1) a) ducand inaltimile AM si BN (M,N∈DC) se obtin Δ dreptunghice AMD si BNC
daca - <BCN=60 atunci <CBN=30, deci cateta opusa lui e jumatate din ipotenuza:
NC=BC/2=12√2/2=6√2
la fel, avem DM=AD/2=12√2/2=6√2
DC=DM+NC+MN=3*6√2=18√2
(MN=AB=6√2)
b) pitagora in ΔBNC: BC²=BN²+CN² BN²=(12√2)²-(6√2)²=216=2² * 3² * 6, deci BN=6√6
Pitagora in ΔBND>: BD²=BN²+DN²
BD²=216+(DM+MN)²=216+(6√2+6√2)²=216+288=504=2² * 3² * 2 * 7
BD=6√14 BD=AC (diagonalele in trapez isoscel sunt egale)
c) DX e perpendiculara pe AC
in ΔAXD: AD²=AX²+DX² DX²=(12√2)²-AX²=288-AX² - rel.1
in ΔDXC: DC²=DX²+XC²=DX²+(AC-AX)²=DX²+AC²-2*AC*AX+AX²,
deci DX²=DC²-AC²+2*AC*AX-AX²=(18√2)²-(6√14)²+2*6√14*AX-AX²=
=648-505+12√14*AX-AX²=144+12√14*AX-AX² - rel.2
din rel. 1 si 2: 288-AX²=144+12√14*AX-AX² 144=12√14*AX AX=12/√14=6√14/7
DX²=288-(6√14/7)²=288-504/49=13608/49=(2² * 3² * 3² * 2*3*7)/49 DX=18√42/7
2) a) intr-un trapez isoscel cu diagonalele perpendiculare, inaltimea trapezului h=(b+B)/2=(12+28)/2=20
aria trapezului: (b+B)*h/2=(12+28)*20/2=400
b) perimetrul trapezului: AB+BC+DC+AD=12+2*AD+28=40+2*AD
daca AH e inaltimea trapezului (H∈DC), AH=h=20, avem AD²=AH²+DH²
conform problemei anter. DH=(DC-AB)/2 - in trapezul isoscel inaltimile formeaza un dreptunghi
deci, DH=(28-12)/2=8 AD²=20²+8²=2² * 2² * 29 AD=4√29
PERIMETRUL=40+2*4√29=40+8√29
c) AC²=AH²+HC², si HC=DC-DH=28-8=20, deci AC²=20²+20², AC=20√2
IMI CER SCUZE, NU MAI AM TIMP SA STAU SA REZOLV 3 SI 4
