Question

dimensions of a tile is 1.6 into 9 cm what how many times should be required to cover a square surface of 18 m ​

Answer 1

$\huge\mathbb{\underline{SOLUTION:-}}$

$\sf\underline{\purple{\:\:\:\: Given\:\:\:\:}}$

$\sf\bullet Dimensions\:of\:tiles \\ \\ \sf\bullet \:\:\: 1.6\times 9cm\\ \\ \sf\bullet Side\:of\: square=18m\\ \\ \sf\bullet or\:\:1800cm$

$\sf\underline{\blue{\:\:\:\: Solution\:\:\:\:}}$

• First the Area of square

$\sf\implies Area\:of\:Square=side{}^{2}\\ \\ \sf:\implies Area=(1800){}^{2}\\ \\ \sf:\implies Area=3240,000cm{}^{2}$

• Now the Area of tiles

$\sf\implies Area\:of\: rectangle=l\times b\\ \\ \sf:\implies Area\:of\:tiles = 1.6\times 9\\ \\ \sf:\implies Area=14.4cm{}^{2}$

Now the no. of tiles

$\sf\:\:No.\:of\:tiles =\dfrac{Area\:of\:square}{Area\:of\: tiles}\\ \\ \sf:\implies Tiles=\cancel{\dfrac{3240,000}{14.4}}\\ \\ \sf:\implies Tiles =225000$

$\boxed{\sf{ Tiles\: required=225,000}}$

Answer 2

Answer:

$\large\boxed{\sf{225000}}$

Step-by-step explanation:

Dimensions of the tiles.

• Length, l = 1.6 cm
• Breadth, b = 9 cm

Therefore, Area, a = l × b

=> a = 1.6 × 9 sq. cm

Now, side of square surface, s = 18 m = 1800 cm

.°. Area of surface, A = (1800×1800) sq. cm

.°. No. of tiles required = A/a

$= \dfrac{1800 \times 1800}{1.6 \times 9} \\ \\ = \frac{1800 \times 200}{1.6} \\ \\ = \frac{1800 \times 50}{0.4} \\ \\ = \frac{1800 \times 500}{4} \\ \\ = 900 \times 250 \\ \\ = 225000$

Hence, no. tiles required = 225000