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Oznaczam :
przekatne:d i d₁ oraz bok - x
Z treści zadania:
d=4√2
to
d₁=4√2:2=4=2√2
Z tw. Pitagorasa:
(½d)²+(½d₁²)=x²
(2√2)²+(√2)²=x²
8+2=x²
x²=10
x=√10
Bok rombu ma √10