 1350001
Therefore, the cuboidal tank holds 135000 liters of water.

Question 3.
A cuboidal vessel is 10 m long and 8 m wide. How high must it be made to hold 380 cubic meters of a liquid?
Solution:
We have given that
length of cuboidal vessel = 10 m
Breadth of cuboidal vessel = 8 m
and volume of cuboidal vessel = 380 m³
We know that,
Volume of cuboidal Vessel = l × b × h
⇒ 380 = 10 × 8 × h
⇒ h = 4.75 m
Therefore, the height of required cuboidal vessels = 4.75m.

Question 4.
Find the cost of digging a cuboidal pit 8m long 6 m broad and 3 m deep at the rate of Rs. 30 per m³.
Solution:
We have given that length of cuboidal pit = 8m

Breadth of cubodical pit = 6 m
and height of cubodical pit = 4 m
Volume of cubodical pit = l × b × h
= 8 × 6 × 3
= 144 m³
Rate od digging = Rs. 30 per m³
Total cost of digging tire cubodical pit = 30 × 144 = Rs. 4,320

Question 5.
The capacity of a cuboidal tank is 50,000 liters of water. Find the breadth of the tank, if its length and depth are respectively 2.5 m and 10 m.
Solution:
We know that
1000 liters = 1 m³
50,000 liters = 50 m³

Now, we have given that length of cuboidal tank = 2.5m
and height of cuboidal tank = 10m
and volume of cuboidal tank = 50 m³
We know that
The volume of cuboidal tank = l × b × h
⇒ 50 = 2.5 × b × 10
⇒ b = 2 m
The breadth of the required tank = 2m.

Question 6.
A village, having a population of 4000, requires 1500 liters of water per head per day. It has a tank measuring 20m × 15m × 6m. For how many days will the water of this tank last.
Solution:
Volume of water tank = l × b × h
= 20 × 15 × 6
= 1800 m³

We know that
1 m³ = 1000 litres
1800 m³ = 1800 × 1000 = 18,00,000 litres
Therefore, capacity of water tank = 18,00,000 litres.
Now, One person requires 150 litres per day
4000 person requires = 4000 × 150 = 6,00,000 litres per day
No. of days will the water of tank last = \(\frac{18,00,000}{6,00,000}\) = 3 days.

Question 7.
A godown measures 40 m × 25 m × 10 m. Find the maximum number of wooden crates each measuring 1.5 m × 1.25 m × 0.5 m that can be stored in the godown.
Solution:
Volume of godown = 40 × 25 × 10 = 10,000 m³
Therefore, total number of crates can be stored in the godown = \(\frac{10,000}{0.9375}\) = 10666.66 or 10,666.

Question 8.
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 larger cube = (side)³
= (12)³
= 1728 cm³
Now, volume of each smaller cube = \(\frac {1}{8}\) × 1728
(side)³ = 216
side = \(\sqrt[3]{216}\) = 6 cm
The side of’new cube = 6 cm
Again, Surface area of larger cube = 6 × (side)²
= 6 × (12)²
= 864 cm²
and surface area of smaller cube = 6 × (side)²
= 6 × (6)²
= 216 cm²
Ratio between them is = 984 : 216 = 4 : 1

Question 9.
A river 3 m deep and 40 m wide is flowing at the rate of 2 km per hour. How much water will fall into the sea in a min?
Solution:

We have given that
Rate of flowing = 2 km/h 100 m
= \(\frac{2 \times 1000}{60}\) m/min
= \(\frac{100}{3}\) m/min.
∴ Volume of water will fall into the sea in one min is = 40 × 3 × \(\frac{100}{3}\) = 4,000 m³.

These NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.4 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Exercise 13.4

Question 1.
Find the surface area of a sphere of radius.
(i) 10.5 cm
(ii) 5.6 cm
(iii) 14 cm
Solution:
(i) We have given that

Radius of sphere = 10.5 cm
Surface area of sphere = 4πr²
= 4 × \(\frac {22}{7}\) × 5.6 × 5.6
= 394.24 cm²

(ii) We have given that

Radius of sphere = 5.6 cm
Surface area of sphere = 4πr²
= 4 × \(\frac {22}{7}\) × 5.6 × 5.6
= 394.24 cm²

(iii) We have given that

Radius of sphere = 14 cm
Surface area of sphere = 4πr²
= 4 × \(\frac {22}{7}\) × 14 × 14
= 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) We have given that

Diameter of sphere = 14 cm
Radius of sphere = 7 cm
Surface area of sphere = 4πr²
= 4 × \(\frac {22}{7}\) × 7 × 7
= 616 cm²

(ii) We have given that

Diameter of sphere = 21 cm
Radius of sphere = 10.5 cm
Therefore,
Surface area of sphere = 4πr²
= 4 × \(\frac {22}{7}\) × 10.5 × 10.5
= 1386 cm²

(iii) We have given that

Diameter of sphere = 3.5 m
Radius os sphere = 1.75 m
Therefore,
Surface area of sphere = 4πr²
= 4 × \(\frac {22}{7}\) × 1.75 × 1.75
= 38.5 m²

Question 3.
Find the total surface area of a hemisphere of radius 10 cm.
Solution:
We have given that
Radius of hemisphere = 10 cm

Therefore, Total surface area of hemisphere = 3πr²
= 3 × 3.14 × 10 × 10 (∴ π = 3.14)
= 942 cm²

Question 4.
The radius of the spherical balloon increases from 7 cm to 14 cm as air is being pumped into it. Find the ratio of surface areas of the balloon in the two cases.
Solution:
In first case,

Radius of spherical balloon = 7 cm
Surface area of spherical balloon = 4πr²
= 4 × \(\frac {22}{7}\) × 7 × 7
= 616 cm²
In second case,

radius of spherical balloon = 14 cm
surface area of spherical balloon = 4πr²
= 4 × \(\frac {22}{7}\) × 14 × 14
= 2464 cm²
Therefore the ratio of the surface area of balloons in two cases is 616 : 2464 = 1 : 4

Question 5.
A hemispherical bowl made of brass has an inner diameter of 10.2. Find the cost of tin plating it on the inside at the rate of Rs. 16 per 100 cm².
Solution:
We have given that
Diameter of hemispherical bowl = 10.5 cm
The radius of hemispherical bowl = 5.25 cm

Curved surface area of hemispherical bowl = 2πr²
= 2 × \(\frac {22}{7}\) × 5.25 × 5.25
= 173.25 cm²
Rate of tin plating = Rs. 16 per 100 cm²
Total cost of tin plating the hemispherical bowl is = 173.25 × \(\frac{16}{100}\) = Rs. 27.72

Question 6.
Find the radius of a sphere whose surface area is 154 cm².
Solution:
We have given that

Surface area of sphere = 154 cm²
But, Surface area of sphere = 2πr²
⇒ 154 = 4 × \(\frac {22}{7}\) × r²
⇒ r² = \(\frac{154 \times 7}{22 \times 4}\) = 12.25
⇒ r = √12.25 = 3.5 cm
Therefore, the radius of the sphere whose surface area is 154 cm² is 3.5 cm.

Question 7.
The diameter of the moon is approximately one-fourth of the diameter of the earth. Find the ratio of their surface areas.
Solution:
Let the diameter of earth = r cm
Diameter of moon = \(\frac{r}{4}\) cm
Again, radius earth = \(\frac{1}{2}\) cm
and radius of moon = \(\frac{r}{8}\) cm
Surface area of earth = 4πr²
= 4 × π × \(\left(\frac{r}{2}\right)^{2}\)
= πr²
and surface area of moon = 4π \(\left(\frac{r}{8}\right)^{2}\) = \(\frac{\pi r^{2}}{16}\)
Ratio of their surface areas = \(\frac{\pi r^{2}}{\frac{16}{\pi r^{2}}}\) = \(\frac{1}{16}\) = 1 : 16
Ratio of moon and earth surface area = 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 curves surface area of the bowl.
Solution:

We have given tha
Inner radius of hemispherical bowl = 5 cm
and thickness of steel = 0.25 cm.
Outer curved surface area of hemispherical bowl = 5.25 cm
Therefore, Outer curved surface area of hemisperical bowl = 2πr²
= 2 × \(\frac {22}{7}\) × 5.25 × 5.25
= 173.25 cm²

Question 9.
A right circular cylinder just encloses a sphere of radius r (see fig. 13.22) Find.
(i) surface area of the sphere.
(ii) Curved surface area of the cylinder
(iii) ratio of areas obtained in (i) and (ii)

Solution:
(i) We have given that

Radius of sphere = r
Surface area of sphere = 4πr²
(ii) Surface area of cylinder = 4πrh
= 2 × π × r × 2r (∵ h = 2r)
= 4πr²
(iii) Ratio of the areas obtaining in (i) and (ii) is
\(\frac{4 \pi r^{2}}{4 \pi r^{2}}\) = 1 : 1

These NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.3 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Exercise 13.3

Question 1.
The diameter of the base of a cone is 10.5 cm and its slant height is 10 cm. Find the curved surface area.
Solution:

We have given that,
Diameter of the base of a cone = 10.5 cm
The radius of the base of a cone = 5.25 cm
and slant height of the cone (l) = 10 cm
∴ The curved surface area of cone = πrl
= \(\frac {22}{7}\) × 5.25 × 10
= 165 cm²

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

We have given,
Slant height of the cone = 21 m
and diameter of its base = 24 m
∴ radius of its base = 12 m
∴ Total surface area of a cone = πr(l + r)
= \(\frac {22}{7}\) × 12 (21 + 12)
= \(\frac {22}{7}\) × 12 × 33
= 1244.57 m² (approx)

Question 3.
The curved surface area of a cone is 308 cm² and its slant height is 14 cm. Find
(i) radius of the base
(ii) Total surface area of the cone.
Solution:
(i) We have given that
The curved surface area of cone = 308 cm²

and slant height of the cone = 14 cm.
We know that,
Curved surface area of cone = πrl
308 = \(\frac {22}{7}\) × r × 14
r = 7 cm
Therefore, radious of base (r) = 7 cm

(ii) Total surface area of cone = πr(l + r)
= \(\frac {22}{7}\) × 7(14 + 7)
= 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 1 m² canvas is Rs. 70.
Solution:
(i) We have given that
height of the cone = 10 m
and radious of the cone = 24 m
By Pythagoras theorem we know that
h² = p² + b²
⇒ AC² = AB² + BC²
⇒ AC² = (10)² + (24)²
⇒ AC² = 100 + 576
⇒ AC² = 676
⇒ AC = √676 = 26 m

Therefore slant height of the cone = 26 m.
(ii) Canvas required to make the tent = curved surface area of cone
or, Curved surface area of cone = πrl
= \(\frac {22}{7}\) × 24 × 26
= \(\frac{13728}{7}\) m²
Cost of canvas = Rs. 70 per m²
∴ Total cost of canvas required to make the tent = \(\frac{13728}{7} \times 70\) = Rs. 1,37,280.

Question 5.
What length of tarpaulin 3 m wide will be required to make a conical tent of height 8 m and base radius 6m? 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:
We have given that
The radius of base of cone = 6 m
Height of conical tent = 8 m

By Pythagoras theorem,
AC² = AB² + BC²
⇒ AC² = (9)² + (6)²
⇒ AC² = 64 + 36
⇒ AC² = 100
⇒ AC = 10 cm
∴ Curved surface area of cone = πrl
= \(\frac {22}{7}\) × 6 × 10
= \(\frac{1320}{7}\) m²
According to question
Curved surface area of conical tent = Area of rectangular tarpaulin.
\(\frac{1320}{7}\) = x × 3
x = \(\frac{1320}{7 \times 3}\) = 62.87 m (approx)
Again, stitching margins and wastage in cutting is 20 cm approximately.
Total length of tarpaulin = 62.87 + 0.20 = 63m (approx).

