Mensuration - 8th math

Mensuration Class Eighth Math NCERT 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?

 

Volume of given cuboids will give the amount material required to make them.

For Cuboid A

Length (l) = 60cm

Width (b) = 40cm

And, height (h) = 50cm

Thus volume = ?

We know that, volume of a cuboid = Length × Width × Height

Thus, volume of the given cuboid = 60cm × 40cm × 50cm

= 120000 cm³

Thus, volume of the cuboid A = 120000cm³

Thus, 120000cm³ material will be required to make cuboid A

For Cuboid B

Length (l) = 50cm

Width (b) = 50cm

And, height (h) = 50cm

Thus volume = ?

We know that, volume of a cuboid = Length × Width × Height

Thus, volume of the given cuboid = 50cm × 50cm × 50cm

= 125000 cm³

Thus, volume of the cuboid A = 125000cm³

Thus, 125000cm³ material will be required to make cuboid A

Thus, cuboid which width is equal to 40cm will required lesser amount of material to make Answer

Mensuration Class Eighth Math NCERT Exercise 11.3 Question (2) A suitcase with measures 80cm × 48cm × 24cm is to be covered with a tarpaulin cloth. How many meters of tarpaulin of width 96cm is required to cover 100 such suitcases?

Solution

The total surface area of suitcase divided by width of cloth will given the length of cloth required to cover one given suitcase.

Given, Length (l) of the suitcase = 80cm

= 80/100 m = 0.8m

And breadth (b) of the suitcase = 48cm

= 48/100 m = 0.48m

And height (h) of the suitcase = 24cm

= 24/100m = 0.24m

And width of tarpaulin cloth = 96cm

= 96/100m = 0.96m

Thus, total length of tarpaulin cloth required to cover 100 such suitcase = ?

We know that, total surface area of a cuboid = 2(lb + bh + hl)

Where l = length, b = breadth and h = height of the cuboid

Thus, total surface area of given suitcase = 2[(0.8 × 0.48) + (0.48 × 0.24) + (0.24 × 0.8)] square meter

= 2(0.384 + 0.1152 + 0.192) m²

= 2 × 0.6912 m²

= 1.3824 m²

Thus, total surface area of suitcase = 1.3824 m²

Now, since total surface area of suitcase = 1.3824 m²

Thus, 1.3824 m² of cloth will be required to cover one such suitcase.

Now, given width of cloth = 0.96 m

Let length of the cloth = k

And we know that, area of a rectangle = length × width

Thus, area of cloth = length × width

⇒ 1.3824 m² = k × 0.96m

`=>k = (1.3824m^2)/(0.96m) `

⇒ Length of cloth (k) = 1.44m

Thus, 1.44m length of cloth of given width will be required to cover one suitcase.

Now, since to cover 1 suitcase length of cloth required = 1.44m

Thus, to cover 100 suitcase length of cloth required = 1.44m × 100 = 144m

Thus, length of cloth required to cover given 100 suitcase = 144m Answer

Mensuration Class Eighth Math NCERT Exercise 11.3 Question (3) Find the side of a cube whose surface area is 600cm²

Solution

Given, surface area of a cube = 600cm²

Thus, side of the given cube = ?

Let side of the cube = s

We know that, Total surface area of a cube = 6 × side²

Thus, surface area of given cube = 6 × side²

⇒ 600cm² = 6 × s²

⇒ 6 × s² = 600cm²

`=>s^2=(600cm^2)/6`

⇒ s² = 100 cm²

`=>s=sqrt(100cm^2)`

⇒ s = 10cm

Thus, side of the given cube = 10cm Answer

Mensuration Class Eighth Math NCERT Exercise 11.3 Question (4) Rukhsar painted the outside of the cabinet of measure 1m × 2m × 1.5m. How much surface area did she cover if she painted all except the bottom of the cabinet.

Solution

Given, length (l) of the cabinet = 2m

Breadth (b) of the cabinet = 1m

And height (h) of the cabinet = 1.5m

Thus, total surface area except bottom = ?

We know that, total surface area of a cuboid = 2(lb + bh + hl)

Where l = length, b = breadth and h = height of the cuboid

Thus, total surface area of given cabinet having cuboidal shape = 2[(2m × 1m) + (1m × 1.5m) + (1.5m × 2m)]

= 2(2m² + 1.5m² + 3m²)

= 2 × 6.5m²

= 13m²

Thus, total surface area of cuboidal shaped cabinet = 13m²

Now, the base of the cabinet is of rectangular shape

In which length = 2m and breadth = 1m

Now, we know that, area of a rectangle = length × breadth

Thus, area of base of the cabinet = 2m × 1m

⇒ Area of base of the cabinet = 2m²

Now, surface area of cabinet except base = total surface area of cabinet – area of base of the cabinet

= 13m² – 2m²

= 11m²

Thus, area of cabinet to cover while painting except base of the cabinet by Rkhsar = 11m² Answer

Mensuration Class Eighth Math NCERT Exercise 11.3 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 100m² of area is painted. How many such cans of paint will she need to paint the room?

Solution

The inner surface area except floor of the cuboidal hall will give the area divided by area can be painted in one can will give the number of can required.

Given, length of a cuboidal hall (l) = 15m

Breadth of the cuboidal hall (b) = 10m

And, height of the cuboidal hall (h) = 7m

Form each can area is painted = 100m²

Thus, number of can required to paint the hall except floor = ?

We know that, total surface area of a cuboid = 2(lb + bh + hl)

Where l = length, b = breadth and h = height of the cuboid

Thus, surface area of the given cuboidal hall = 2[(15m × 10m) + (10m × 7m) + (7m × 15m)]

= 2(150m² + 70m² + 105m²)

= 2 × 325m²

= 650 m²

Thus, surface area of cuboidal hall = 650m²

Now, we know that, Area of a rectangle = Length × Breadth

Thus, area of floor of the hall = 15m × 10m

⇒ Area of floor of the hall = 150m²

Now, surface area of the cuboidal hall to be painted = Surface area of hall – Area of floor

= 650m² – 150m²

⇒ Area of hall to be painted = 500m²

Calculation of number of can required to paint the hall

∵ To paint 100m² number of can required = 1

∴ To paint 1m² number of can required `=1/100`

∴ To paint 500m² number of can required `=1/100xx500=5`

Thus, number of can required to paint the given hall = 5 Answer

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