A=[tex] \frac{(B+B) * h}{2} [/tex]
[tex]42 \sqrt{3}= \frac{(13 \sqrt{3 } + \sqrt{3}*h }{2} [/tex]
[tex]14 \sqrt{3} *h=84 \sqrt{3} [/tex]
h=[tex] \frac{84 \sqrt{3} }{14 \sqrt{3} } [/tex]
h=6 (INALTIMEA TRAPEZULUI eu am notat-o cu CE)
In ΔCEB dr. in E
⇒T.P. [tex]CE ^{2} +EB ^{2} =BC ^{2} [/tex]
[tex]6 ^{2} +(6 \sqrt{3} ) ^{2} =BC ^{2} [/tex]
36+108=[tex]BC ^{2} [/tex]
[tex]BC ^{2} =144[/tex]
BC=12
Perimetrul=12*2+[tex]13 \sqrt{3} + \sqrt{3} [/tex]
=24+[tex]14 \sqrt{3} [/tex]
=2(12+[tex]7 \sqrt{3} [/tex]
ABCD trapez isoscel⇒ ungh.A≡ ungh. B
in ΔCEB ungh.E=90 CE=6 CB=12⇒ UNGH. B=30 (cateta ce se opune unghiului de 30 este 1/2 din ipotenuza)
ungh.B≡ ungh. A =30
A+B+C+D=360
30+30+2*C=360
2C=360-60
2C=300
C=150
deci ungh. A=B=30
C=D=150
