Question

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$\huge\mathcal{\fcolorbox{lime}{black}{\orange{QuestioN}}}$

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​

Answer 1

Answer:

Answer:

$\huge{\mathfrak{\green{☆elloh}}}$

\huge\mathcal{\fcolorbox{lime}{black}{\orange{QuestioN}}}

QuestioN

\huge\boxed{\fcolorbox{pink}{red}{\blue{Please AnsweR\:it\:!}}}

Please AnsweRit!

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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.

ʜᴏᴘᴇ ʏᴏᴜ ᴜɴᴅᴇʀsᴛᴀɴᴅ♥!

Answer 2

$\large\green{\textsf{✩ Verified Answer ✓ }}$

Mass of the packet = 500 g

Volume of the packet = 350 cm³

Density of the packet is given by

As the density of the packet is less than that of water it will sink in water.

Volume of the water displaced by the packet = volume of the packet = 350 cm³

Mass of the water displaced by the packet = Volume of the water displaced by the packet X Density of water

= 350 X 1

= 350 g

Mass of water displaced is less than the mass of packet, so the packet will sink.

$\bf\pink{\textsf{Answered By MrAkdu}}$