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 ..? 

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 × 12 cm × 8 cm, How many bricks would be required?

Approximately 4167 bricks would be required to build the wall.

To calculate the number of bricks required to build the wall, we need to find out how many bricks can fit into the volume of the wall.

First, let's calculate the volume of the wall:

1. The length of the wall = 10 m
2. The height of the wall = 4 m
3. The thickness of the wall = 24 cm = 0.24 m (since 1 cm = 0.01 m)

So, the volume of the wall is: Volume = length × height × thickness Volume = 10 m × 4 m × 0.24 m Volume = 9.6 cubic meters

Now, let's calculate the volume of one brick:

1. Length of the brick = 24 cm = 0.24 m
2. Width of the brick = 12 cm = 0.12 m
3. Height of the brick = 8 cm = 0.08 m

So, the volume of one brick is: Volume of one brick = length × width × height Volume of one brick = 0.24 m × 0.12 m × 0.08 m Volume of one brick = 0.002304 cubic meters

Now, to find the number of bricks required, we divide the volume of the wall by the volume of one brick:

Number of bricks = Volume of wall / Volume of one brick Number of bricks = 9.6 cubic meters / 0.002304 cubic meters Number of bricks ≈ 4166.67

Since we can't have a fraction of a brick, we round up to the nearest whole number.

So, approximately 4167 bricks would be required to build the wall.

Total Surface Area of a Cuboid

The total surface area A of a cuboid can be calculated using the formula:

Total Surface Area (A) = 2lw + 2lh + 2wh

Where:

• l is the length of the cuboid.
• w is the width of the cuboid.
• h is the height of the cuboid.

You add together the areas of all six faces: two times the area of the length times width, two times the area of the length times height, and two times the area of the width times height.