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d₁=10 Obw=? P=d₁*d₂/2
P=40 Obw=4a
Obliczenia:
5d₂=40/:5
d₂=8
Z pitagorasa obliczymy bok a:
(d₁/2)²+(d₂/2)²=a²/√
a=√(5²+4²)=√49=7
Obw=4a=4*7=28
Odp. Obwód tego rombu wynosi 28.
40=1/2*10*f
40=5f /:5
f=8
b - bok
(1/2e)²+(1/2f)²=b²
(1/2*10)²+(1/2*8)²=b²
5²+4²=b²
b² = 25+16
b²=41
b=√41
b=41
Obw=4b
Obw=4*√41
Obw=4√41 [j]