Trapezul este isoscel , deci AD=BC=5 dam .
a) P ABCD = 30 dam = AB+AD+BC+CD = 7 dam+5 dam +5 dam + CD = 17 dam+ CD , de unde rezulta ca : 17 dam + CD=30 dam => CD= 30 dam - 17 dam = 13 dam .
b) Ducem inaltimi din A si din B , notand intersectia cu CD cu E si F .
AB=EF=7dam
CD= 13 dam = EF+DE+FC , unde DE=FC (trapez isoscel ) => 13 dam = 7 dam + 2 FC => 2 FC = 13 dam - 7 dam= 6 dam => FC = 3 dam = DE
Triunghiul BFC dreptunghic => prin Teorema lui Piagora ca : BC^2 =FC^2 +BF^2 . Facand calculele => BF^2 = 16 => BF(inaltime) = 4 dam .
c ) A= (B+b) * h / 2 ( B= baza mare , b= baza mica , h= inaltime )
A = (13 dam + 7 dam ) * 4 dam / 2 => A = 10 dam * 4 dam / 2 => A = 40 dam^2 / 2 => A= 20 dam ^2
Am pus si figura ^.^
