Question

Se da trapezul ABCD cu ADparalelCB m A=90grade,m D=30grade,AB=4cm,Bc=4 radical din 3.Aria trapezului-?

Answer 1

Construiesti inaltimea CM, CM⊥ AD
CM//AB⇒ CM=4 cm
In Δ CMD, m<M=90°,m<D=30°⇒Th 30-60-90 CD=8 cm

Δ CMD, m<M=90°, CM=4 cm, CD=8 cm⇒Th Pitagora MD²=CD²-CM²
MD²=64-16=48
MD=√48
MD=4√3
AD=AM+MD
AD=4√3+4√3=8√3 cm
A=$\frac{(BC+AD)*CM}{2} = \frac{(4 \sqrt{3} +8 \sqrt{3})*4 }{2} =24 \sqrt{3}$