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  1. Asked: In:

    Daniel is paining 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?

    Best Answer
    Kanita

    Here, Length of wall (l) = 15 m, Breadth of wall (b) = 10 m and Height of wall (h) = 7 m ∴ Total Surface area of classroom = lb + 2(bh+hl) = 15 x 10 + 2 (10 x 7 + 7 x 15) = 150 + 2 (70 + 105) = 150 + 350 = 500 m² Now Required number of cans = Area of hall/Area of one can =500/100=5cms Hence, 5 cansRead more

    Here,
    Length of wall (l) = 15 m, Breadth of wall (b) = 10 m and Height of wall (h) = 7 m
    ∴ Total Surface area of classroom = lb + 2(bh+hl)
    = 15 x 10 + 2 (10 x 7 + 7 x 15)
    = 150 + 2 (70 + 105) = 150 + 350 = 500 m²
    Now Required number of cans = Area of hall/Area of one can =500/100=5cms
    Hence, 5 cans are required to paint the room.

    Class 8 Maths Chapter 11 Exercise 11.3 Solution in Video

    for more answers vist to:
    https://www.tiwariacademy.com/ncert-solutions/class-8/maths/chapter-11/

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  2. Asked: In:

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

    Kanita

    Given: Length of suitcase box (l) = 80 cm, Breadth of suitcase box (b) = 48 cm And Height of cuboidal box (h) = 24 cm ∴ Total surface area of suitcase box = 2(lb+bh+hl) = 2 (80 x 48 + 48 x 24 + 24 x 80) = 2 (3840 + 1152 + 1920) = 2 x 6912 = 13824 cm² Area of Tarpaulin cloth = Surface area of suitcasRead more

    Given: Length of suitcase box (l) = 80 cm,
    Breadth of suitcase box (b) = 48 cm
    And Height of cuboidal box (h) = 24 cm
    ∴ Total surface area of suitcase box = 2(lb+bh+hl)
    = 2 (80 x 48 + 48 x 24 + 24 x 80)
    = 2 (3840 + 1152 + 1920)
    = 2 x 6912 = 13824 cm²
    Area of Tarpaulin cloth = Surface area of suitcase
    ⇒ lxb = 13824
    ⇒ lx96 = 13824
    ⇒ l= 13824×96=144cm
    Required tarpaulin for 100 suitcases = 144 x 100 = 14400 cm = 144 m
    Hence, the tarpaulin cloth required to cover 100 suitcases is 144 m.

    Class 8 Maths Chapter 11 Exercise 11.3 Solution in Video

    Class 8 Maths Chapter 11 Exercise 11.4 Solution in Video

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  3. Asked: In:

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

    Best Answer
    Kanita

    (a) Given: Length of cuboidal box (l) = 60 cm Breadth of cuboidal box (b) = 40 cm Height of cuboidal box (h) = 50 cm ∴ Total surface area of cuboidal box = 2(lb+bh+hl) = 2 (60 x 40 + 40 x 50 + 50 x 60) = 2 (2400 + 2000 + 3000) = 2 x 7400 = 14800 cm² (b) Given: Length of cuboidal box (l)= 50 cm BreadRead more

    (a) Given: Length of cuboidal box (l) = 60 cm
    Breadth of cuboidal box (b) = 40 cm
    Height of cuboidal box (h) = 50 cm
    ∴ Total surface area of cuboidal box = 2(lb+bh+hl)
    = 2 (60 x 40 + 40 x 50 + 50 x 60)
    = 2 (2400 + 2000 + 3000)
    = 2 x 7400 = 14800 cm²

    (b) Given: Length of cuboidal box (l)= 50 cm
    Breadth of cuboidal box (b) = 50 cm
    Height of cuboidal box (h) = 50 cm
    ∴ Total surface area of cuboidal box = 2(lb+bh+hl)
    = 2 (50 x 50 + 50 x 50 + 50 x 50)
    = 2 (2500 + 2500 + 2500)
    = 2 x 7500 = 15000 cm²

    Hence, the cuboidal box (a) requires the lesser amount of material to make, since
    surface area of box (a) is less than that of box (b).

    Class 8 Maths Chapter 11 Exercise 11.3 Solution in Video

    for more answers vist to:
    https://www.tiwariacademy.com/ncert-solutions/class-8/maths/chapter-11/

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