The product of the arithmetic mean of the len | vedclass

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(B) Let the sides of the triangle be $a, b, c$ and the corresponding altitudes be $h_1, h_2, h_3$.We know that $a h_1 = b h_2 = c h_3 = 2 \Delta$.Thus

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$2 \Delta$

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(B) Let the sides of the triangle be $a, b, c$ and the corresponding altitudes be $h_1, h_2, h_3$.We know that $a h_1 = b h_2 = c h_3 = 2 \Delta$.Thus,$h_1 = \frac{2 \Delta}{a}$,$h_2 = \frac{2 \Delta}{b}$,and $h_3 = \frac{2 \Delta}{c}$.The arithmetic mean of the sides is $AM = \frac{a + b + c}{3}$.The harmonic mean of the altitudes is $HM = \frac{3}{\frac{1}{h_1} + \frac{1}{h_2} + \frac{1}{h_3}}$.Substituting the values of $h_1, h_2, h_3$:$\frac{1}{h_1} + \frac{1}{h_2} + \frac{1}{h_3} = \frac{a}{2 \Delta} + \frac{b}{2 \Delta} + \frac{c}{2 \Delta} = \frac{a + b + c}{2 \Delta}$.Therefore,$HM = \frac{3}{\frac{a + b + c}{2 \Delta}} = \frac{6 \Delta}{a + b + c}$.The product $AM 	imes HM = \left(\frac{a + b + c}{3}ight) 	imes \left(\frac{6 \Delta}{a + b + c}ight) = 2 \Delta$.

The product of the arithmetic mean of the lengths of the sides of a triangle and the harmonic mean of the lengths of the altitudes of the triangle is equal to: [where $\Delta$ is the area of the triangle $ABC$]

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