Question 6.
The slant height and base diameter of a Conical tomb is 25 m and 14 m respectively. Find the white-washing its curved surface at the rate of Rs. 210 per 100 m².
Solution:
We have given that
The slant height of cone = 25 m
Diameter of its base = 14 m
The radius of its base = 7 m

∴ Curved surface area of cone = πrl
= \(\frac {22}{7}\) × 7 × 25
= 550 m²
Now, rate of white washing = Rs. 210 per 100 m²
∴ Total cost of white washing the conical tomb = 550 × \(\frac {210}{100}\) = 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:
We have given that
Radius of conical cap = 7 cm
and height of conical cap = 24 cm
By Pythagoras theorem,
AC = \(\sqrt{(\mathrm{AB})^{2}+(\mathrm{BC})^{2}}\)
⇒ AC = \(\sqrt{(24)^{2}+(7)^{2}}\)
⇒ AC = \(\sqrt{576+49}\)
⇒ AC = √625
⇒ AC = 25 cm

∴ Curved surface area of one cap = πrl
= \(\frac {22}{7}\) × 7 × 25
= 550 cm²
Sheet required to make one jocker cap = 550 cm²
Sheet required to make ten such caps = 550 × 10 = 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 one has a base diameter of 40 cm and height lm. 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?
(Use π = 3.14 and take √1.04 = 1.02)
Solution:
We have given that
Height of the cone = 1 m
and diameter of its base = 40 cm = 0.40 m
radius = \(\frac{0.40}{2}\) = 0.20 m

By Pythagoras theorem
AC = \(\sqrt{(\mathrm{AB})^{2}+(\mathrm{BC})^{2}}\)
= \(\sqrt{(1)^{2}+(0.20)^{2}}\)
= \(\sqrt{100400}\) (∴ √1.04 = 1.02 Given)
= 1.02m
Now, curved surface area of one cone = πrl
= 3.14 × 0.20 × 1.02
= 0.640 m² (approx)
Therefore, the curved surface area of 50 such cones = 50 × 0.640
= 32.02 m²
Rate of painting the cone = Rs. 12 per m²
Total cost for painting fifty cones is 32.02 × 12 = Rs. 384.24 (approx)

These NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.2 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Exercise 13.2

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:
We know that

Curved surface area of cylinder = 2πrh
⇒ 88 = 2 × \(\frac{22}{7}\) × 14 × r
⇒ 88 = 88 × r
⇒ r = 1 cm
Diameter of cylinder = 2r = 2 × 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 is required for the same?
Solution:
Height of the cylinder = 1 m
and diameter of cylinder = 140 cm = 1.4 m

∴ radius of cylinder = \(\frac{14}{2}\) = 0.7 m
∴ Total surface are of cylinder = 2r(h + r)
= 2 × \(\frac{22}{7}\) × 0.7(1 + 0.7)
= 4.4 × 1.7
= 7.48 m²
Therefore, 7.48 m² of sheets are required to make the closed circular cylinder.

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. 13.11) find its.
(i) inner curved surface area.
(ii) Outer curved surface area.
(iii) Total surface area.

Solution:
(i) We have given,
Height of cylinder = 77 cm

and inner diameter = 4 cm
inner radius = \(\frac{4}{2}\) = 2 cm
∴ Inner curved surface area = 2πrh
= 2 × \(\frac{22}{7}\) × 2 × 77
= 968 cm².
(ii) We have given
Height of cylinder = 77 cm
and Outer diameter = 4.4 cm
∴ Outer curved surface area of cylinder = 2πrh
= 2 × \(\frac{22}{7}\) × 2.2 × 77
= 1064.8 cm².
(iii) Total surface ara of pipe = Inner curved surface area + Outer curved surface area + Area of the two bases.
= 2πrh + 2πRh + 2π(R² – r²)
= 2 × π × 2 × 77 + 2 × π × 2.2 × 77 + 2π((2.2)² – (2)²)
= 2 × π × 77(2 + 2.2) + 2 × π × (4.84 – 4)
= 2 × \(\frac{22}{7}\) × 88 × 4.2 + 2 × \(\frac{22}{7}\) × 0.84
= 2032.8 + 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:
Clearly, the roller is a right circular cylinder of height h = 120 cm = 1.2 m
and diameter of its base = 84 cm = 0.84 m
∴ radius of its base = \(\frac{84}{2}\) = 0.42 m.
∴ Area covered by roller in one revolution = Curved surface area of the roller = 2πrh

= 2 × \(\frac{22}{7}\) × 0.42 × 1.2
= 3.168 m²
So, Area covered by roller in 500 revolutions = 3.168 × 500 = 1584 m².
Hence, Area of playground = 1584 m².

