promien koregu wpisanego r=4√3
r=⅓h
4√3=⅓ ·a√3/2
4√3=(a√3)/6
a√3=24√3 /:√3
a=24--->dl,boku Δ
pole PΔ=(a²√3)/4 =(24²√3)/4=(576√3)/4=144√3 j²
r = 4√3
r=a√3/6
4√3 = a√3/6 /*6
24√3 = a√3 /:√3
a = 24
P =a²√3/4
P =24² √3/4 = 576√3/4 = 144√3
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promien koregu wpisanego r=4√3
r=⅓h
4√3=⅓ ·a√3/2
4√3=(a√3)/6
a√3=24√3 /:√3
a=24--->dl,boku Δ
pole PΔ=(a²√3)/4 =(24²√3)/4=(576√3)/4=144√3 j²
r = 4√3
r=a√3/6
4√3 = a√3/6 /*6
24√3 = a√3 /:√3
a = 24
P =a²√3/4
P =24² √3/4 = 576√3/4 = 144√3