Trapezul dreptunghic ABCD cu AB paralel cu DC, m (A)=90 de grade, m(<B)=60 grade are baza mica CD=12√3 cm si latura AD=18 cm. Se cere sa se afle:a)perimetrul trapezuluib) AC= BD=
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sin 60⁰=√3/2
CF/BC=18/BC=√3/2
BC=18×2/√3=36√3/3=12√3
FB=√[(12√3)²-18²]=√(432-324)=√108=6√3
AB=12√3+6√3=18√3
P=AB+BC+CD+AD=18√3+12√3+12√3+18=42√3+18
AC=√[(12√3)²+18²]=√(432+324)=√756=6√21
BD=√[(18√3)²+18²]=√(972+324)=√1296=36