trapez równoramienny

[talybytomy]

w trapezie równoramiennym ABCD |AB|=12 |CD|=6 a przekątna |AC|=\(\displaystyle{ 6\sqrt{3}}\)
oblicz pole i obwód

[silicium2002]

Wysokość z trójkąta \(\displaystyle{ 90 \ \ 60 \ \ 30}\) (przekątna wysokość \(\displaystyle{ \frac{3}{4}}\) dolnej podstawy) wyjdzie nam \(\displaystyle{ 3 \sqrt{3}}\)

Natomiast potem ramię również z trójkąta \(\displaystyle{ 90 \ \ 60 \ \ 30}\) (tylko tego wysokość \(\displaystyle{ \frac{1}{4}}\) dolnej podstawy i ramię) czyli: \(\displaystyle{ 6}\)

A więc mając wszystkie boki i wysokość łatwo już policzyć pole i obwód

[talybytomy]

a bardziej jasno?

[silicium2002]

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}}\)
Tak lepiej?

[talybytomy]

<czasowy zanik myslenia D: >

[silicium2002]

nie ogarniam dlaczego takie katy

hmm a własności trójkąta 30 60 90 znasz? (Na pewno miałeś w szkole! jeśli masz takie zdanie...)
to właśnie wynika z tych własności...