How many tiles whose length and breadth are 12 cm and 5 cm respectively will be needed to cover a rectangular region whose length and breadth are respectively:
ii) 144 cm and 1 m.

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Solution

No. of tiles $= \frac{A r e a o f r e g i o n}{A r e a o f 1 t i l e}$
$= \frac{144 \times 100}{12 \times 5}$
$= 240$ tiles.

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Q. How many tiles whose length and breadth are 12 cm and 5 cm respectively will be needed to fit in a rectangular region whose length and breadth are 100 cm and 144 cm respectively?

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Question 12(a)

How many tiles whose length and breadth are 12 cm and 5 cm respectively will be needed to fit in a rectangular region whose length and breadth are 100 cm and 144 cm respectively?

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(i) 100 cm and 144 cm
(ii) 70 cm and 36 cm

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How many tiles whose length and breadth are 12 cm and 5 cm respectively will be needed to cover a rectangular region whose length and breadth are respectively:ii) 144 cm and 1 m.

No. of tiles =AreaofregionAreaof1tile=144×10012×5=240 tiles.

How many tiles whose length and breadth are 12 cm and 5 cm respectively will be needed to cover a rectangular region whose length and breadth are respectively:
ii) 144 cm and 1 m.

No. of tiles $= \frac{A r e a o f r e g i o n}{A r e a o f 1 t i l e}$
$= \frac{144 \times 100}{12 \times 5}$
$= 240$ tiles.