Ducem DE║BC
IN ΔADE AD=3cm , DE=4cm , AE =5cm care sunt nr pitagoreice
5²= 4² +3² ⇒ ΔADE dreptunghic cu ∡D=90°
Duc DF inaltimea Δ ADE care este si inaltimea trapezului
In ΔADE ⇒TC AD²=AE·AF
3²=5·AF
AF=9/5
In ΔADF cu ∡F=90°⇒ TP DF²= AD²- AF²
DF²= 3² - 81/25= 9(1-9/25)=9(25-9)/25= 9·16/25=144/25
DF= √144/23=12/5
Atrapez (B+b)·i/2=(7+2)12/5/2= 9·12/5/2=108/5/2=108/5·1/2=54/5cm²
InΔDBF cu F=90° ⇒ TP DB²= DF²+BF²= 144/25 +(7-9/5)²=144/25+26/25= 170/25
DB =√170/5
