Question

a park is (25)4/5m long and (20) 2/3m wide. find it's perimeter.if a walkway is to be constructed inside the park which is (2) 3/8m wide, what would be the inside perimeter of walkway​

Answer 1

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Step-by-step explanation:

a park is 25 4/5 m long and 20 2/3 m wide . a walkway is to be constructed inside the park which is 2 3/8m wide,

To find : perimeter of park  

inside perimeter of the walkway

Solution:

Park length  25 4/5 m  = 25 + 4/5  m =  (125 + 4)/5 = 129/5 m

Park Width =  20 2/3 m = 20 + 2/3 m = (60 + 2)/3  = 62/3  m

Park perimeter = 2 ( Length + Width )

= 2 ( 129/5  + 62/3)

= 2( 387  + 310)/15

= 2(697)/15

= 1394 / 15

= 92  14/15  m

walkway is to be constructed inside the park which is 2 3/8m wide

2  + 3/8  m   = 19/8  Meter

8 times the width will be reduced  

8 * 19/8   = 19 m

inside perimeter of the walkway  = 92  14/15  - 19

= 75 14/15 m

Park perimeter =  92  14/15  m

inside perimeter of the walkway  = 75 14/15 m

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Answer 2

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

Let assume that the rectangular park as ABCD and when the path is constructed inside the rectangular park, let the inner rectangular region be PQRS.

Dimensions of ABCD

$\rm :\longmapsto\:Length = 25 \dfrac{4}{5} = \dfrac{129}{5} \: m$

$\rm :\longmapsto\:Breadth = 20 \dfrac{2}{3} = \dfrac{62}{3} \: m$

We know,

Perimeter of Rectangular Park ABCD is

$\bf :\longmapsto\:Perimeter_{ABCD} \: = 2(Length + Breadth)$

$\rm \: \: = \: \: 2\bigg(\dfrac{129}{5} + \dfrac{62}{3} \bigg)$

$\rm \: \: = \: \: 2\bigg(\dfrac{129 \times 3 + 62 \times 5}{15} \bigg)$

$\rm \: \: = \: \: 2\bigg(\dfrac{387 + 310}{15} \bigg)$

$\rm \: \: = \: \: 2\bigg(\dfrac{697}{15} \bigg)$

$\rm \: \: = \: \: \dfrac{1394}{15}$

$\rm \: \: = \: \: 92 \dfrac{14}{15} \: m$

Dimensions of PQRS

$\bf :\longmapsto\:Width \: of \: path = 2 \dfrac{3}{8} = \dfrac{19}{8} \: m$

$\rm :\longmapsto\:Length = \dfrac{129}{5} - \dfrac{19}{8} - \dfrac{19}{8}$

$\rm \: \: = \: \: \dfrac{129 \times 8 - 19 \times 5 - 19 \times 5}{40}$

$\rm \: \: = \: \: \dfrac{1032 - 95 - 95}{40}$

$\rm \: \: = \: \: \dfrac{842}{40}$

$\rm \: \: = \: \: \dfrac{421}{20} \: m$

$\rm :\longmapsto\:Breadth = \dfrac{62}{3} - \dfrac{19}{8} - \dfrac{19}{8}$

$\rm \: \: = \: \: \dfrac{62 \times 8 - 19 \times 3 - 19 \times 3}{24}$

$\rm \: \: = \: \: \dfrac{496 - 57 - 57}{24}$

$\rm \: \: = \: \: \dfrac{382}{24}$

$\rm \: \: = \: \: \dfrac{191}{12} \: m$

Now,

Perimeter of rectangular park PQRS is

$\bf :\longmapsto\:Perimeter_{PQRS} \: = 2(Length + Breadth)$

$\rm \: \: = \: \: 2\bigg(\dfrac{421}{20} + \dfrac{191}{12} \bigg)$

$\rm \: \: = \: \: 2\bigg(\dfrac{421 \times 3 + 191 \times 5}{60} \bigg)$

$\rm \: \: = \: \: 2\bigg(\dfrac{1263 + 955}{60} \bigg)$

$\rm \: \: = \: \: 2\bigg(\dfrac{2218}{60} \bigg)$

$\rm \: \: = \: \: \bigg(\dfrac{1109}{15} \bigg)$

$\rm \: \: = \: \: 73 \: \dfrac{14}{15} \: m$

Additional Information :-

$\rm :\longmapsto\:Area_{(rectangle)} = Length \times Breadth$

$\rm :\longmapsto\:Area_{(square)} = {side}^{2}$

$\rm :\longmapsto\:Perimeter_{(square)} = 4 \times side$

$\rm :\longmapsto\:Perimeter_{(rhombus)} = 4 \times side$

$\rm :\longmapsto\:Area_{(rhombus)} = base \times height$

$\rm :\longmapsto\:Area_{(circle)} = \pi \: {r}^{2}$