ABCD e trapez, AB||CD
Unghiul A=120 de grade Unghiul B=150 de grade BC=12 cm AB= 12 cm AD=? Linia mijlocie=? ducem AE si BF perpendiculare pe DC m(<FBC)= 150⁰-90⁰=60⁰ m(<BCF)= 90⁰-60⁰=30⁰ m(<DAE)= 120⁰-90⁰=30⁰ m(<ADE)= 90⁰-30⁰=60⁰ in triunghiul BFC sin 30⁰= BF/BC 1/2=BF/12 BF=6 sin 60⁰= FC/BC √3/2=FC/12 FC=6√3
in triunghiul ADE sin 60⁰= AE/AD √3/2=6/AD AD=4√3 sin 30⁰= DE/AD 1/2=DE/4√3 DE=2√3 DC=2√3+12+6√3=12+8√3 Linia mijlocie=(AB+DC)/2=(12+12+8√3)/2=12+4√3
