Question

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 :
(i) 70 Cm And 36 Cm (ii) 144 Cm And 1 M

Answer 1

⇝Given :-

✧ How many tiles whose length and Breadth are 12cm and 5cm respectively will be needed to cover a rectangular region whose length and breadth are :

1. 70cm and 36cm

2.144cm and 1m

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⇝Formula Used :-

✧ Area of rectangle :

${\large{\red{\bigstar \: \: {\blue{\underbrace{\underline{\orange{\bf{Area = Length \times Breadth }}}}}}}}}$

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⇝Solution :-

❒ Area of tiles :

✧ Area of tiles :

${\large{:{\longmapsto{\bf{Area = Length \times Breadth}}}}}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: {\large{:{\longmapsto{\bf{12 \times 5}}}}}$

$\: \: \: {\large{\red{\dashrightarrow}} \: \: {\underline{\underline{\orange{\bf{60 {cm}^{2} }}}}}}$

❒ 1st Case :

✧ Area of rectangular region :

${\large{:{\longmapsto{\bf{Area = Length \times Breadth}}}}}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: {\large{:{\longmapsto{\bf{70 \times 36}}}}}$

$\: \: \: {\large{\red{\dashrightarrow}} \: \: {\underline{\underline{\orange{\bf{2520 cm}}}}}}$

✧ Tiles required :

${\large{:{\longmapsto{\bf{2520 \div 60}}}}}$

${\large{\purple{:{\longmapsto{\underline{\overline{\boxed{\bf{42 \: tiles}}}}}}}}}$

❒ 2nd Case :

✧ Area of rectangular region :

${\large{:{\longmapsto{\bf{Area = Length \times Breadth}}}}}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: {\large{:{\longmapsto{\bf{144 \times 100}}}}}$

$\: \: \: {\large{\red{\dashrightarrow}} \: \: {\underline{\underline{\orange{\bf{14400 \: cm}}}}}}$

✧ Tiles Required :

${\large{:{\longmapsto{\bf{14400 \div 60}}}}}$

${\large{\purple{:{\longmapsto{\underline{\overline{\boxed{\bf{240 \: tiles}}}}}}}}}$

❒ Hence :

${\large{\green{1.}{\underline{\pink{\pmb{\underline{\mathfrak{ \: 42 Tiles}}}}}}}}$

${\large{\green{2.}{\underline{\pink{\pmb{\underline{\mathfrak{ \: 240 \: Tiles}}}}}}}}$

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Answer 2

Answer:

⇝Formula Used :-

✧ Area of rectangle :

{\large{\red{\bigstar \: \: {\blue{\underbrace{\underline{\orange{\bf{Area = Length \times Breadth }}}}}}}}}★

Area=Length×Breadth

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⇝Solution :-

❒ Area of tiles :

✧ Area of tiles :

{\large{:{\longmapsto{\bf{Area = Length \times Breadth}}}}}:⟼Area=Length×Breadth

\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: {\large{:{\longmapsto{\bf{12 \times 5}}}}}:⟼12×5

\: \: \: {\large{\red{\dashrightarrow}} \: \: {\underline{\underline{\orange{\bf{60 {cm}^{2} }}}}}}⇢

60cm

2

❒ 1st Case :

✧ Area of rectangular region :

{\large{:{\longmapsto{\bf{Area = Length \times Breadth}}}}}:⟼Area=Length×Breadth

\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: {\large{:{\longmapsto{\bf{70 \times 36}}}}}:⟼70×36

\: \: \: {\large{\red{\dashrightarrow}} \: \: {\underline{\underline{\orange{\bf{2520 cm}}}}}}⇢

2520cm

✧ Tiles required :

{\large{:{\longmapsto{\bf{2520 \div 60}}}}}:⟼2520÷60

{\large{\purple{:{\longmapsto{\underline{\overline{\boxed{\bf{42 \: tiles}}}}}}}}}:⟼

42tiles

❒ 2nd Case :

✧ Area of rectangular region :

{\large{:{\longmapsto{\bf{Area = Length \times Breadth}}}}}:⟼Area=Length×Breadth

\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: {\large{:{\longmapsto{\bf{144 \times 100}}}}}:⟼144×100

\: \: \: {\large{\red{\dashrightarrow}} \: \: {\underline{\underline{\orange{\bf{14400 \: cm}}}}}}⇢

14400cm

✧ Tiles Required :

{\large{:{\longmapsto{\bf{14400 \div 60}}}}}:⟼14400÷60

{\large{\purple{:{\longmapsto{\underline{\overline{\boxed{\bf{240 \: tiles}}}}}}}}}:⟼

240tiles

❒ Hence :

{\large{\green{1.}{\underline{\pink{\pmb{\underline{\mathfrak{ \: 42 Tiles}}}}}}}}1.

42Tiles

42Tiles

{\large{\green{2.}{\underline{\pink{\pmb{\underline{\mathfrak{ \: 240 \: Tiles}}}}}}}}2.

240Tiles

240Tiles

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Step-by-step explanation:

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