**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 = 6l^{2} = 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 cm^{2}

**Solution 7(3):**

c. Rs. 60,000

Lateral surface area of the cuboidal tank

= 2h( l + b)

= 2 × 5(30 + 10)

= 10 × 40

= 400 m^{2}

Expense of painting = Rs. (150 × 400) = Rs. 60,000

**Solution 7(4):**

c. 4πx^{2}

Total surface area of a cylinder

= 2πr(h + r)

= 2πx(x + x)

= 2πx(2x)

= 4πx^{2} cm^{2}

**Solution 7(5):**

**Solution 7(6):**

**Solution 7(7):**

d. 10π

Curved surface area of a cone

= πrl

= π × 2 × 5

= 10π cm^{2}

**Solution 7(8):**

a. 2πx^{2}

Total surface area of a cone

= πr(l + r)

= π × x(x + x)

= π × 2x^{2}

= 2πx^{2}

**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

∴ 4πr^{2} = 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:**

- 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 cm^{2}

Total surface area of a cuboid

= 2(lb + bh + hl)

= 2(18 × 10 + 10 × 5 + 5 × 18)

= 2(180 + 50 + 90)

= 2(320)

= 640 cm^{2} - For a given cube, l = b = h = 3 m

Lateral surface area of a cube = 4l^{2}

= 4(3)^{2}

= 36 m^{2}

Total surface area of a cube = 6l^{2}

= 6(3)^{2}

= 54 m^{2} - 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 cm^{2}

Total surface area of a cuboid

= 2(lb + bh + hl)

= 2(100 × 75 + 75 × 50 + 50 × 100)

= 2(7500 + 3750 + 5000)

= 2(16250)

= 32500 cm^{2}

**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 m^{2}

Cost of whitewashing 1m^{2} area = Rs. 15

∴ Cost of whitewashing 260 m^{2} area = Rs. (15 × 260) = Rs. 3900

Thus, the area of the four walls plus the ceiling is 260 m^{2} 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 m^{2}

For the rectangle doors, l = 5 m and b = 3 m

Area of two doors

= 2(l × b)

= 2(5 × 3)

= 30 m^{2}

Area of four rectangular windows

= 4(l × b)

= 4(3 × 1.5)

= 18 m^{2}

Now, area to be painted

= Area of four walls including doors and windows – Area of door – Area of windows

= 3000 – 30 – 18

= 2952 m^{2}

Cost of painting 1 m^{2} area = Rs. 30

∴ Cost of painting 2952 m^{2} 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

= 4l^{2}

= 4(15)^{2}

= 4 × 225

= 900 cm^{2}

Total surface area of cubical box

= 6l^{2}

= 6(15)^{2}

= 6 × 225

= 1350 cm^{2}

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 cm^{2}

Total surface area of cuboidal box

= 2(lb + bh + hl)

= 2(25 × 20 + 20 × 10 + 10 × 25)

= 2(500 + 200 + 250)

= 2(950)

= 1900 cm^{2}

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

**Exercise – 15.2**

**Solution 1:**

**Solution 2:**

**Solution 3:**

**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 cm^{2}

Thus, the total surface area of the cylinder is 31400 cm^{2}.

**Solution 6:**

**Solution 7:**

**Exercise – 15.3**

**Solution 1:**

**Solution 2:**

**Solution 3:**

**Solution 4:**

**Solution 5:**

**Exercise – 15.4**

**Solution 1:**

**Solution 2:**

**Solution 3:**

**Solution 4:**

**Solution 5:**

**Exercise – 15.5**

**Solution 1:**

**Solution 2:**

**Solution 3:**

**Solution 4:**

**Solution 5:**

**Solution 6:**

**Solution 7:**

**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 cm^{3}

Thus, the volume of the box is 6000 cm^{3}.

**Exercise – 15.6**

**Solution 1:**

**Solution 2:**

**Solution 3:**

**Solution 4:**

**Solution 5:**

**Solution 6:**

**Solution 7:**

**Solution 8:**

**Solution 9:**

**Exercise – 15.7**

**Solution 1:**

**Solution 2:**

**Solution 3:**

**Solution 4:**

**Solution 5:**

**Solution 6:**

**Exercise – 15.8**

**Solution 1:**

**Solution 2:**

**Solution 3:**

**Solution 4:**

**Solution 5:**