Here we will learn how to solve the application problems on Volume of cuboid using the formula.
Formula for finding the volume of a cuboid
Volume of a Cuboid (V) = l × b × h;
Where l = Length, b = breadth and h = height.
1. A field is 15 m long and 12 m broad. At one corner of this field a rectangular well of dimensions 8 m × 2.5 m × 2 m is dug, and the dug-out soil is spread evenly over the rest of the field. Find the rise in the level of the rest of the field.
Solution:
The volume of soil removed = The Volume of the Well
= 8 m × 2.5 m × 2 m
= 8 × 2.5 × 2 m^{3}
= 40 m^{3}
Let the level of the rest of the field be raised by h.
The volume of the soil spread evenly on the field
= Volume of the cuboid of dimensions + Volume of the cuboid of dimensions
= 2.5 m × 4 m × h + 12.5 m × 12 m × h
= (2.5 m × 4 m × h + 12.5 m × 12 m × h)
= (10h + 150h) m\(^{2}\)
= 160h m\(^{2}\)
Therefore, 160h m\(^{2}\) = 40 m^{3}
⟹ h = \(\frac{40}{160}\) m
⟹ h = \(\frac{1}{4}\) m
Therefore, the rise in the level = \(\frac{1}{4}\) m
= 25 cm
2. Squares each side 8 cm are cut off from the four corners of a sheet of tin measuring 48 cm by 36 cm. The remaining portion of the sheet is folded to form a tank open at the top. What will be the capacity of the tank?
Solution:
To make the tank, NGHP has to folded up along NP, LMQK along MQ, EFNM along MN and IJQP.
Now, MN = QP = (48 - 2 × 8) cm = 32 cm, and
NP = MQ = (36 - 2 × 8) cm = 20 cm.
EM = KQ = IP = GN = 8 cm.
Therefore, the capacity of the tank = 32 × 20 × 8 cm^{3}
= 5120 cm^{3}
= 5.12 litres [Since, 1 litre = 1000 cm^{3}]
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