Question

10. Two plots of land have the same perimeter. One is a square of side 64 m and the other is a
rectangle of length 70 m. Find the breadth of the rectangular plot. Which plot has the
greater area and by how much?
JENO​

Answer 1

$\huge\bold{answer - }$

$\color{blue}according \: to \: the \: question \: there \: are \: 2 \: plots - \: \\ \pink{\star \: square}\\ \pink{\star \: rectangle}$

$\huge\bold \purple{given - }$

$\color{red}\dagger \: perimeter \: of \: square = perimeter \: of \: rectangle \\ \color{red}\dagger \: one \: side \: of \: square = 64 \: m \\ \color{red}\dagger \: length \: of \: rectangle \: = 70 \: m$

$\huge\bold\purple{find - }$

$\orange{\leadsto \:the \: breadth \: of \: the \: rectangle}\\ \orange{\leadsto \: which \: plot \: is \: greater \: and \: by \: how \: much \: ? }$

$\huge \sf{solution}$

$\circ \green{first \: we \: will \: solve \: square \: plot \: measurements}$

$\pink{as \: 1 \: side \: of \: square = 64} \\ \color{blue} so \: perimeter = 4 \times s = 4 \times 64 \\ = \fbox{256}$

$\dagger \: as \: perimeter \: of \: square \: and \: rectangle \: are \: equal \\ \therefore \: perimeter \: of \: rectangle = 256$

$\green{now \: lets \: solve \: rectangle \: measurement's }$

$\pink{\star \: length \: = 70m}\\ \pink{\star \: breadth = b}\\ \pink{\star \: perimeter = 256}$

$p = 2(70 + b) \\ p = 140 + 2b \\ 256 = 140 + 2b \\ 2b = 256 - 140 \\ 2b = 116 \\ \color{blue}b = \frac{116}{2} = \fbox{ 58}$

$\green{\hookrightarrow \: now \: lets \: find \: area \: of \: both \: plots}$

$\blacktriangleright \: area \: of \: rectangle = l \times b \\ = 70 \times 58 \\ = \color{blue}{4060 \: m}^{2}$

$\blacktriangleright \:area \: of \: square = s \times s \\ = 64 \times 64 \\ = \color{blue} {4096 \: m}^{2}$

$\color{purple}\therefore \: square \: plot \: is \: larger \: by \: 36 \: m$

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Answer 2

Step-by-step explanation:

llxXItzAshXxll

$\huge\bold{answer - }$

$\color{blue}according \: to \: the \: question \: there \: are \: 2 \: plots - \: \\ \pink{\star \: square}\\ \pink{\star \: rectangle}$

$\huge\bold \purple{given - }$

$\color{red}\dagger \: perimeter \: of \: square = perimeter \: of \: rectangle \\ \color{red}\dagger \: one \: side \: of \: square = 64 \: m \\ \color{red}\dagger \: length \: of \: rectangle \: = 70 \: m$

$\huge\bold\purple{find - }$

$\orange{\leadsto \:the \: breadth \: of \: the \: rectangle}\\ \orange{\leadsto \: which \: plot \: is \: greater \: and \: by \: how \: much \: ? }$

$\huge \sf{solution}$

$\circ \green{first \: we \: will \: solve \: square \: plot \: measurements}$

$\pink{as \: 1 \: side \: of \: square = 64} \\ \color{blue} so \: perimeter = 4 \times s = 4 \times 64 \\ = \fbox{256}$

$\dagger \: as \: perimeter \: of \: square \: and \: rectangle \: are \: equal \\ \therefore \: perimeter \: of \: rectangle = 256$

$\green{now \: lets \: solve \: rectangle \: measurement's }$

$\pink{\star \: length \: = 70m}\\ \pink{\star \: breadth = b}\\ \pink{\star \: perimeter = 256}$

$p = 2(70 + b) \\ p = 140 + 2b \\ 256 = 140 + 2b \\ 2b = 256 - 140 \\ 2b = 116 \\ \color{blue}b = \frac{116}{2} = \fbox{ 58}$

$\green{\hookrightarrow \: now \: lets \: find \: area \: of \: both \: plots}$

$\blacktriangleright \: area \: of \: rectangle = l \times b \\ = 70 \times 58 \\ = \color{blue}{4060 \: m}^{2}$

$\blacktriangleright \:area \: of \: square = s \times s \\ = 64 \times 64 \\ = \color{blue} {4096 \: m}^{2}$

$\color{purple}\therefore \: square \: plot \: is \: larger \: by \: 36 \: m$

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