RS Aggarwal Class 9 Solutions Volume And Surface Area

Solution Of RS Aggarwal For Class 9 Chapter 13

Volume and Surface Area Exercise 13.1
Volume and Surface Area Exercise 13.2
Volume and Surface Area Exercise 13.3

Q.1: Find the volume, lateral surface area and the total surface area of the cuboid whose dimensions are:

(i)  length = 12 cm, breadth = 8 cm and height = 4.5 cm.

∴ Volume of cuboid = l × b × h

= (12 ×8 ×4.5) cm3

∴ Lateral surface area of a cuboid = 2(l + b) × h

= [2(12 + 8) × 4.5] cm2

= ( 2 × 20 × 4.5) cm2 = 180 cm2

∴ Total surface area of cuboid = 2(lb + bh + lh)

= 2(12 × 8) = (8 × 4.5) + (4.5 × 12) cm2

= 2(96 + 36 + 54) cm2 = 2(186) cm2 = 372 cm2

 

(ii) Length = 26 m, breadth = 14 m and height = 6.5 m

∴ Volume of cuboid = l × b × h = (26 × 14 × 6.5) m3 = 2366 m3

∴ Lateral surface area of a cuboid = 2(l + b) × h

= [2(26 + 14) × 6.5] m2 =(2 × 40 × 6.5) m2 = 520 m2

∴Total surface area = 2(lb + bh + hl)

= 2((26 × 14) + (14 × 6.5) + (6.5 × 26)) m2 = 2(364 + 91 + 169) m2 = (2 × 624) m2 = 1248 m2

 

(iii) Length = 15 m, breadth = 6 m and height = 5 dm = 0.5 m.

∴ Volume of the cuboid = l × b × h = (15 × 6 × 0.5) m3

∴ Lateral surface area of a cuboid = 2(l + b) × h

= [2(15 + 6) × 0.5] m2 = (2 × 21 × 0.5) m2 = 21 m2

∴Total surface area = 2(lb + bh + hl)

= 2((15 × 6) + (6 × 0.5) + (0.5 × 15)) m2 = 2(90 + 3 + 7.5) m2 = (2 × 100.5) m2 = 201 m2

 

(iv)  Length = 24 m, breadth = 0.25 m and height = 6 m.

∴ Volume of the cuboid = l × b × h = (24 × 0.25 × 6) m3

∴ Lateral surface area of a cuboid = 2(l + b) × h

= [2(24 + 0.25) × 6] m2 = (2 × 24.25 × 6) m2 =291 m2

∴Total surface area = 2(lb + bh + hl)

= 2((24 × 0.25) + (0.25 × 6) + (24 × 6)) m2 = 2(6 + 1.5 + 144) m2 = (2 × 151.5) m2 = 303 m2

 

Q.2: Find the capacity of a closed rectangular cistern whose length is 8 m, breadth 6 m and depth 2.5 m. Also, find the area of the iron sheet requires making the cistern.

Solution:

Length of Cistern = 8 m

Breadth of Cistern = 6 m

Height of Cistern = 2.5 m

Therefore, the capacity of the cistern = Volume of the cistern

Volume of cistern = l × b × h = (8 × 6 × 2.5) m3 =120 m3

Area of the iron sheet required = Total surface area of the cistern.

∴Total surface area = 2(lb + bh + hl)

= 2((8 × 6) + (6 × 2.5) + (2.5 × 8)) m2

= 2(48 + 15 + 20) m2 = (2 × 83) m2 = 166 m2

 

Q.3: The dimensions of a room are (9 m × 8 m × 6.5 m). It has one door of dimensions (2 m × 1.5 m) and two windows, each of dimensions (1.5 m × 1 m). Find the cost of whitewashing the walls at Rs. 6.40 per square meter.

Solution:

Length of a room = 9 m

Breadth of a room = 8 m

Height of a room = 6.5 m

Therefore, area of 4 walls =Lateral surface area

Lateral surface area of the room = 2(l + b) × h

= [2(9 + 8) × 6.5] m2 = (2 × 17 × 6.5) m2 = 221 m2

∴ Area not white washed = (area of one door) + (area of 2 windows)

= (2 × 1.5) m2 + (2 × 1.5 × 1) m2 = 3 + 3 = 6 m2

∴ Area whitewashed = (221 – 6) m2 = 215 m2

∴ Cost of whitewashing the walls at the rate of Rs.6.40 per square meter = Rs. (6.40 × 215) = Rs. 1376

 

Q.4: How many planks of dimensions ( 5 m × 25 cm × 10 cm) can be stored in a pit which is 20 m long, 6 m wide and 80 cm deep?

