A wall of length 10, m was to be built across | vedclass

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(B) The wall is a cuboid. First,convert all dimensions to centimeters:Length of the wall $= 10\, m = 1000\, cm$.Thickness of the wall $= 24\, cm$.Heig

Vedclass · Accepted Answer

$4167$

Vedclass · Answer

(B) The wall is a cuboid. First,convert all dimensions to centimeters:Length of the wall $= 10\, m = 1000\, cm$.Thickness of the wall $= 24\, cm$.Height of the wall $= 4\, m = 400\, cm$.Volume of the wall $= 1000\, cm 	imes 24\, cm 	imes 400\, cm = 9,600,000\, cm^3$.Volume of one brick $= 24\, cm 	imes 12\, cm 	imes 8\, cm = 2304\, cm^3$.Number of bricks required $= \frac{	ext{Volume of the wall}}{	ext{Volume of one brick}} = \frac{9,600,000}{2304} \approx 4166.67$.Since we cannot have a fraction of a brick,we round up to the nearest whole number,which is $4167$ bricks.

$A$ wall of length $10\, m$ was to be built across an open ground. The height of the wall is $4\, m$ and thickness of the wall is $24\, cm$. If this wall is to be built up with bricks whose dimensions are $24\, cm \times 12\, cm \times 8\, cm$,how many bricks would be required?

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