oblicz promien okregu wpisanego w trojkat rownoramienny o bokach majacych dlugosc 13 13 24 i to samo dla 17 17 16
13² = 12² + h²
h = 5
P = 5*24*½ = 60
r = 120 / 24+13*2 = 2,4
17² = 8² + h²
h = 15
P= 15*16*½ = 120
r =2*120/17*2+16 = 4,8
rozwiazanie w zalaczniku
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13² = 12² + h²
h = 5
P = 5*24*½ = 60
r = 120 / 24+13*2 = 2,4
17² = 8² + h²
h = 15
P= 15*16*½ = 120
r =2*120/17*2+16 = 4,8
rozwiazanie w zalaczniku