Physics, asked by arnavvats2020, 3 months ago

Please solve past
Resistance between A and b
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Answered by BRAINLYking024
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Answer:

❶ No. of tiles in 100cm and 144cm \large\leadsto\boxed{\tt\purple{240 \: tiles}}⇝240tiles

❷ No. of tiles in 70cm and 36cm \large\leadsto\boxed{\tt\pink{42 \: tiles}}⇝42tiles

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• GIVEN:-

Dimension of tiles is 5cm and 12cm

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• To Find:-

Number of tiles needed to fit in region of

100cm and 144cm

70cm and 36cm

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• Solution:-

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1.)

Given that,

Length of the tiles = 5cm

Length of the tiles = 5cmBreadth of the tiles = 12cm

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\mathrm{Therefore,}Therefore,

\large\underline{\boxed{\bf\green{Area = length \times breadth}}}Area=length×breadth

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➪ \sf Area = 12 \times 5Area=12×5Area=12×5Area=12×5

★ \bf\red{Area = 60 \: cm^2}Area=60cm2

Also,

Length of the rectangular region = 100cm

Length of the rectangular region = 100cmBreadth of the rectangular region = 144cm

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\mathrm{Therefore,}Therefore,

\large\underline{\boxed{\bf\green{Area = length \times breadth}}}Area=length×breadth

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➪ \sf Area = 100 \times 144Area=100×144Area=100×144Area=100×144

★ \bf\red{Area = 14400 \: cm^2}Area=14400cm2

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\mathrm{Hence,}Hence,

The tiles required to fit in the given region will be,

No. of tiles = Area of the region / Area of the one tile

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➪ \sf No. \: of \: tiles = \dfrac{14400}{60}No.oftiles=6014400

★ \large{\bf\pink{No. \: of \: tiles = 240 \: tiles}}No.oftiles=240tiles

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2.)

Length of the rectangular region = 70cm

Length of the rectangular region = 70cmBreadth of the rectangular region = 36cm

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\large\underline{\boxed{\bf\green{Area = length \times breadth}}}Area=length×breadth

➪ \sf Area = 70 \times 36Area=70×36Area=70×36Area=70×36

★ \bf\red{Area = 2520 \: cm^2}[tex]⠀⠀⠀⠀⠀⠀⠀⠀ < /p > < p > < /p > < h3 > Now, < /h3 > < p > < /p > < p > < strong > We know the area of 1 tile = 60 cm² < /strong > < /p > < p > < /p > < p > < /p > < p > [tex]\red\mathmm{Therefore,}Area=2520cm2[tex]⠀⠀⠀⠀⠀⠀⠀⠀</p><p></p><h3>Now,</h3><p></p><p><strong>Weknowtheareaof1tile=60cm²</strong></p><p></p><p></p><p>[tex]\mathmmTherefore,

No. of tiles = Area of the region / Area of the one tile

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➪ \sf No. \: of \: tiles = \dfrac{2520}{60}No.oftiles=602520

★ \large{\bf{\blue{No. \: of \: tiles = 42 \: tiles}}}No.oftiles=42tiles

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