Odpowiedź:
P = [tex]\frac{1}{3} \pi ^2 = \pi *r^2[/tex] / : π
r² = [tex]\frac{1}{3}[/tex] π ⇒ r = [tex]\sqrt{\frac{1}{3}*\pi }[/tex]
więc h = 3*r = 3*[tex]\sqrt{\frac{1}{3} *\pi } =[/tex][tex]\sqrt{3*\pi }[/tex]
oraz h = a*[tex]\frac{\sqrt{3} }{2}[/tex] = [tex]\sqrt{3*\pi }[/tex] / * 2
a*[tex]\sqrt{3} = 2*\sqrt{3*\pi }[/tex] / : √3
a = 2*[tex]\sqrt{\pi }[/tex]
Obwód Δ L = 3*a
L = 3*2*√π = 6 √π
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Odpowiedź:
P = [tex]\frac{1}{3} \pi ^2 = \pi *r^2[/tex] / : π
r² = [tex]\frac{1}{3}[/tex] π ⇒ r = [tex]\sqrt{\frac{1}{3}*\pi }[/tex]
więc h = 3*r = 3*[tex]\sqrt{\frac{1}{3} *\pi } =[/tex][tex]\sqrt{3*\pi }[/tex]
oraz h = a*[tex]\frac{\sqrt{3} }{2}[/tex] = [tex]\sqrt{3*\pi }[/tex] / * 2
a*[tex]\sqrt{3} = 2*\sqrt{3*\pi }[/tex] / : √3
a = 2*[tex]\sqrt{\pi }[/tex]
Obwód Δ L = 3*a
L = 3*2*√π = 6 √π
=====================
Szczegółowe wyjaśnienie: