**.**

**NCERT Solutions for Class 8 Maths Chapter 9 ****Mensuration**** Ex 9.2**

**Ex 9.2 Class 8 Maths Question 1.**

### There are two cuboidal boxes as shown in the figure. Which box requires the lesser amount of material to make?

**Solution:**

(a) Volume of the cuboid = l × b × h = 60 × 40 × 50 = 1,20,000 cm

^{3}

(b) Volume of the cube = (Side)

^{3}= (50)

^{3}= 50 × 50 × 50 = 1,25,000 cm

^{3}

Cuboidal box (a) requires the lesser amount of material.

**Ex ****9.2**** Class 8 Maths Question 2.**

### A suitcase with measures 80 cm × 48 cm × 24 cm is to be covered with a tarpaulin cloth. How many metres of tarpaulin of width 96 cm is required to cover 100 such suitcases?

**Solution:**Dimensions of the suitcase = 80 cm × 48 cm × 24 cm

l = 80 cm, b = 48 cm and h = 24 cm

Total surface area of the suitcase = 2[lb + bh + hl]

= 2[80 × 48 + 48 × 24 + 24 × 80]

= 2[3840 + 1152 + 1920]

= 2 × 6912

= 13824 cm

^{2}

Let the length of tarpaulin required be l cm.

Area of tarpaulin required = length × breadth = l × 96 = 96l cm^{2}

Area of tarpaulin required = Total surface area of 100 suitcase

96l = 100 × 13824

l = 100 × 144 = 14400 cm = 144 m

Hence, the required length of the tarpaulin is 144 m.

**Ex ****9.2**** Class 8 Maths Question 3.**

### Find the side of a cube whose surface area is 600 cm^{2}.

**Solution:**Total surface area of a cube = 6l

^{2}

Therefore, 6l

^{2}= 600

l

^{2}= 100

l = √100 = 10 cm

Hence, the required length of a side of the cube is 10 cm.

**Ex ****9.2**** Class 8 Maths Question 4.**

### Rukhsar painted the outside of the cabinet of measure 1 m × 2 m × 1.5 m. How much surface area did she cover if she painted all except the bottom of the cabinet?

**Solution:**

Given: l = 2 m, b = 1 m, h = 1.5 m

Area of the surface to be painted = Total surface area of box – Area of base of box

= 2[lb + bh + hl] – lb

= 2[2 × 1 + 1 × 1.5 + 1.5 × 2] – 2 × 1

= 2[2 + 1.5 + 3] – 2

= 2[6.5] – 2

= 13 – 2

= 11 m

^{2}

Hence, the required area to be painted is 11 m

^{2}.

**Ex ****9.2**** Class 8 Maths Question 5.**

### Daniel is painting the walls and ceiling of a cuboidal hall with length, breadth and height of 15 m, 10 m and 7 m respectively. From each can of paint 100 m^{2} of the area is painted. How many cans of paint will she need to paint the room?

**Solution:**Surface area of a cuboidal hall without bottom = Total surface area – Area of base

= 2[lb + bh + hl] – lb

= 2[15 × 10 + 10 × 7 + 7 × 15] – 15 × 10

= 2[150 + 70 + 105] – 150

= 2[325] – 150

= 650 – 150

= 500 m

^{2}

Area painted by one can of paint = 100 m

^{2}

Number of cans required to paint 500 m

^{2}= 500/100 = 5 cans.

**Ex ****9.2**** Class 8 Maths Question 6.**

### Describe how the two figures given below are alike and how they are different. Which box has a larger lateral surface area?

**Solution:**

The two given figures are cylinder and cube.

The two figures are alike in respect of their same height.

Cylinder: d = 1 cm, h = 7 cm

Cube: Length of each side (a) = 7 cm

Both the figures are different in respect of their shapes.

Lateral surface area of a cylinder = 2Ï€rh

= 2 × 22/7 × 7/2 × 7 = 154 cm

^{2}

Lateral surface area of a cube = 4l

^{2}= 4 × (7)

^{2}= 4 × 49 = 196 cm

^{2}

So, the cube has the larger lateral surface area.

**Ex ****9.2**** Class 8 Maths Question 7.**

### A closed cylindrical tank of radius 7 m and height 3 m is made from a sheet of metal. How many sheets of metal is required?

**Solution:**Area of required metal sheet = Total surface area of the cylindrical tank = 2Ï€r(h + r)

= 2 × 22/7 × 7(3 + 7)

= 2 × 22/7 × 7 × 10

= 440 m

^{2}

Hence, the required area of metal sheet is 440 m

^{2}.

**Ex ****9.2**** Class 8 Maths Question 8.**

### The lateral surface area of a hollow cylinder is 4224 cm^{2}. It is cut along its height and formed a rectangular sheet of width 33 cm. Find the perimeter of the rectangular sheet.

**Solution:**Breadth of the rectangular sheet (b) = 33 cm

Lateral surface area of the hollow cylinder = Area of the rectangular sheet

4224 = l × b

4224 = l × 33

l = 4224/33

l = 128 cm

Perimeter of the rectangular sheet = 2(l + b) = 2(128 + 33) = 2 × 161 = 322 cm

Hence, the required perimeter is 322 cm.

**Ex ****9.2**** Class 8 Maths Question 9.**

### A road roller takes 750 complete revolutions to move once over to level a road. Find the area of the road if the diameter of a road roller is 84 cm and length is 1 m.

**Solution:**Radius of road roller = 84/2 = 42 cm and height = 1 m = 100 cm

The lateral surface area of the road roller = 2Ï€rh

^{2}.

**Ex ****9.2**** Class 8 Maths Question 10.**

### A company packages its milk powder in a cylindrical container whose base has a diameter of 14 cm and height 20 cm. The company places a label around the surface of the container (as shown in the figure). If the label is placed 2 cm from top and bottom, what is the area of the label?

**Solution:**Here, r = 14/2 = 7 cm

Height of the cylindrical label = 20 – (2 + 2) = 16 cm

Surface area of the cylindrical-shaped label = 2Ï€rh

= 2 × 22/7 × 7 × 16

= 704 cm

^{2}

Hence, the required area of the label is 704 cm

^{2}.

**Related Links:**

**NCERT Solutions for Maths Class 9**

**NCERT Solutions for Maths Class 10**

**NCERT Solutions for Maths Class 11**