Question

Pinky runs around a square field of side 75m, Rosy runs around a rectangular field with length 160m and breadth 105m. Who covers more distance by how much? plz plz answer for this question plz.....​

Answer 1

Given:

• Side of square field = 75 m
• Length of Rectangular field = 160 m
• Breadth of Rectanglular field = 105 m

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To find:

• Who cover more distance and by how much?

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$\underline{\bigstar\:\boldsymbol{According\:to\:the\:question\::}}$

• It is given that, both Pinky and Rosy ran around a square and rectangular field. So, In order to calculate Who cover more distance and by how much? We need to find Perimeter of both square and rectangular field. After that subtract the lesser value from the greater one.

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PINKY:

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$\setlength{\unitlength}{0.8cm}\begin{picture}(0,0)\thicklines\multiput(0,0)(4,0){2}{\line(0,1){4}}\multiput(0,0)(0,4){2}{\line(1,0){4}}\put(-0.5,-0.5){\bf D}\put(-0.5,4.2){\bf A}\put(4.2,-0.5){\bf C}\put(4.2,4.2){\bf B}\put(1.5,-0.6){\bf\large 75\ m}\put(4.4,2){\bf\large 75\ m}\end{picture}$

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Given that,

• Side of square field = 75 m

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$\dag\;{\underline{\frak{As\;we\;know\;that,}}}$

$\star\;{\boxed{\sf{\pink{Perimeter_{\;(square)} = 4 \times side}}}}$

Therefore,

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$:\implies\sf Perimeter_{\;(square\:field)} = 4 \times 75$

$:\implies{\underline{\boxed{\frak{\purple{Perimeter_{\;(square\:field)} = 300\:m}}}}}\;\bigstar$

$\therefore\:{\underline{\sf{Distance\:covered\:by\:Pinky\:is\: {\textsf{\textbf{300\:m}}}.}}}$

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ROSY:

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$\setlength{\unitlength}{0.9cm}\begin{picture}(0,0)\thicklines\multiput(0,0)(5,0){2}{\line(0,1){3}}\multiput(0,0)(0,3){2}{\line(1,0){5}}\put(0.03,0.02){\framebox(0.25,0.25)}\put(0.03,2.75){\framebox(0.25,0.25)}\put(4.74,2.75){\framebox(0.25,0.25)}\put(4.74,0.02){\framebox(0.25,0.25)}\multiput(2.1,-0.7)(0,4.2){2}{\sf\large 160 m}\multiput(-1.4,1.4)(6.8,0){2}{\sf\large 105 m}\put(-0.5,-0.4){\bf A}\put(-0.5,3.2){\bf D}\put(5.3,-0.4){\bf B}\put(5.3,3.2){\bf C}\end{picture}$

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Given that,

• Length and breadth of rectanglular is 160 m and 105 m respectively.

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$\dag\;{\underline{\frak{Perimeter\:of\:rectangle\:is\;given\:by,}}}$

$\star\;{\boxed{\sf{\pink{Perimeter_{\;(rectangle)} = 2(length + breadth)}}}}$

Therefore,

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$:\implies\sf Perimeter_{\;(Rectanglular\:field)} = 2(160 + 105)$

$:\implies\sf Perimeter_{\;(Rectanglular\:field)} = 2 \times 265$

$:\implies{\underline{\boxed{\frak{\purple{Perimeter_{\;(Rectanglular\:field)} = 530\:m}}}}}\;\bigstar$

$\therefore\:{\underline{\sf{Distance\:covered\:by\:Rosy\:is\: {\textsf{\textbf{530\:m}}}.}}}$

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Hence, Rosy covered more distance than Pinky.

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☯ Now, Finding their diffrence,

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$:\implies\sf Distance\:covered\:by\:Rosy - Distance\:covered\:by\;Pinky$

$:\implies\sf 530 - 300$

$:\implies{\underline{\boxed{\frak{\pink{230\:m}}}}}\;\bigstar$

$\therefore\:{\underline{\sf{Rosy\:ran\:more\:distance\;than\;Pinky\;by\; {\textsf{\textbf{230\:m}}}.}}}$

Answer 2

Answer:

Given :-

• Pinky cover a square field of side = 75 m
• Rosy cover a rectangular field of length 160 m and breadth 105 m

To Find :-

• Cover more distance
• By how much

Solution :-

As we know that

$\mathfrak \red {Perimeter (rectangle) = 2(l + b)}$

$\mathfrak \red {Perimeter \: (square) = 4 \times side}$

Distance covered by Pinky

$\tt \: Perimeter = 4 \times side$

$\tt \: Perimeter = 4 \times 75$

$\tt \: Perimeter = 300 \: m$

Pinky covers a distance of 300 m

$\tt \: Perimeter = 2(l + b)$

$\tt \: Perimeter = 2(160 + 105)$

$\tt \: Perimeter = 2(265)$

$\tt \: Perimeter = 530 \: m$

$\tt \therefore \: Rosy \: cover \: more \: distance \: by(530 - 300) = 230 \: m$