Question

How many tiles whose length and breadth are 10 cm and 5 cm respectively will be needed to fit in a rectangular region whose length and breadth are respectively:
(a) 40 cm and 20 cm

(b) 70 cm and 60 cm​

Answer 1

Answer:

Area of a rectangle = L×B (As the tiles are rectangle in shape)

L = 10 cm

B = 5 cm

Area = 10 × 5 = 50 cm²

1. Area of region = L×B

L = 40 cm

B = 20 cm

40 × 20 = 800 cm²

800 cm² ÷ 50 cm² = 80÷5 (Cutting of zero)

= 16 tiles are required

2. Area of region = L×B

L = 70 cm

B = 60 cm

70 × 60 = 4200 cm²

4200 cm² ÷ 50 cm ² = 420÷5 (Cutting of zero)

= 84 tiles are required

Answer 2

Information provided with us:

• The length and breadth of tiles are 10 cm and 5 cm respectively and will be needed to fit in a rectangular region whose length and breadth are 40 cm and 20 cm , 70 cm and 60 cm respectively .

What we have to calculate༄

➪ The required number of tiles

$\rm \bigstar \: a)$

✰Now :

➪ Area of rectangular region

$\rm\implies \:length \times breadth$

➡ Substitute the given values in above formula and solve

$\rm \implies \: 40 \: cm \times 20 \: cm$

$\bf \implies \: 800 \: c {m}^{2}$

✰So ༄

➪ Area of one tile

$\rm \implies \: 10 \: cm \times 5 \: cm$

$\bf \implies \: 50 \: c {m}^{2}$

✰Therefore ༄

➪Numbers of tiles

$\rm \mapsto \: \dfrac{Area \: of \: rectangular \: region}{Area \: of \: one \: tile}$

➡ Substitute the given values in above formula and solve

$\rm \implies \: \dfrac{800}{50}$

$\bf \implies \: 16$

$\rm \therefore \: Thus \: required \: numbers \: of\:tiles \: are$

$\implies\bigstar \: \: \sf\boxed{\bold{\gray{16}}}\: \: \: \bigstar$

$\rm \bigstar \: b)$

✰Now :

➪ Area of rectangular region

$\rm\implies \:length \times breadth$

➡ Substitute the given values in above formula and solve

$\rm \implies \: 70\: cm \times 60 \: cm$

$\bf \implies \: 4200 \: c {m}^{2}$

✰So ༄

➪ Area of one tile

$\rm \implies \: 10 \: cm \times 5 \: cm$

$\bf \implies \: 50 \: c {m}^{2}$

✰Therefore ༄

➪Numbers of tiles

$\rm \mapsto \: \dfrac{Area \: of \: rectangular \: region}{Area \: of \: one \: tile}$

➡ Substitute the given values in above formula and solve

$\rm \implies \: \dfrac{4200}{50}$

$\bf \implies \: 84$

$\rm \therefore \: Thus \: required \: numbers \: of\:tiles \: are$

$\implies\bigstar \: \: \sf\boxed{\bold{\gray{84}}}\: \: \: \bigstar$