GSEB Solutions for Class 9 Mathematics – Surface Area and Volume

GSEB Solutions for Class 9 Mathematics – Surface Area and Volume (English Medium)

GSEB SolutionsMathsScience
Exercise – 15

Solution 1:

Solution 2:

Solution 3:

Solution 4:

Solution 5:

Solution 6:

Solution 7(1):

c. 24
Surface area of a cube = 6l² = 6(2)² = 6(4) = 24 cm

Solution 7(2):

d. 94
Surface area of a cuboid
= 2(lb + bh + hl)
= 2(5 × 4 + 4 × 3 + 3 × 5)
= 2(20 + 12 + 15)
= 2 × 47
= 94 cm²

Solution 7(3):

c. Rs. 60,000
Lateral surface area of the cuboidal tank
= 2h( l + b)
= 2 × 5(30 + 10)
= 10 × 40
= 400 m²
Expense of painting = Rs. (150 × 400) = Rs. 60,000

Solution 7(4):

c. 4πx²
Total surface area of a cylinder
= 2πr(h + r)
= 2πx(x + x)
= 2πx(2x)
= 4πx² cm²

Solution 7(5):

Solution 7(6):

Solution 7(7):

d. 10π
Curved surface area of a cone
= πrl
= π × 2 × 5
= 10π cm²

Solution 7(8):

a. 2πx²
Total surface area of a cone
= πr(l + r)
= π × x(x + x)
= π × 2x²
= 2πx²

Solution 7(9):

Solution 7(10):

Solution 7(11):

Solution 7(12):

Solution 7(13):

a. 2
Curved surface area of a sphere = Curved surface area of a cylinder
∴ 4πr² = 2πrh
∴ h = 2r

Solution 7(14):

Solution 7(15):

Solution 7(16):

Solution 7(17):

Solution 7(18):

Solution 7(19):

Solution 7(20):

Solution 7(21):

Solution 7(22):

Exercise – 15.1

Solution 1:

1. For a given cuboid, l = 18 cm, b = 10 cm and h = 5 cm
   Lateral surface area of a cuboid
   = 2h(l + b)
   = 2 × 5(18 + 10)
   = 10 × 28
   = 280 cm²
   Total surface area of a cuboid
   = 2(lb + bh + hl)
   = 2(18 × 10 + 10 × 5 + 5 × 18)
   = 2(180 + 50 + 90)
   = 2(320)
   = 640 cm²
2. For a given cube, l = b = h = 3 m
   Lateral surface area of a cube = 4l²
   = 4(3)²
   = 36 m²
   Total surface area of a cube = 6l²
   = 6(3)²
   = 54 m²
3. For a given cuboid, l = 1 m = 100 cm, b = 75 cm and h = 50 cm
   Lateral surface area of a cuboid
   = 2h(l + b)
   = 2 × 50(100 + 75)
   = 100 × 175
   = 17500 cm²
   Total surface area of a cuboid
   = 2(lb + bh + hl)
   = 2(100 × 75 + 75 × 50 + 50 × 100)
   = 2(7500 + 3750 + 5000)
   = 2(16250)
   = 32500 cm²

Solution 2:

Solution 3:

For the given cuboidal room,
l = 10 m, b = 8 m and h = 5 m
Area of the four walls plus the ceiling
= Lateral surface area of a cuboid +  Area of top of a cuboid
= 2h(l + b) + (l × b)
= 2 × 5(10 + 8) + (10 × 8)
= 10(18) + 80
= 180 + 80
= 260 m²
Cost of whitewashing 1m² area = Rs. 15
∴ Cost of whitewashing 260 m² area = Rs. (15 × 260) = Rs. 3900
Thus, the area of the four walls plus the ceiling is 260 m² and the cost of whitewashing them Rs. 3900.

Solution 4:

For the given cuboidal hall,
Perimeter of the base = 300 m and height, h = 10 m
Area of the four walls of the hall including doors and windows
= Lateral surface of a cuboid
= Perimeter of the base × height
= 300 × 10
= 3000 m²
For the rectangle doors, l = 5 m and b = 3 m
Area of two doors
= 2(l × b)
= 2(5 × 3)
= 30 m²
Area of four rectangular windows
= 4(l × b)
= 4(3 × 1.5)
= 18 m²
Now, area to be painted
= Area of four walls including doors and windows – Area of door – Area of windows
= 3000 – 30 – 18
= 2952 m²
Cost of painting 1 m² area = Rs. 30
∴ Cost of painting 2952 m² area = Rs. (30 × 2952) = Rs. 88560
Thus, the cost of painting the four walls of the hall is Rs. 88,560.

Solution 5:

For the given cubical box, l = 15 cm
Lateral surface area of cubical box
= 4l²
= 4(15)²
= 4 × 225
= 900 cm²
Total surface area of cubical box
= 6l²
= 6(15)²
= 6 × 225
= 1350 cm²
For the given cuboidal box, l = 25 cm, b = 20 cm and h = 10 cm
Lateral surface area of the cuboidal box
= 2h(l + b)
= 2 × 10(25 + 20)
= 20 × 45
= 900 cm²
Total surface area of cuboidal box
= 2(lb + bh + hl)
= 2(25 × 20 + 20 × 10 + 10 × 25)
= 2(500 + 200 + 250)
= 2(950)
= 1900 cm²

1. Both the boxes have equal lateral surface areas.
2. The cuboidal box has a greater total surface area than the cubical box by (1900 – 1350) = 550 cm².

Exercise – 15.2

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Solution 4:

Solution 5:

For the given cylinder,
Radius, r = 50 cm and Height, h = 50 cm
Total surface area of a cylinder
= 2πr(h + r)
= 2 × 3.14 × 50(50 + 50)
= 314 × 100
= 31400 cm²
Thus, the total surface area of the cylinder is 31400 cm².

Solution 6:

Solution 7:

Exercise – 15.3

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Solution 5:

Exercise – 15.4

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Exercise – 15.5

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Solution 8:

Length of the box made
= Length of the sheet – (2 × 5)
= 50 – 10
= 40 cm
Breadth of the box made
= Breadth of the sheet – (2 × 5)
= 40 – 10
= 30 cm
Height of the box made
= Side of the square cut off
= 5 cm
∴ Volume of the cuboidal box made
= l × b × h
= 40 × 30 × 5
= 6000 cm³
Thus, the volume of the box is 6000 cm³.

Exercise – 15.6

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Exercise – 15.7

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Exercise – 15.8

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Solution 5: