Question

Se da trapezul ABCD cu AD||CB,m(

Answer 1

$\rm Fie ~E \in (AD)~astfel~incat~CE \perp AD. \\ \\ ABCE-dreptunghi \Rightarrow BC=AE=4 \sqrt 3~cm~si~AB=CE=4~cm. \\ \\ In ~ \Delta CED,~dreptunghic~in~E,~avem: \\ \\ tg~D= \frac{CE}{DE} \Rightarrow DE= \frac{CE}{tg ~D} = \frac{4}{tg 30 \textdegree}= \frac{4}{ \frac{\sqrt{3}}{3} }= \frac{12}{\sqrt{3}} = \frac{12 \sqrt{3}}{3}=4 \sqrt{3}~(cm). \\ \\ A_{ABCD}= \frac{(BC+AD) \cdot AB}{2}= \frac{(BC+AE+DE) \cdot AB}{2}= \frac{(4 \sqrt{3} + 4 \sqrt{3}+ 4 \sqrt{3}) \cdot 4}{2}=$

$=\boxed{24 \sqrt{3} ~(cm)^2 }$