Math, asked by shahadxx711, 11 months ago

ples help i don’t no how to solve that


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Answered by shadowsabers03
6

1.  The geometric mean of two numbers a and b is \sf{\sqrt{ab}.}

Hence the geometric mean of 2 and 50 is,

\longrightarrow\sf{\sqrt{2\times50}=\sqrt{100}}

\longrightarrow\sf{\underline{\underline{\sqrt{2\times50}=10}}}

2.  We see that the triangles JKM, KLM and JLK are similar.

  • ∠JKM = ∠JLK

  • ∠MKL = ∠KJL

  • ∠JMK = ∠KML = ∠JKL

From ΔJKM and ΔKLM, we have,

\longrightarrow\sf{\dfrac{JM}{MK}=\dfrac{MK}{ML}}

\longrightarrow\sf{(MK)^2=JM\times ML}

\longrightarrow\sf{e^2=6\times24}

\longrightarrow\sf{e^2=144}

\longrightarrow\sf{\underline{\underline{e=12}}}

In ΔJKM,

\longrightarrow\sf{d=\sqrt{e^2+6^2}}

\longrightarrow\sf{d=\sqrt{12^2+6^2}}

\longrightarrow\sf{d=\sqrt{144+36}}

\longrightarrow\sf{d=\sqrt{180}}

\longrightarrow\sf{\underline{\underline{d=6\sqrt5}}}

In ΔKLM,

\longrightarrow\sf{c=\sqrt{e^2+24^2}}

\longrightarrow\sf{c=\sqrt{12^2+24^2}}

\longrightarrow\sf{c=\sqrt{144+576}}

\longrightarrow\sf{c=\sqrt{720}}

\longrightarrow\sf{\underline{\underline{c=12\sqrt5}}}

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