Question

the length of a regular hexagon playground is 13 cm. if it takes 100 rs. per metre to fence it, then how much does it cost to fence four rounds​

Answer 1

Length if regular hexagon = 13 by 6
= 78
If it takes 100rs to fence it = 78 by 100
= 7800
Cost of four rounds = 7800 by 4
= 31200

Answer 2

$\large\underline{\sf{Solution-}}$

Given that,

Length of regular hexagon, a = 13 cm

We know,

$\boxed{ \rm{ \:Perimeter_{(hexagon)} \: = \: 6 \times side \: }}$

So,

$\rm \: Perimeter_{(hexagon)} = 6 \times a$

$\rm \: Perimeter_{(hexagon)} = 6 \times 13$

$\rm\implies \:Perimeter_{(hexagon)} = 78 \: cm$

Now,

$\rm \: Length \: required \: to \: fence \: in \: 1 \: round = Perimeter_{(hexagon)}$

$\rm \: Length \: required \: to \: fence \: in \: 1 \: round = 78 \: cm$

$\rm \: Length \: required \: to \: fence \: in \:4 \: round = 78 \times 4 = 312 \: cm$

Now, further given that,

$\rm \: Cost \: of \: fencing \: 1 \: m = Rs \: 100$

That means,

$\rm \: Cost \: of \: fencing \: 100 \:cm = Rs \: 100$

$\rm \: Cost \: of \: fencing \: 1 \:cm = Re \: 1$

So,

$\rm \: Cost \: of \: fencing \: 312 \:cm = 312 \times 1 = Rs \: 312$

$\rule{190pt}{2pt}$

Additional Information :-

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