Question

The room of length 12 m and breadth 8 m needs to be covered with tiles. What is the maximum
size of the square tile required to cover the floor?​

Answer 1

Answer:

${ {3}^{2} }^{3} \times {(2 \times {3}^{5}) }^{ - 2} \times {18}^{2 } \\ \\ = {3}^{6} \times \frac{1}{4 \times {3}^{10} } \times {18}^{2} \\ \\ = \frac{ {18}^{2} }{4 \times {3}^{4} } \\ \\ = \frac{ \cancel{18 } \: ^{ \cancel{6}} \: ^{ \cancel{2}}\times { \cancel{18}} \: ^{ \cancel{6 }} \: ^{ \cancel{2}} \: ^1 }{ { \cancel{4 }\:_1}\times { \cancel{3} \: _1} \times { \cancel{3 } \: _1}\times { \cancel{3} \:_1} \times { \cancel{3} \: _1} } \\ \\ = { \huge{ \red{ \boxed{1}}}}$

Answer 2

Answer:

600 tile

Step-by-step explanation:

Length of floor = 6 m = 600 cm

Breadth of floor = 4 m = 400 cm

Area of floor = Length \times BreadthLength×Breadth

Area of floor = 600 \times 400600×400

Area of floor = 240000 cm^2240000cm

2

Area requires by 1 tile = 400 sq.cm.

No. of tiles required for the floor = \frac{240000}{400}

400

240000

= 600