Question

courtyard 20 m long and 15 m broad is to be paved with bricks of length 25 cm
and breadth 12 cm. Find the number of bricks required.​

Answer 1

$\large \pink{{Question:-}}$

A courtyard 20 m long and 15m broad is to be paved with bricks of length 25cm and breadth 12cm. Find the number of bricks required.

$\large \pink{{To\: Find:-}}$

Number of bricks required

$\large \pink{{Given:-}}$

• Length of the courtyard = 20 m

• Breadth of the courtyard = 15 m

• Length of the brick = 25 cm

• Breadth of the brick = 12 cm

$\large \pink{{Solution:-}}$

$Length \: of \: the \: courtyard \: = \: 20 \: m \: = \: ( \: 20 \: \times \: 100) cm \:$

$Breadth \: of \: the \: courtyard \: = \: 15m \: = (15 \times 100)cm$

$Area \: of \: the \: courtyard \: = (l ength \: \times breadth)$

$= (20 \times 100 \times 15 \times 100)sq \: cm$

$Length \: of \: the \: brick \: = 25cm$

$Breadth \: of \: the \: brick \: = 12 \: cm$

$Area \: of \: the \: brick \: = (25 \times 12)sq \: cm$

$⇒Required \: number \:of \: bricks \: = {(Area \: of \: the \: courtyard) \: }÷ (Area \: of \: a \: brick)$

$= \binom{20 \times 100 \times 15 \times 100}{25 \times 12} = 10000$

$Hence, the \: number \: of \: bricks \: required \: is \: 10000$

$\large \pink{{Additional\: information:-}}$

• Area of a rectangle = (length × breadth) sq units
• Area of a square = ( side × side) sq units
• Perimeter of a rectangle = 2(length × breadth) units.
• Perimeter of a square = 4 × (side)

Answer 2

Answer:

A courtyard 20 m long &15 m broad. To find number of Bricks. Required no of bricks = area of courtyard/ area of bricks. ... Number of bricks is 10000