Znajdz wszystkie wysokosci trojkata o bokach:
a) 16, 12 i 8
b) 13, 10 i 9
Prosze o szybka odpowiedz
a)
P=1/2ah1=1/2bh2=1/2ch3= √p(p-a)(p-b)(p-c) (wzor Herona, gdzie p - polowa obwodu)
p=(16+12+8)/2=18
P=√18*(18-16)(18-12)(18-8)=√18*2*60=6√60=12√15
1/216*h1=12√15
h1=1,5√15
h2=2√15
h3=3√15
b)
P=√16*(16-13)(16-10)(16-9)=√16*18*7=12√14
1/2ah1=12√14
h1=12√14/6,5=24√14/13
h2=12√14/5
h3=12√14/4,5=8√14/3
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a)
P=1/2ah1=1/2bh2=1/2ch3= √p(p-a)(p-b)(p-c) (wzor Herona, gdzie p - polowa obwodu)
p=(16+12+8)/2=18
P=√18*(18-16)(18-12)(18-8)=√18*2*60=6√60=12√15
1/216*h1=12√15
h1=1,5√15
h2=2√15
h3=3√15
b)
P=√16*(16-13)(16-10)(16-9)=√16*18*7=12√14
1/2ah1=12√14
h1=12√14/6,5=24√14/13
h2=12√14/5
h3=12√14/4,5=8√14/3