The length and breadth of a rectangular hall are 40 m and 18 m respectively. Find the number of tiles $1 m \times 50 c m$ that will cover the floor of the ball.

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Solution

Length of a rectangular hall $L = 40 m$
Breadth of the rectangular hall $B = 18 m$
So, the area of hall
$= L \times B = 40 \times 18 = 720 m^{2}$
Since,
Length of tiles $l = 1 m$
Breadth of tiles $b = 50 c m = 0.5 m$
So, the area of each tile
$= l \times b = 1 \times 0.5 = 0.5 m^{2}$
Therefore, the numbers of tiles are required
$= \frac{720}{0.5} = 1440$

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The length and breadth of a rectangular hall are 40 m and 18 m respectively. Find the number of tiles 1m×50cm that will cover the floor of the ball.

Length of a rectangular hall L=40 mBreadth of the rectangular hall B=18 mSo, the area of hall=L×B=40×18=720 m2Since,Length of tiles l=1 mBreadth of tiles b=50 cm=0.5 mSo, the area of each tile=l×b=1×0.5=0.5 m2Therefore, the numbers of tiles are required=7200.5=1440

The length and breadth of a rectangular hall are 40 m and 18 m respectively. Find the number of tiles $1 m \times 50 c m$ that will cover the floor of the ball.