Question

exercitiul 5 dau coroana!!

Answer 1

a) Fie CM ⊥ AB => tr. BMC - dreptunghic,

∠B = 60° => ∠MCB = 30° => (conform T. ∠

de 30°)

$MB = \frac{ BC }{2}$

=> $MB= \frac{24}{2} = 12 \: cm$

=> CM (din Pitagora) = 12√3 cm

AMCD - dreptunghi => AM = CD 8 cm

=> AB = AM + MB = 8 + 12 = 20 cm

=> AD = h = 12√3 cm

Aria trapezului = $\frac{AB + CD}{2} \times h$

=> aria trapezului = $\frac{20 + 8}{2 } \times 12√3 = 14 \times 12 √3 =$$168√3$

♫ dia ♫

Answer 2

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