Solution:

Length of plank = 5 m = 500 cm

Breadth of plank = 25 cm

Height of plank = 10 cm

Hence, volume = l × b × h = (500 × 25 × 10) cm3

Now,

Length of pit = 20 m = 2000 cm

Breadth of pit = 6 m = 600 cm

Height of pit = 80 cm

Hence, volume = l × b × h = ( 2000 × 600 × 80) cm3

∴ Number of planks that can be stored = VolumeofpitVolumeofplank

= (2000×600×80)(500×25×10)= 768

 

Q.5: How many bricks will be required to construct a wall 8 m long, 6 m high and 22.5 cm thick if each brick measures (25 cm × 11.25 cm × 6 cm)?

Solution:

Length of wall = 8 m = 800 cm

Breadth of wall = 6 m = 600 cm

Height of wall = 22.5 cm

Hence, volume = l × b × h = (800 × 600 × 22.5) cm3

Length of brick = 25 cm

Breadth of brick = 11.25 cm

Height of brick = 6 cm

Hence, volume = l × b × h = (25 × 11.25 × 6) cm3

∴ Number of bricks required = VolumeofwallVolumeofbrick

= (800×600×22.5)(25×11.25×6)= 6400

 

Q.6: A wall 15 m long, 30 cm wide and 4 m high is made of bricks, each measuring (22 cm × 12.5 cm × 7.5 cm). If (1/12) of the total volume of the wall consists of the motor, how many bricks are there in the wall?

Solution:

Length of wall = 15 m

Breadth of wall = 0.3 m

Height of wall = 4 m

Hence, volume = l × b × h = (15 × 0.3 × 4) m3 = 18 m3

Volume of motor = 112×18=1.5m3

Volume of wall = (18 – 1.5) m3 = 16.5 = 332m3

Length of brick = 22 cm

Breadth of brick = 12.5 cm

Height of brick = 7.5 cm

Hence, volume = l × b × h

= [22100×12.5100×7.5100]m3 = 3316000m3

Therefore, the number of bricks = VolumeofbricksVolumeof1brick = 332×1600033 = 8000

 

Q.7: An open rectangular cistern when measured from outside is 1.35 m long, 1.08 m broad and 90 cm deep. It is made up of iron, which is 2.5 cm thick. Find the capacity of the cistern and the volume of the iron used.

Solution:

External Length of Cistern = 1.35 m = 135 cm

External Breadth of Cistern = 1.08 m = 108 cm

External Height of Cistern = 90 cm

External Volume of cistern = l × b × h = (135 × 108 × 90) cm3 =1312200 cm3

Internal Length of Cistern = (135 – 2 * 2.5) cm = 130 cm

Internal Breadth of Cistern = (108 – 2 * 2.5) cm = 103 cm

Internal Height of Cistern = (90 – 2.5) cm = 87.5 cm

Therefore, the Internal capacity of the cistern = Volume of the Internal cistern

Volume of cistern = l × b × h

= (130 × 103 × 87.5) cm3 =1171625 cm3

Volume of iron used = External Volume of the cistern. – Internal Volume of the citern.

= (1312200 – 1171625) cm3 = 140575 cm3

 

Q.8: A river 2 m deep and 45 m wide is flowing t the rate of 3 km per hour. Find the volume of water that runs into the sea per minute.

Solution:

Depth of the river = 2 m

Breadth of the river = 45 m

Length of the river = 3 km/h = 3×100060m/min = 50 m/min.

∴ Volume of water running into the sea per minute = (50 × 45 × 2) m3 = 4500 m3

 

Q.9: A box made of sheet metal costs Rs 1620 at Rs. 30 per square metre. If the box is 5 m long and 3 m wide, find its height.

Solution:

Total cost of sheet = Rs. 1620

Cost of metal sheet per square meter = Rs. 30

∴ Area of the sheet required = [Totalcostrate/m2]sq.m. = [162030]sq.m. = 54sq.m.

Length of box = 5 m

Breadth of box = 3 m

Now, let the height of the box be x meters.

∴ Area of the sheet = Total surface area of the box =  2(lb + bh + hl)

54 = 2((5 × 3) + (3 × x) + (x × 5)) m2 = 2(15 + 3x + 5x)

54 = (2 × (15 + 18x))

Solving for x, we get, x = 1.5 m

∴ The height of the box = 1.5 m.

 

Q.10: Find the length of the longest pole that can be put in a room of dimensions (10 m × 10 m × 5 m).

Solution:

Length of room = 10 m

Breadth of room = 10 m

Height of room = 5 m

∴ Length of the longest pole = Length of diagonal

= l2+b2+h2 = 102+102+52

= 100+100+25 = 225 = 15 m

∴ The length of the longest pole that can be put in a room with given dimensions = 15 m.

 

Q.11: How many persons can be accommodated in a dining hall of dimensions (20 m × 16 m × 5 m)

Solution:

Length of hall = 20 m

Breadth of hall = 16 m

Height of hall = 4.5 m

∴ Volume of hall = l × b × h = (20 × 16 × 4.5) m3

Volume of air needed per person = 5 m3

∴ Number of persons = VolumeofthehallVolumeofairneededperperson

= [(20×16×4.5)5]=288

 

Q.12: A classroom is 10 m long, 6.4 m wide and 5m high. If each student is given 1.6 m2 of the floor area, how many cubic meters of air would each student get?

