Question

un trapez isoscel, ABCD, avand bazele AB si CD, m( <A)= 30  grade
Stiind ca DC= 12√3 cm si BC =8 cm calculati:
- inaltimea trapezului
- lungimile diagonalelor

Answer 1

sin 30 grade = 1 /2
sin A = h / AD => h / 8 = 1 / 2 => h = 4

DE - inaltime
=> 2 * AE = 12 sqrt(3) - DC
cos A = AE / 8
sqrt(3) / 2 = AE / 8
=> AE = 4 sqrt(3)

DC= 12 sqrt(3) - 2*4sqrt(3) = 4 * sqrt(3)

in triunghiul BED dreptunghic, cf teoremei lui pitagora 
=> BD2=BE2+h2
BE = 12sqrt(3)-4sqrt(3)=8sqrt(3)
BD2 = 64 *3 + 16 = 208
BD = 4 sqrt(13)

BD = AC intr-un trapez isoscel