Question 5.
A cylindrical pillar is 50 cm in diameter and 3.5 m in height. Find the cost of painting the curved surface area of the pillar at the rate of Rs. 12.50 per m².
Solution:
We have given that,
the height of pillar = 3.5m
and the diameter of pillar = 50 cm = 0.5 m
∴ Radious of the pillar = \(\frac{0.5}{2}\) = 0.25 m
∴ Curved surface area of pillar = 2πrh
= 2 × \(\frac{22}{7}\) × 0.25 × 3.5
= 5.5 m²
Now, Rate of painting the pillar = Rs. 12.50 per m².
∴ Total cost of painting = 5.5 × 12.50 = Rs. 68.75.

Question 6.
The 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:
We have given that
The curved surface area of cylinder = 4.4 m²
and radius of its base = 0.7 m
Height =?
We know that
Curved surface area of cylinder = 2πrh
⇒ 4.4 = 2 × \(\frac{22}{7}\) × 0.7 × h (∵ π = \(\frac {22}{7}\))
⇒ h = \(\frac{4.4 \times 7}{2 \times 22 \times 0.7}\) = 1 m
∴ Height of cylinder = 1 m.

Question 7.
The inner diameter of the circular well is 3.5 m. It is a 10 m deep find.
(i) It’s inner curved surface area.
(ii) The cost of plastering this curved surface at the rate of Rs. 40 per m².
Solution:
(i) We have given that
The inner diameter of circular well = 3.5 m

Inner radius circular well = \(\frac{3.5}{2}\) m
and height of circular well = 10 m.
Now, we know that,
Curved surface area of well = 2πrh
= \(2 \times \frac{22}{7} \times \frac{3.5}{2} \times 10\)
= 110 m²
So, inner curved surface area of well = 110 m²
(ii) Rate of plastering = Rs. 40 per m²
Total cost of plastering its curved surface area = 110 × 40 = Rs. 4400

Question 8.
In a hot water heating system, there is a cylindrical pipe of length 28m and diameter 5 cm. Find the total radiating surface in the system.
Solution:
We have given that
Length of the cylindrical pipe (h) = 28m
and diameter of its base = 5 cm = 0.05 m
Radius of its base = \(\frac{.05}{2}\) m

∴ Curved surface area of cylindrical pipe = 2πrh
= \(2 \times \frac{22}{7} \times \frac{.05}{2} \times 28\)
= 4.4 m²

Question 9.
Find
(i) The lateral or curved surface area of a cylindrical petrol Storage tank that is 4.2 m in diameter and 1.5 m high.
(ii) How much steel was actually used if \(\frac{1}{12}\) of the steel actually used was wasted in making the closed tank.
Solution:
(i) We have given that,
Height of cylindrical tank = 4.5 m

Diameter of cylindrical tank = 4.2 m
Radius of cylindrical tank = 2.1 m
So, Curved surface area of tank = 2πrh
= 2 × \(\frac{22}{7}\) × 4.5 × 21
= 59.4 m²
(ii) Let the actual area of steel used be x m².
Since \(\frac{1}{12}\) of the actual steel used was wasted = \(\frac{1}{12}\) of x = \(\frac{x}{12}\)
∴ Area of steel which has gone into the tank = x – \(\frac{x}{12}\) = \(\frac{11x}{12}\)
Total surface area of tank = 2πr(r + h)
= 2 × \(\frac{22}{7}\) × 2.1(2.1 + 4.5)
= 87.12 m²
The actual area of steel used be
\(\frac{11x}{12}\) = 87.12
⇒ x = \(\frac{87.12 \times 12}{11}\)
⇒ x = 95.04 m²
Hence, total steel was used in tank = 95.04 m²

Question 10.
In Fig. 13.12, you see the frame of a lampshade. It is to be covered with a decorative cloth. The frame has a base diameter of 20 cm and a height of 30 cm. A margin of 2.5 cm is to be given for folding it over the top and button of the frame. Find how much cloth is required for covering the lampshade.

