Se da:
In ΔABC se cunosc laturile:
a = 6 cm
b = 3 cm
Semisuma inaltimilor corespunzatoare acestor laturi = inaltimea a 3-a.
adica: (ha + hb)/2 = hc
Se cere:
A treia latura:
c = ? cm
.
[tex]\displaystyle\bf\\h_c=\frac{h_a+h_b}{2}\\\\\implies~\boxed{\bf~h_a,~h_c, h_b~~(in~aceasta~ordine)~~sunt~in~progresie~aritmetica.}\\\\ \textbf{Relatia dintre laturi si inaltimi este data de aria triunghiului.}\\\\A_{\Delta ABC}=\frac{a\times h_a}{2}=\frac{c\times h_c}{2}=\frac{b\times h_b}{2}~~\Big|\times2\\\\2A_{\Delta ABC}=a\times h_a=c\times h_c=b\times h_b[/tex]
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[tex]\displaystyle\bf\\Avem~relatia:\\\\a\times h_a=c\times h_c=b\times h_b\\\\Echivalenta~cu~relatia:\\\\\frac{h_a}{\dfrac{1}{a}}=\frac{h_c}{\dfrac{1}{c}}=\frac{h_b}{\dfrac{1}{b}}\\\\\\De~aici~rezulta~ca:\\\\\boxed{\bf\Big(h_a:~h_c;~h_b\Big)~sunt~direct~proportionale~cu~\Big(\frac{1}{a};~\frac{1}{c};~\frac{1}{b}\Big)}\\\\Stim~ca~\boxed{\bf\Big(h_a:~h_c;~h_b\Big)~sunt~in~progresie~aritmetica.}\\\\Rezulta~ca~si~\boxed{\bf\Big(\frac{1}{a};~\frac{1}{c};~\frac{1}{b}\Big)~sunt~in~progresie~aritmetica.}[/tex]
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[tex]\displaystyle\bf\\\implies~~\frac{1}{c}=\frac{\dfrac{1}{a}+\dfrac{1}{b}}{2}\\\\ a=6~cm;~b=3~cm\\\\\frac{1}{c}=\frac{\dfrac{1}{6}+\dfrac{1}{3}}{2}=\frac{\dfrac{1}{6}+\dfrac{2}{6}}{2}=\frac{1}{c}=\frac{\dfrac{3}{6}}{2}=\frac{\dfrac{1}{2}}{2}=\frac{1}{4}\\\\\frac{1}{c}=\frac{1}{4}\\\\\implies~~\boxed{\boxed{\bf~c=4~cm}}[/tex]
