Oblicz dlugosc promienia okregu wpisanego w trojkat o bokach 13 cm, 13 cm i 10 cm.
zad
a=10cm
b=13
c=13
liczymy h Δ:
½·10=5cm
5²+h²=13²
25+h²=169
h²=169-25
h=√144=12cm
PΔ=½ah=½·10·12=60cm²
liczymy promień r:
r=2P/(a+b+c)=2·60:(10+13+13)=120:36=3⅓cm
r=
" Life is not a problem to be solved but a reality to be experienced! "
© Copyright 2013 - 2024 KUDO.TIPS - All rights reserved.
zad
a=10cm
b=13
c=13
liczymy h Δ:
½·10=5cm
5²+h²=13²
25+h²=169
h²=169-25
h=√144=12cm
PΔ=½ah=½·10·12=60cm²
liczymy promień r:
r=2P/(a+b+c)=2·60:(10+13+13)=120:36=3⅓cm
r=