Question

Trapezul isoscel ABCD are AB//CD,AB=24 cm,CD=18 cm,iar m(unghiului A)=60 grade.Determina aria trapezului.Explicatie va rog!

Answer 1

Duci inaltimea DE, E∈AB
AE=(AB-DC)/2=(24-18)/2=3cm
A=60, E=90 deci D=30 grade
AE=AD/2 deci AD=3*2=6cm
Pitagora in ADE
AE²+DE²=AD²
DE²=6²-3²
DE=√36-9
DE=√27=3√3cm
     (AB+CD)·DE      42·3√3
A=----------------- = ---------- = 21·3√3=63√3cm²
              2                   2

Answer 2

Daca ducem inaltimile CE si DF atunci avem AF=BE=(24-18):2=3

In tr ADF m(<F)=90, m(<A)=60,

$tg 60= \sqrt{3} = \frac{DF}{AF} = \frac{DF}{3} \\ \\ DF=3 \sqrt{3} \\ \\ A= \frac{(B+b)\cdot h}{2} = \frac{(24+18)\cdot 3 \sqrt{3 }}{2} =63 \sqrt{3}$