ᴛʜᴇ ʟᴇɴɢᴛʜ ᴀɴᴅ ᴛʜᴇ ʙʀᴇᴀᴅᴛʜ ᴏғ ᴀ ʀᴇᴄᴛᴀɴɢᴜʟᴀʀ ᴘᴀʀᴋ ᴀʀᴇ ɪɴ ᴛʜᴇ ʀᴀᴛɪᴏ :,ᴀ . ᴍ ᴡɪᴅᴇ ᴘᴀᴛʜ ʀᴜɴɴɪɴɢ ᴀʟʟ ᴀʀᴏᴜɴᴅ ᴛʜᴇ ᴏᴜᴛsɪᴅᴇ ᴏғ ᴛʜᴇ ᴘᴀʀᴋ ʜᴀs ᴀɴ ᴀʀᴇᴀ ᴏғ sǫ.ᴍ. ғɪɴᴅ ᴛʜᴇ ᴅɪᴍᴇɴsɪᴏɴ ᴏғ ᴛʜᴇ ᴘᴀʀᴋ.
."ᴀɴsᴡᴇʀ ɪs = ᴍ ʙʏ ᴍ."
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Answers
Explanation:
Answer:
Let the length and Breadth of the rectangular park be 5x and 2x respectively.
⠀⠀⠀
\underline{\bf{\dag} \:\mathfrak{As\;we\;know\: that\: :}}
†Asweknowthat:
†Asweknowthat:
⠀⠀⠀⠀
\star\;\boxed{\sf{Area_{\:(rectangle)} = Length \times Breadth}}⋆
Area
(rectangle)
=Length×Breadth
⋆Area
(rectangle)
=Length×Breadth
⠀
Therefore,
\begin{gathered}\begin{gathered}:\implies\sf 5x \times 2x \\\\\\:\implies\sf 10x^2\end{gathered}\end{gathered}
:⟹5x×2x
:⟹10x
2
:⟹5x×2x
:⟹10x
2
⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━⠀⠀⠀
⠀⠀⠀⠀⠀⠀⠀
\underline{\boldsymbol{According\: to \:the\: Question :}}
AccordingtotheQuestion:
According to the Question:
⠀
A 2.5 m wide path running all around the outside of the park has an area of 305 m².
⠀
Therefore,
Length which is including path,
\begin{gathered}\begin{gathered}:\implies\sf (5x + 2.5m + 2.5m) \\\\\\:\implies\sf (5x + 5)m \end{gathered}\end{gathered}
:⟹(5x+2.5m+2.5m)
:⟹(5x+5)m
:⟹(5x+2.5m+2.5m)
:⟹(5x+5)m
⠀
Similarly,
Breadth which is including path,
\begin{gathered} \begin{gathered}:\implies\sf (2x + 2.5 m + 2.5 m)\\\\\\:\implies\sf (2x + 5)m \end{gathered} \end{gathered}
:⟹(2x+2.5m+2.5m)
:⟹(2x+5)m
:⟹(2x+2.5m+2.5m)
:⟹(2x+5)m
⠀⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━⠀⠀⠀⠀⠀⠀⠀⠀
⠀⠀
Now, Area of the rectangular park:
⠀
\begin{gathered}\begin{gathered}:\implies\sf (5x + 5)m \times (2x + 5)m\\\\\\:\implies\sf (10x^2 + 25x + 10x + 25) m^2\\\\\\:\implies\sf (10x^2 + 35x + 25)m^2\\\\\\:\implies\sf (\cancel{10x^2}\; + 35x + 25) - \cancel{10x^2}\\\\\\:\implies\sf 35x + 25 = 305\\\\\\:\implies\sf 35x = 305 - 25\\\\\\:\implies\sf 35x = 280 \\\\\\:\implies\sf x = \cancel\dfrac{280}{35}\\\\\\:\implies{\underline{\boxed{\frak{\pink{x = 8}}}}}\end{gathered}\end{gathered}
:⟹(5x+5)m×(2x+5)m
:⟹(10x
2
+25x+10x+25)m
2
:⟹(10x
2
+35x+25)m
2
:⟹(
10x
2
+35x+25)−
10x
2
:⟹35x+25=305
:⟹35x=305−25
:⟹35x=280
:⟹x=
35
280
:⟹
x=8
:⟹(5x+5)m×(2x+5)m
:⟹(10x
2
+25x+10x+25)m
2
:⟹(10x
2
+35x+25)m
2
:⟹(
10x
2
+35x+25)−
10x
2
:⟹35x+25=305
:⟹35x=305−25
:⟹35x=280
:⟹x=
35
280
:⟹
x=8
⠀⠀
Hence, dimensions of the park are:
Length of the park, 2x = 2(8) = 16m
Breadth of the park, 5x = 5(8) = 40m
⠀
\therefore{\underline{\sf{Hence, \; Length \; and \; Breadth \: of \; the \: park \; are \: \bf{16m\: and \; 40m }.}}}∴
Hence,LengthandBreadthoftheparkare16mand40m.
∴
Hence,Length and Bread thof the park are16m and 40m.
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