1) fie trapezul ABCD , mas < A = 60* , BD _|_AB
daca BB' _|_ AD mas < ABB' = 30* ⇒ AB' = AB/2
AD = baza mare (B = BC + 2AB') B = b + AB
in ΔABD mas <BDA = 30* ⇒AB = B/2
B = b + B/2 B/2 = 4 B= AD = 8cm AB= CD = 4cm
BB' ² = AB² - AB'² = 16 - 4 BB' = 2√3 cm
2) in trapezul ABCD mas < A = mas < D = 45* , BC= 7cm, AD = 13cm
AB' = (B-b)/2 = 3cm
in Δ ABB' (dreptunghic isoscel) AB' = BB'= h = 3cm
A = (B+ b)·h/2 = 20·3/2 = 30cm²