Question

A park 80m by 65m has a 3m wide track all around it. A wall has been constructed around the track leaving a space for gate which is 2m. what is the length of the outer boundary leaving the gate which has a wall? Find the perimeter of the park without track ​

Answer 1

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

Given that,

A park 80m by 65m has a 3m wide track all around it. A wall has been constructed around the track leaving a space for gate which is 2m.

Let assume that ABCD represents the park such that AB = 80 m and CD = 65 m

Let assume that EFHG be the outer part, which is constructed 3 m wide track all around it.

So, EF = AB + 3 + 3 = 80 + 6 = 86 m

and HG = CD + 3 + 3 = 65 + 6 = 71 m

Now, we have to find the length of the outer boundary leaving the space of 2 m for the gate which has a wall.

So, Length of the outer boundary wall

$\rm \: = \: EF + FG + GH + HE - 2$

$\rm \: = \: 81 + 76 + 81 + 76 - 2$

$\rm \: = \: 312 \: m$

Now, Perimeter of the park ABCD

$\rm \: = \: AB + BC + CD + DA$

$\rm \: = \: 80 + 65 + 80 + 65$

$\rm \: = \: 290 \: m$

Hence,

Length of boundary wall leaving the space for gate = 312 m

and

Perimeter of the park without track = 290 m

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

Additional Information :-

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