Solution:
We have given that
Diameter of lampshade = 20 cm
Radius of lampshade = 10 cm
and height of lampshade = 30 + 2.5 + 2.45 = 35 cm
Therefore, Curved surface area of cylindrical lampshade = 2πrh
= 2 × \(\frac{22}{7}\) × 10 × 35
= 2200 cm²
Therefore, the cloth is required for covering the lampshade is 2200 cm².

Question 11.
The students of a Vidyalaya were asked to participate in a competition for making and decorating penholders in the shape of a cylinder with a base, using cardboard. Each penholder was to be of radius 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:
Total surface area of one penholder is 2πrh + πr²
= 2 × \(\frac{22}{7}\) × 3 × 10.5 + \(\frac{22}{7}\) × 3 × 3
= 198 + 28.28
= 226.28 cm²

∴ Total cardboard is used to making one penholder is 226.28 cm²
Therefore, total cardboard is used to making 35 penholders is = 226.28 × 35 = 7919.8 cm².

These NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Ex 13.1 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes Exercise 13.1

Question 1.
A plastic box 13 m long, 1.25 wide and 65 cm deep are to be made. It is to be open at the top. Ignoring the thickness of a plastic sheet, determine.
(i) The area of the sheet required for making the box.
(ii) The cost of the sheet for it, if a sheet measuring 1m costs Rs. 20.
Solution:
(i) We have given,
length of box = 1.5 m
width of the box = 1.25 m
height of the box = 65 cm = 0.65 m

∴ Surface area of box = 2(lb + bh + lh)
= 2(1.5 × 1.25 + 1.25 × 0.65 + 0.65 × 1.5)
= 2(1.875 + 0.8125 + 0.975)
= 2 × 3.6625
= 7.325 m²
But we have given box is open at the top.
Area of top of the box = 1.875 m²
∴ Surface area of required box = 7.325 – 1.875 = 5.45 m²
So, the area of sheet required for making the box is 5.45 m²
(ii) Rate of sheet = Rs. 20/cm²
∴ Total cost of sheet for box = 20 × 5.45 = Rs. 109.

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

We have given
length of room = 5 m
breadth of a room = 4 m
height of a room = 3 m
Surface area of 4 walls and ceiling of the room
= 2(l + b)h + lb
= 2(5 + 4)3 + 5 × 4
= 54 + 20
= 74 m²
Cost of whitewashing the walls and ceiling of the room at Rs. 7.50 per m² is = 74 × 7.50 = Rs. 555.

Question 3.
The floor of the rectangular hall has a perimeter of 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:
We have given that
Perimeter of floor = 250 m
2(l + b) = 250
l + b = 125 m
Now,
Area of four walls = 2 (l + b)h
= 2 × 125 × h
= 250h m².
Rate of painting the four walls is Rs. 10 per m² is Rs. 250h × 10
But we have given that the total cost of painting of four walls = 15000
i.e. 2500h = 15000
or, h = \(\frac{15000}{2500}\) = 6 m
∴ Height of the hall = 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 dimension 22.5 cm × 10 cm × 7.5 cm can be painted out of this container?
Solution:
We have given that the container contains sufficient paint for the area of 9.375 m².

Again
length of the bricks = 22.5 cm = 0.225 m
breadth of the bricks = 10 cm = 0.10 m
height of the bricks = 7.5 cm = 0.075 m
∴ surface area of one brick = 2(lb + bh + lh)
= 2(0.225 × 0.10 + 0.10 × 0.075 + 0.075 × 0.225)
= 2(0.0225 + 0.0075 + 0.016875)
= 0.09375 m².
Area of one brick = 0.09375 m².
Let x bricks can be painted out of this container
∴ 0.9375 × x = 9.375
or, x = \(\frac{9.375}{0.09375}\) = 100
Total number of bricks can be painted out of this container is 100.

