Question

the length and the breadth of rectangular Hall is 15.5m and 6.4m. it is to be paved by square tiles each of side 80cm. how many such tiles will be required​

Answer 1

Answer:

The number of tiles required is 155.

Step-by-step-explanation:

We have given that,

For rectangular hall,

• Length = 15.5 m
• Breadth = 6.4 m

Side of square tile = 80 cm = ( 80 / 100 ) m = 0.8 m

The rectangular hall is to be paved by the square tiles.

We have to find the number of tiles required.

Now,

No. of tiles required = Area of rectangular hall / Area of square tile

We know that,

• Area of rectangle = Length * Breadth

• Area of square = ( side )²

⇒ No. of tiles required = ( 15.5 * 6.4 ) / ( 0.8 )²

⇒ No. of tiles required = ( 15.5 * 6.4 ) / 0.64

⇒ No. of tiles required = 15.5 * 6.4 ÷ 0.64

⇒ No. of tiles required = 15.5 * 10

⇒ No. of tiles required = 155

∴ The number of tiles required is 155.

Answer 2

Answer:

Given:

• Length of the rectangular hall is 15.5m.
• Breadth of the rectangular hall is 6.4m.
• Side of Square tile is 80cm.

To Find:

• Number of tiles required to pave the hall.

Solution:

${ \pink{\large{Area \: of \: Square ={ (Side)}^{2} }}}$

Here, Side of Square is 80cm

In m :

• 100cm = 1m
• 80cm = 0.8m

➝ (0.8)²

➝ 0.64

Therefore, Area of Square tile is 0.64m²

$\: \:$

${\large{ \pink{Area \: of \: Rectangle = l \times b}}}$

Here,

• Length (l) => 15.5m
• Breadth (b) => 6.4m

➝ 15.5 × 6.4

➝ 99.2

Therefore, Area of Rectangular hall is 99.2m²

$\: \:$

_______________________

Now,

${ \green{ \boxed{No. \: of \: tiles \: required = \frac{Area \: of \: Rectangular \: hall}{Area \: of \: Square \: tile } }}}$

${ \dashrightarrow{ \blue{\large{ \frac{99.2{m}^{2} }{0.64 {m}^{2} } ={ \red{ \boxed{155 \: tiles}}}}}}}$

Therefore, 155 tiles required to pave the rectangular hall.

________________________

${ \bold{\large{ \underline{ \sf{ \purple{Additional \: Information }}}}}}$

$\: \: \: \: \: { \pink{ \bf{Formula's \: for \: Area }}}$

★ Triangle => ½bh

★ Rectangle => Length × Breadth

★ Circle => πr²

★ Square => Side × Side

★ Parallelogram => Base × Height

★ Sphere => 4πr²