Question

The length of a rectangle sharped park exceeds its breadth by 17 meters if the perimeter of the park is 178 meters find the dimensions of the park?​

Answer 1

Answer:

Let the breadth of the park be = x metres

Then the length of the park be = x±17 metres

perimeter of the park=2( length ±breadth)

=2(x±17±x) metres

=2(2x±17) metres

But it is given that the perimeter of the rectangle is 178 metres.

2(2x ±17) =178

4x±34=178

4x =178-34

4x=144

x=144/4=36

Therefore breadth of the park=36 metres

Length of the park=36±17=53 metres.

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Answer 2

Step-by-step explanation:

$⍟ \: \large\sf{Given }$

• Length of a rectangular park exceeds it's breadth by 17m
• Perimeter = 178m

$⍟ \: \large\sf{To \: Find}$

• Length and Breadth of the rectangular park

$⍟ \: \large\sf{Explanation}$

Let us assume

Breadth = x metres

Length = (x + 17) metres

Then,

Perimeter = 2(Length + breadth)

178 = 2(x + 17 + x)

→ 178 = 2(2x + 17)

→ 4x + 34 = 178

→ 4x = 178 - 34

→ 4x = 144

$\small\rm{x = \frac{144}{4} = 36 }$

$\small\boxed{ \rm{x = 36}}$

Breadth = 36metres

Length = (x + 17) = (36 + 17) metres = 53 metres