Class 8 Maths


Mensuration

Exercise 11.3

Question 1: There are two cuboidal boxes as shown in the adjoining figure. Which box requires the lesser amount of material to make?

Cuboid and Cube

Answer: Volume of the Cuboid `= l xx \b\ xx \h`
`= 60 xx 40 xx 50 = 12000` cubic cms
Volume of Cube = Side3
`=50^3 = 125000`
As the volume of the cube is greater than that of the cuboid so it will require more material, and the cuboid will require less material

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?

Answer: Surface Area of Cuboid `= 2(lb + lh + bh)`
`= 2 (80 xx 48 + 80 xx 24 + 48 xx 24)`
`= 2(3840 + 1920 + 1152)`
`= 2 xx 6912 = 13824`
Hence, Surface Area of 100 suitcases = 1382400 sq cms
Required tarpaulin will have same area

So, required length of tarpaulin `=text(Area)/text(Width)`

`=(1382400)/(96)=14400` cm = 14.4 m

Question 3: Find the side of a cube whose surface area is 600 cm2.

Answer: Surface Area `= 6 xx` side2
Or, `600 = 6 xx` side2
Or, Side2 `= 100`
Or, Side = 10 cm

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.

Answer: Surface Area of Cabinet `= 2(lb + lh + bh)`
`= 2(1 xx 2 + 1 xx 1.5 + 1.5 xx 2)`
`= 2(2 + 1.5 + 3) = 2 xx 6.5 = 13`
Area of Bottom Surface `= 1 xx 2 = 2` sq m
Hence, Area Covered by painting `= 13-2 = 11` sq m

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 m² of area is painted. How many cans of paint will she need to paint the room?

Answer: Area of All walls, ceilings and floor `= 2(lb + lh + bh)`
`= 2(15 xx 10 + 15 xx 7 + 10 xx 7)`
`= 2(150 + 105 + 70)`
`= 2 xx 325 = 650`
Area of Floor `= 15 xx 10 = 150`
Hence, Painted Area `=650-150=500`
As, one can of paint is enough for 100 sq m
So, `(500)/(100) = 5` cans are needed to paint the hall.

Alternate Method: Area of Walls of a Room `= 2 xx \h \xx (l + b)`
`= 2 xx 7 xx (15 + 10)`
`= 14 xx 25 = 350`
Area of Roof = Area of Floor = 150
Total Area to be painted `= 350+150=500`

Question 6: Describe how the two figures given below are alike and how they are different. Which box has larger lateral surface area?

Cylinder and Cube

Answer:
Similarity:
Their dimensions are equal
Difference: The figure on the left is cylindrical and that on the right is cuboidal.

Curved surface area of cylinder `=2πrh`

`=2xx(22)/(7)xx7/2xx7`

`=22xx7=154` sq cm

Surface area of cube `=4xx\text(Side)^2`

`=4xx7^2=4xx49=196` sq cm

It is clear that the cuboid is having a larger surface area.

Question 7: A closed cylindrical tank of radius 7 m and height 3 m is made from a sheet of metal. How much sheet of metal is required?

Answer: Surface area of cylinder `=2πr(r+h)`

`=2xx(22)/(7)xx7xx(7+3)`

`=44xx10=440` sq m

Question 8: The lateral surface area of a hollow cylinder is 4224 cm2. It is cut along its height and formed a rectangular sheet of width 33 cm. Find the perimeter of rectangular sheet?

Answer: Area of Rectangular Sheet = Curved Surface Area of the Cylinder
Area of Rectangle `= l \xx \b`
Or, `l\ xx \b = 4224`
`l \xx 33 = 4224`

Or, `l=(4224)/(33)=128`

Perimeter of rectangular sheet = 2 × (128 + 32) = 2 × 160 = 320 cm

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.

Answer: In one revolution the wheel of the roller will cover an area equal to its surface area.

Curved surface area of cylinder `=2πrh`

`=2xx(22)/(7)xx42xx100`
=26400 sq cm = 2.64 sq m

Area covered after 750 revolutions = 2.64 × 750 = 1980 sq m

Question 10: A company packages its milk powder in cylindrical container whose base has a diameter of 14 cm and height 20 cm. 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.

Cylinder

Answer: The cylindrical area covered by the label will have
height `= 20-4 = 16` cms

Curved surface area of cylinder `=2πrh`

`=2xx(22)/(7)xx7xx16=704` sq cm