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
(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
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?
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:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-8/maths/chapter-11/
There are two cuboidal boxes as shown in the adjoining figure. Which box requires the lesser amount of material to make?
(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:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-8/maths/chapter-11/