If out of the six faces of a cuboid, we only find the sum of the areas of four faces leaving the bottom and top faces. This sum is called the Lateral surface area of the cuboid. Consider a cuboid of the length l, breadth b, and height h Lateral surface area of the cuboid = Area of face ABGF + Area of face BCHG + Area of face EFGH + Area of face ADEF = 2(b x h) + 2(l x h) = 2(l + b) x h = Perimeter of base x height Note: Base of a cuboid is a rectangle. |
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Illustration: A cubical box has each edge 10cm and another cuboidal box is 12.5cm long, 10cm wide, and 8cm high. Which box has the greater lateral surface area and by how much?
Solution: Let the lateral surface areas of the cubical and cuboidal boxes be Area1 and Area2 respectively. Then, Area 1 = = 4 x 10 x 10 =
Area 2 = 2 (Length + Breadth) x Height = 2 (12.5 + 10) x 8 =
Area of cube is more than the area of cuboid = 400 - 360 =
Illustration: Find the area of the four walls of a room whose length is 6 m, breadth 5 m and height 4 m.Also find the cost of white-washing the walls, if the rate of whitewashing is Rs 5 per square metre.
Solution: Given: l = 6 m, b = 5 m, and h = 4 m
Area of the walls = 2h(l + b)
Cost of white washing of 1 square metre = Rs. 5
So, Cost of white washing the walls = Rs(5 X 88) = Rs 440
The floor of a rectangular hall has a perimeter 250m. If the cost of painting the four walls at the rate of Rs.10 per sq m is Rs 15000, find the height of the wall. | |||
Right Option : A | |||
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The length and the height of a cuboidal box are 10 cm and 15 cm respectively. If its lateral surface area is 1200 | |||
Right Option : D | |||
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The length, breadth and height of a cuboidal tank are 1.2 m, 1.5 m and 1 m respectively. The lateral surface area of the tank is equal to _______________ | |||
Right Option : B | |||
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