0

Dimension of garden:

Length = 120 metre

Breadth = 90 metre

Dimension of brick:

Length = .25 metre

Breadth = .125 metre

Thickness= .08 metre

Dimension of wall:

Height = 2 metre

Thickness = .25 metre

[P.S. This not a homework rather a mathematical problem related to Solid Geometry.]

Jayed Yeameen
  • 11
  • Let's start with the garden. Do you need to lay only one layer of bricks, or ... ? And then: consider a $1~\text{m} \times 1~\text{m}$ area. How many bricks are needed to cover that? – Matti P. Sep 04 '19 at 12:43
  • 1
    Did you try finding volume of a brick and whole wall. Volume=Area . height – Rishi Sep 04 '19 at 12:48
  • Brick Thickness= .08 metre but Wall Thickness is 0.25 meters, This means you would have multiple layers of bricks and (may have to break some). – NoChance Sep 04 '19 at 13:00
  • @NoChance: Perhaps you could arrange the bricks so that the 'length' of the brick corresponded to the 'thickness' of the wall. – paw88789 Sep 04 '19 at 13:02
  • @paw88789 Thank you for your comment, I think the question got to be clear on this unless it is obvious...For example the Brick Length =Wall Thickness= 0.25 meter but I guess the OP should state the constraints (unless someone is willing to solve for all different ways, which would be an interesting problem) – NoChance Sep 04 '19 at 13:18

1 Answers1

1

The number of bricks required to make the wall along the full breadth is $$ \frac{90}{0.125} = 720 $$ Similarly for bricks along the length we need $$ \frac {120}{0.125} = 960$$ Now we have 4 overlapping areas so we need to deduct them $$ 4* \frac {0.25}{0.125}=8$$ Therefore in one layer we have $$ (720+960)-9 = 3352 $$ Therefore for the height of the wall which is 2m we will need 25 such layers So, We need $$3352*25 = 83800 $$ Therefore, we need a total of 83800 bricks in all such that the length of the bricks correspond to the thickness of the wall.

P.S. this problem can be solved by another possible arrangement in which the breadth of the brick corresponds to the thickness of the wall by fortune of co-incidence ;-) the answer of both the cases is the same This is because in the 2nd arrangement we cut one half of the brick and place it next to the other, i.e in the net case we just change the orientation of 2 bricks and since the no. of bricks in each row is the same it does'n change the no. of bricks.

Hope it helps :-)

rockin numbers
  • 61