Math, asked by someonebutnotmeowo, 4 days ago

A rectangular park of dimensions 50 m by 40 m  is surrounded by a path of width 2m internally . A circular pond of radius 3.5 m is in the center of the park and the remaining area is covered by grass. find the area covered by grass.   ​

Answers

Answered by mathdude500
8

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

Given that,

A rectangular park of dimensions 50m by 40m is surround ed by a path of width 2m internally.

Let assume that

ABCD be the rectangular park and PQRS be the area left after the path of 2 m wide internally surrounded.

So, Dimensions of ABCD

  • Length, AB = 50 m

  • Breadth, BC = 40 m

Now, Dimensions of PQRS

  • Length, PQ = AB - 2 × 2 = 50 - 4 = 46 m

  • Breadth, QR = BC - 2 × 2 = 40 - 4 = 36 m

So,

\rm \: Area_{(PQRS)} = 46 \times 36 \\

\rm\implies \:\boxed{ \rm{ \:Area_{(PQRS)} = 1656 \:  {m}^{2}  \: }} \\

Now, Further given that, A circular pond of radius, r = 3.5 m is in the center of the park.

\rm \: Area_{(Pond)} = \pi \:  {r}^{2}  \\

\rm \:  =  \: \dfrac{22}{7} \times 3.5 \times 3.5 \\

\rm \:  =  \: 38.5 \:  {m}^{2}  \\

\rm\implies \:\boxed{ \rm{ \:Area_{(Pond)} \:  =  \: 38.5 \:  {m}^{2}  \: }} \\

Now,

\rm \: Area_{(covered \: by \: grass)} \:  \\

\rm \:  =  \: Area_{(PQRS)} - Area_{(Pond)} \\

\rm \:  =  \: 1656 - 38.5 \\

\rm \:  =  \: 1617.5 \:  {m}^{2}  \\

Hence,

\rm\implies \:\boxed{ \rm{ \:\rm \: Area_{(covered \: by \: grass)} \: =  \: 1617.5 \:  {m}^{2} \:  \: }}   \\

\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}

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