Question 5.
A cubical box has each edge 10 cm and another cuboidal box is 12.5 cm long, 10 cm wide and 8 cm high.
(i) Which box has a greater lateral surface area and by how much?
(ii) Which box has the smaller total surface area and by how much?
Solution:
(i) Length of edge of cubical box = 10 cm
The lateral surface area of cubical box = 4a²
= 4 × (10)²
= 400 cm²

Again, length of cuboidal box = 12.5 cm
breadth of cuboidal box = 10 cm
and height of cuboidal box = 8 cm.
∴ Lateral surface area of cubodical box = 2(l + b)h
= 2(12.5 + 10)8
= 360 cm²
Lateral surface area of cubical box is greater than cubodical box by = (400 – 360) = 40 cm²
Now,
Total surface ara of cubical box = 6a²
= 6 × (10)²
= 600 cm²
and total surface area of cubodical box = 2 (lb + bh + lh)
= 2(12.5 × 10 + 10 × 8 + 8 × 12.5)
= 2(125 + 80 + 100)
= 610 cm²
Total surface area of cubodical box is greater than cubical box by = (610 – 600) = 10 cm².

Question 6.
A small indoor greenhouse (herbarium) is made entirely of glass panes (including base) held together with tape. It is 30 cm long, 25 cm wide and 25 cm high.
(i) What is the area of glass?
(ii) How much tape is needed for all the 12 edges?
Solution:
(i) We have given that
length of small indoor greenhouse = 30 cm.
the breadth of small indoor greenhouse = 25 cm.
and height of small indoor greenhouse = 25 cm.

Total surface area of indoor greenhouse (herbarium)
= 2(lb + bh + lh)
= 2(30 × 25 + 25 × 25 + 25 × 30)
= 2(750 + 625 + 750)
= 4250 cm²
∴ Area of glass to make herbarium is 4250 cm².
(ii) Total length of tape is needed = 4(l + b + h)
= 4(30 + 25 + 25)
= 320 cm

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 × 20 cm × 5 cm and the smaller of dimensions 15 cm × 12 cm × 5 cm. 5% of the total surface area is required extra, for all the overlaps. If the cost of cardboard is Rs. 4 for 1000 cm², find the cost of cardboard required for supplying 250 boxes of each kind.
Solution:
Dimensions of the bigger box are 25 cm × 20 cm × 5 cm
Total surface ara of one big box = 2(lb + bh + lh)
= 2(25 × 20 + 20 × 5 + 5 × 25)
= 2(500 + 100 + 125)
= 1450 cm².
∴ Total surface area of 250 such boxes are 250 × 1450 = 3,62,600 cm².
Again,
Dimension of smaller box is 15 cm × 12 cm × 5 cm.
∴ Total surface area of one small box = 2(lb + bh + lh)
= 2(15 × 12 + 12 × 5 + 5 × 15)
= 2(180 + 60 + 75)
= 630 cm².
∴ Total surface area of 250 such boxes = 250 × 630 = 1,57,500 cm².
∴ Total surface area of both type of boxes are = 3,62,500 + 1,57,500 = 5,20,000 cm²
Now, 5% of the total surface area is required for the overlaps.
5% of 5,20,000 = \(\frac{5}{100}\) × 5,20,000 = 26,000
∴ Total area of card board required = 5,20,000 + 26000 = 5,46,000
∴ Total cost of cardboard at the rate of Rs. 4 per 1000 cm² = \(\frac{546000}{1000} \times 4\) = Rs. 2184

Question 8.
Praveen wants to make a temporary shelter for her car, by making a box-like structure with a tarpaulin that covers all 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 many tarpaulins would be required to make the shelter of height 2.5m, with base dimensions 4 m × 3 m?
Solution:
We have given that,
length of temporary shelter for car = 4 m
the breadth of temporary shelter for car = 3 m
and height of temporary shelter for car = 2.5 m

∴ Total tarpaulin be required to make the shelter = Area of 4 wall + Area of roof
= 2(l + b)h + l × b
= 2(4 + 3) × 2.5 + 4 × 3
= 2 × 7 × 2.5 + 4 × 3
= 35 + 12
= 47 m²
Therefore total tarpaulin be required to make temporary shelter for the car is 47 m².