Here we will learn how to solve the problems on Volume and Surface Area of Cube and Cuboid:
1. Two cubes of edge 14 cm each are joined end to end to form a cuboid. Find the volume and the total surface area of the cuboid.
Solution:
The volume of the cuboid = 2 × volume of one cube
= 2 × 14\(^{3}\) cm\(^{3}\)
= 5488 cm\(^{3}\)
The total surface area of the cuboid = 2(28 × 14 + 14 × 14 + 14 × 28) cm\(^{2}\)
= 2(28 + 14 + 28) × 14 cm\(^{2}\)
= 2(2 × 14 + 1 × 14 + 2 × 14) × 14 cm\(^{2}\)
= 2(2 + 1 + 2) × 14 × 14 cm\(^{2}\)
= 2(5) × 14 × 14 cm\(^{2}\)
= 10× 14 × 14 cm\(^{2}\)
= 1960 cm\(^{2}\)
2. The dimensions of the base of a rectangular vessel are 60 cm × 45 cm. Its height is 20 cm. The vessel is half-filled with water. What should be the size of a solid iron cube which when dropped into the vessel will raise the water level up to the brim?
Solution:
The volume of the vessel = 60 × 45 × 20 cm\(^{3}\)
= 54000 cm(^{3}\)
It is half-filled with water
So, the volume of the empty portion of the vessel = \(\frac{1}{2}\) × 54000 cm\(^{3}\)
= 27000 cm\(^{3}\)
The volume of the iron cube should be 27000 cm\(^{3}\) so that it displaces this amount of water and the water level comes up to the brim.
If the edge of the cube is x then,
x\(^{3}\) = 27000 cm\(^{3}\)
⟹ x\(^{3}\) = (30)\(^{3}\) cm\(^{3}\)
⟹ x = 30 cm
Therefore, the edge of the iron cube = 30 cm
The size of the cube should be 30 cm × 30 cm × 30 cm.
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