Solution:

Length of classroom = 10 m

Breadth of classroom = 6.4 m

Height of classroom = 5 m

Each student is given 1.6 m2 of the floor area.

Number of students = Areaoftheroom1.6 = (10×6.4)1.6=641.6=40

Therefore, the number of students = 40

Hence, air required by each student = Volumeoftheroomnumberofstudentsm3

= 10×6.4×540=32040=8m3

 

Q.13: The volume of a cuboid is 1536 m3. Its length is 16 m, and its breadth and height are in the ratio 3 : 2. Find the breadth and height of the cuboid.

Solution:

Volume of a cuboid = 1536 m3

Length of the cuboid = 16 m

Let the breadth and height of the cuboid be 3x and 2x.

Therefore, the volume of the cuboid =   l × b × h

1536 = (16 × 3x × 2x)

1536 = 96x2

x2=153696=16

x=16=4m

∴ Breadth of the cuboid = 3x = 3 × 4 = 12 m

And height of the cuboid = 2x = 2 × 4 = 8 m

 

Q.14: The surface area of a cuboid is 758 cm2. Its length and breadth are 14 cm and 11 cm respectively. Find its height.

Solution:

Surface area of a cuboid = 758 cm2

Length of the cuboid = 14 cm

Breadth of the cuboid = 11 cm

Let the height of the cuboid be h cm

∴ Surface area of cuboid = 2(lb + bh + hl)

758 = 2((14 × 11) + (11 × h) + (h × 14)) cm2

758 = 2(154 + 11h + 14h) cm2

758 = (154 + 25h) cm2

758 = (308 + 50h) cm2

50h = 758 – 308

h=45050=9cm

∴ The height of the cuboid = 9 cm.

 

Q.15: Find the volume, lateral surface area, the total surface area and the diagonal of a cube, each of whose edges measures (a) 9 m, (b) 6.5 cm [Take 3=1.73 ]

Solution:

(a) Each edge of a cube = 9 m

∴ Volume of the cube = a3 = (9)3 m3 = 729 m3

∴ Lateral surface area of the cube = 4a2 = 4 * (9)2 = (4 * 81) = 324 m2

∴ Total surface area of the cube = 6a2 = 486 m2

∴ Diagonal of the cube = 3a=39=15.57m

(b) Each edge of a cube = 6.5 cm

∴ Volume of the cube = a3 = (6.5)3 m3 = 274.625 cm3

∴ Lateral surface area of the cube = 4a2 = 4 * (6.5)2 = (4 * 42.25) = 169 cm2

∴ Total surface area of the cube = 6a2 = 253.5 cm2

∴ Diagonal of the cube = 3a=36.5=11.245cm

 

Q.16: The total surface area of a cube is 1176 cm2. Find its volume.

Solution:

Let each side of the cube be ‘a’ cm.

Then, the total surface area of the cube = 6a2 cm2

Given,

6a2 = 1176

a2=11766=196 a=196=14cm

Volume of the cube = a3 = (14)3 = 2744 cm3

 

Q.17: The lateral surface area of a cube is 900 cm2. Find its volume.

Solution:

Let each side of the cube be ‘a’ cm

Then, the lateral surface area of the cube = 4a2

∴ 4a2 = 900

a2=9004=225 a=225=15cm

Volume of the cube = a3 = (15)3 = 3375 cm3

 

Q.18: The volume of a cube is 512 cm3. Find its surface area.

Solution:

Volume of the cube = 512 cm3 [Volume = a3]

∴ Each edge of the cube = 5123=8cm

∴ Surface area of the cube = 6a2 = 6(8)2 cm2 = 6(64) cm2 = 384 cm2

 

Q.19: Three cubes of metal with edges 3 cm, 4 cm, 5 cm respectively are melted to form a single cube. Find the lateral surface area of the new cube formed.

Solution:

Volume of the new cube = [(3)3 + (4)3 + (5)3] cm3 = [27 + 64 + 125] cm3 = 216 cm3

Now, edge of this cube = a cm

And,                          a3 = 216

Hence,                       a = 6 cm

Lateral surface area of the new cube = 4a2 cm2 = 4 (6)2 cm2 = 144 cm2

∴ The lateral surface area of the new cube formed = 144 cm2

 

Q.20: In a shower, 5 cm of rain falls. Find the volume of water that falls on 2 hectares of ground.

Solution:

1 hectare = 10000 m2

Area = 2 hectares = 2 × 10000 m2

Depth of the ground = 5 cm = 5100m

Volume of water = (area × depth) = (2×10000×5100)m3 = 1000 m3

∴ Volume of water that falls = 1000 m3

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