1) a) in ΔDAC dreptunghic in D:
sin <C=AD/AC=> √2/2=AD/18√2 => AD= 18*2/2= 18
=>cos<C=CDAC=> √2/2=CD18√2 => CD= 18*2/2= 18
in ΔDAB dreptunghic in D, avem
tg <B= AD/BD=> √3/3 = 18/BD=> BD=18√3
BC=BD+DC=18+18√3=18(1+√3)
Aria ABC=AD*BC/2=18*(18+18√3)/2=162+162√3=162(1+√3)
b) in ΔBDA, sin <B=AD/AB=> 1/2=18/AB=> AB=18*2=36
In ΔABC : AD*BC=CF*AB=BE*AC
AD*BC=CF*AB => 18*18(1+√3)=CF*36 => CF= 18*18(1+√3) :36=9(1+√3)
AD*BC=BE*AC => 18*18(1+√3)=BE*18√2 => BE= 18*18(1+√3)/18√2= 9√2(1+√3)
2)Trapezul Isoscel ABCD , AB| |CD , are AB=80 cm , CD=32 cm , BC=AD=30cm .
a)Fie DE_|_ AB=> AE=(AB-CD)/2=24
DE²=AD²-AE²=30²-24²=324 => DE=18
Aria trapezului ABCD=(CD+AB)*DE/2=(80+32)*18/2=1008
b){P}=AD∩BC
fie PF_|_CD=> inaltimea ΔPAB=H =PF+DE
PF/H=CD/AB => PF/(PF+DE)=32/80=>PF/(PF+18)=32/80
32PF+576=80PF=> 48PF=576=> PF=12
Aria triunghiului PCD =PF*CD/2=12*32/2=192
c)Aria triunghiului PAB=(PF+DE)*AB/2=(12+18)*80/2=1200
Aria triunghiului PCD , din aria triunghiului PAB=192*100/1200= 16%
3) tr. dreptunghic ABCD , AB| |CD , si m(∠A)=90, AB=27cm , CD=18cm si AD=12cm.
a)Fie CE_|_AB=> CE=AD=12
In ΔCEB, EB=AB-CD=27-18=9
BC²=CE²+EB²=12²+9²=225=> BC=15
b) fie AF_|_ BC=> AF=Distanta de la punctul A la dreapta BC
in ΔABC avem CE*AB=AF*CB=> 12*27=AF*15=> AF=324/15=108/5=21,6
c) in ΔCAE AC²=AE²+CE²=18²+12²=468 => AC=6√13
d( B ;AC) *AC=CE*AB=>d( B ;AC) =CE*AB/AC= 12*27/6√13=54√13/13
4. romb ABCD , AB=24cm ,
a)m(∠DAC)=30 => <DAB=60=> ΔABD=echilateral=> BD=AB=24
AC=2*inaltime Δechilateral =2/ √3/2 *AB=2*√3/2*24=24√3
b ) Aria rombului =BD*AC/2=24*24√3/2=288√3
c)fie BD∩AC={F} si BF=BD/2=24/2=12
In ΔBAC avem:
BF*AC=d( C ;AB)*AB=>
d( C ;AB)=BF*AC/ AB =12*24√3/24=12√3
