Question

ples help i don’t no how to solve that

Answer 1

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,

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

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

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

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

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