Question

Find the perimeter of a square carrom board of side 41/8m.​

Answer 1

Correct Question-:

• Find the perimeter of a square carrom board of side $\sf{\dfrac{41}{8}m}$.

AnswEr-:

• $\underline{\boxed{\star{\sf{\blue{ Perimeter_{(Square)}  \: = \: \dfrac{41}{2}m }}}}}$

Explanation-:

• $\frak{Given \:\: -:} \begin{cases} \sf{The\:side\:of\:Square \:\:is\:= \frak{ \dfrac{41}{8}m }} \end{cases}$

◇ Figure related to question-:

$\setlength{\unitlength}{1cm}\begin{picture}(0,0)\thicklines\multiput(0,0)(4,0){2}{\line(0,1){4}}\multiput(0,0)(0,4){2}{\line(1,0){4}}\put(-0.5,-0.5){\bf D}\put(-0.5,4.2){\bf A}\put(4.2,-0.5){\bf C}\put(4.2,4.2){\bf B}\put(1.5,-0.6){\bf\large \dfrac{41}{8} \ m}\put(4.4,2){\bf\large \dfrac{41}{8} \ m}\end{picture}$

• $\frak{To \:Find\: -:} \begin{cases} \sf{The\:Perimeter \:of\:Square \:\:. } \end{cases}$

$\dag{\underline {\sf{Solution \:of\:Question \:-:\:}}}$

• $\underline{\boxed{\star{\sf{\blue{ Perimeter_{(Square)}  \: = \: 4 \times side }}}}}$

• $\frak{Here \:\: -:} \begin{cases} \sf{The\:side\:of\:Square \:\:is\:= \frak{ \dfrac{41}{8}m }} \end{cases}$

• $\star{\bigstar {\sf{Now,\:Put\:known\:valued\:in\:Formula-:\:}}}$

• $\longrightarrow {\sf{ Perimeter _{(Square)} \:= \:4 \times \dfrac{41}{8}}}$

• $\longrightarrow {\sf{ Perimeter _{(Square)} \:= \:\cancel {4} \times \dfrac{41}{\cancel {8}\:=2}}}$

• $\longrightarrow {\sf{ Perimeter _{(Square)} \:= \: \dfrac{41}{2}m}}$

$\star{\bigstar {\sf{Hence-:\:}}}$

• $\underline{\boxed{\star{\sf{\blue{ Perimeter_{(Square)}  \: = \: \dfrac{41}{2}m }}}}}$

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$\dag{\underline {\sf{More \:to\:know \:-:\:}}}$

• Formulas of area :

$\boxed{\begin {minipage}{9cm}\\ \dag\quad \Large\underline{\bf Formulas\:of\:Areas:-}\\ \\ \star\sf Square=(side)^2\\ \\ \star\sf Rectangle=Length\times Breadth \\\\ \star\sf Triangle=\dfrac{1}{2}\times Breadth\times Height \\\\ \star \sf Scalene\triangle=\sqrt {s (s-a)(s-b)(s-c)}\\ \\ \star \sf Rhombus =\dfrac {1}{2}\times d_1\times d_2 \\\\ \star\sf Rhombus =\:\dfrac {1}{2}p\sqrt {4a^2-p^2}\\ \\ \star\sf Parallelogram =Breadth\times Height\\\\ \star\sf Trapezium =\dfrac {1}{2}(a+b)\times Height \\ \\ \star\sf Equilateral\:Triangle=\dfrac {\sqrt{3}}{4}(side)^2\end {minipage}}$

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Answer 2

Given : Side of Square Carrom board is $\dfrac{41}{8}$

Exigency To Find : Perimeter of Carrom board.

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❍ Formula for Perimeter of Square is given by :

$\dag\:\:\frak{\underline {As,\:We\:know\:that\::}}$

$\dag\:\:\boxed {\sf{Perimeter _{(Square)} = 4 \times a }}$

Where ,

• a is the Length of side of Square is $\dfrac{41}{8}$.

⠀⠀⠀⠀⠀⠀$\underline {\frak{\star\:Now \: By \: Substituting \: the \: Given \: Values \::}}$

$:\implies \sf{ Perimeter _{(Square)} = 4 \times \dfrac{41}{8} }\\\\ :\implies \sf{ Perimeter _{(Square)} = \cancel {4} \times \dfrac{41}{\cancel {8}} }\\\\\underline {\boxed{\pink{ \mathrm {  Perimeter _{(Square)}= \dfrac{41}{2}\: m}}}}\:\bf{\bigstar}$

Therefore,

⠀⠀⠀⠀⠀$\therefore {\underline{ \mathrm {Hence,\:  Perimeter \:of\:Square \:is\:\bf{\dfrac{41}{2}\: m}}}}$

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$\large {\boxed{\sf{\mid{\overline {\underline {\star More\:To\:know\::}}}\mid}}}$

$\begin{gathered}\boxed{\begin {array}{cc}\\ \dag\quad \Large\underline{\bf Formulas\:of\:Areas:-}\\ \\ \star\sf Square=(side)^2\\ \\ \star\sf Rectangle=Length\times Breadth \\\\ \star\sf Triangle=\dfrac{1}{2}\times Breadth\times Height \\\\ \star \sf Scalene\triangle=\sqrt {s (s-a)(s-b)(s-c)}\\ \\ \star \sf Rhombus =\dfrac {1}{2}\times d_1\times d_2 \\\\ \star\sf Rhombus =\:\dfrac {1}{2}p\sqrt {4a^2-p^2}\\ \\ \star\sf Parallelogram =Breadth\times Height\\\\ \star\sf Trapezium =\dfrac {1}{2}(a+b)\times Height \\ \\ \star\sf Equilateral\:Triangle=\dfrac {\sqrt{3}}{4}(side)^2\end {array}}\end{gathered}$

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