Math, asked by Benjamini, 4 months ago

Find the Area of a Rhombus whose side is 15cm and one Diagonal is 24cm.​

Answers

Answered by itzrithvik
26

Answer:

Rhombus

Solve for area

A=216cm²

a Side

15

cm

p Diagonal

24

cm

Answered by thebrainlykapil
13

Question :-

  • Find the Area of a Rhombus whose side is 15cm and one Diagonal is 24cm.

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Given :-

  • Side of Rhombus = 15cm
  • One Diagonal = 24cm

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To Find :-

  • What is the Area of the Rhombus ?

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Explanation :-

In this question we have to find the Area of the Rhombus , also it is given that one of it's side is 15cm and one Diagonal is 24cm. So, Firstly we will find the Second Diagonal then we will apply the Formula of Area of Rhombus to find the Area.

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Solution :-

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Let the Second Diagonal be D

 \quad {:} \longrightarrow \sf \boxed{\bf{Side \: of \: Rhombus \: = \:   \dfrac{1}{2} \:  \sqrt{d ^{2} \:  +  \:  {D}^{2}  }    }}\\

 \quad {:} \longrightarrow \sf{\sf{15 \: = \:   \dfrac{1}{2} \:  \sqrt{24 ^{2} \:  +  \:  {D}^{2}  }    }}\\

 \quad {:} \longrightarrow \sf{\sf{15 \:  \times  \: 2 \: = \:  \sqrt{24 ^{2} \:  +  \:  {D}^{2}  }    }}\\

 \quad {:} \longrightarrow \sf{\sf{30 \: = \: \sqrt{24 ^{2} \:  +  \:  {D}^{2}  }    }}\\

 \quad {:} \longrightarrow \sf{\sf{30^{2}  \: = \: 24 ^{2} \:  +  \:  {D}^{2}    }}\\

 \quad {:} \longrightarrow \sf{\sf{30^{2} \:  -  {24}^{2}   \: =    \:  {D}^{2}    }}\\

 \quad {:} \longrightarrow \sf{\sf{900\:  -   \: 576   \: =    \:  {D}^{2}    }}\\

 \quad {:} \longrightarrow \sf{\sf{324   \: =    \:  {D}^{2}    }}\\

 \quad {:} \longrightarrow \sf{\sf{ \sqrt{324}  \: =    \:  D  }}\\

 \quad {:} \longrightarrow \sf{\bf{ 18  \: =    \:  D  }}\\

So, Second Diagonal = 18cm.

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 \quad {:} \longrightarrow \sf \boxed{\bf{ Area \: of \: rhombus \:  =  \:  \frac{1}{2} \:  \times  \: (Products \: of \: diagonals)   }}\\

 \quad {:} \longrightarrow \sf{\sf{ Area \: of \: rhombus \:  =  \:  \frac{1}{2} \:  \times  \: 24 \:  \times  \: 18   }}\\

 \quad {:} \longrightarrow \sf{\sf{ Area \: of \: rhombus \:  =  \:  \frac{1}{ \cancel2} \:  \times  \:  \cancel{24} \:  \times  \: 18   }}\\

 \quad {:} \longrightarrow \sf{\sf{ Area \: of \: rhombus \:  =   \: 12\:  \times  \: 18   }}\\

 \quad {:} \longrightarrow \sf{\bf{ Area \: of \: rhombus \:  =   \: 216  }}\\

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\begin{gathered}\begin{gathered}\qquad \therefore\: \sf{  Area \: of \: rhombus    \: = \underline {\underline{ 216cm^{2}}}}\\\end{gathered}\end{gathered}

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