Question

the floor of a room is 8cm 50cm long ad 7m 25cm wide. find the largest size of each tile and hence find the least number of tiles requried​​

Answer 1

Answer:

Length of the floor =8 m 96 cm=896 cm

Breadth of the floor =6 m 72 cm=672 cm

H.C.F. of 896 and 672 is =2×2×2×2×2×7=224

So, the required size of the square tile must be 224×224

Hence, the minimum number of square tiles of the same size needed to cover the floor =

Area of one square tile

Area of the floor

=

(224×224)

(896×672)

=

50176

602112

=12

Hence, 12 square tiles each of 224 cm×224 cm will be needed to cover the entire floor.

Answer 2

Answer:

The largest size of each tile is 625cm² and the least number of tiles required is 986.​​

Step-by-step explanation:

Given

Length of a room(l)= 8m50 cm=800+50=850 cm

Breadth of a room(b)=7 m 25 cm=700+25=725 cm

To find the largest size of each tile and hence find the least number of tiles required​​

Solution:

From the given quantities, it is confirmed that the floor of the room is rectangular

So,

$Area of rectangle(A)=length(l)xbreadth(b)\\\\\\A=850x725\\\\A=616250m^{2}$

Now,

Largest size of each tile= HCF of 850 and 725

850=5x5x34

725=5x5x29

=5x5

=25cm

The largest size of tile = Length = 25x25=625cm²

Again,

Least number of tiles required= Area if room/Area of each tile

Number of tiles=616250/625

Number of tiles=986

Therefore, the largest size of each tile is 625cm² and the least number of tiles required​​ is 986.

To find the size of tiles and area of the floor:

https://brainly.in/question/10752118

https://brainly.in/question/24544