Question

1.Fie triunghiul dreptunghic LPQ cu m(QLP)=90,inaltimea LS de 12 si QS=2 ori SP. Calculati lungimile:LP,LQ,QP,SP SI SQ
2.FIE ROMBUL ABCD CU AB=10 CM SI AC=16 CM
a) CALCULATI ARIA ROMBULUI
b) DACA M APARTINE [AB] SI MN PERPENDICULAR CU DC (N APARTINE DC),CALCULATI LUNGIMEA MN.
URGENT VA ROG,CAT MAI REPEDE imi trebuie peste o jumatate de ora!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! haideti va rog

Answer 1

PROBLEMA 1)
LS=12 si QS=2*SP
Problema 1)
In ΔQLP, dreptunghic in L, aplicam teorema inaltimii: LS²=QS*SP ⇒12²=2*SP*SP⇒SP=√(12²/2)=√(144/2)=√72=8.5
QS=2*8.5=17
QP=QS+SP=17+8.5=25.5

Aplicam teorema catetei:
QL=√(QP*QS)=√(25.5*17)=20.82
LP=√(SP*QP)=√(8.5*25.5)=14.72

PROBLEMA 2)
a) Aria rombului AB*AC = 10*16=160 cm²
b) In rombul ABCD, notam AC∩DB cu O.
AC=16⇒AO=OB=8
In ΔAOB dreptunghic in O, calculam cateta OB=√(AB²-AO²)=√(10²-8²)=6
In ΔAOB, dreptunghic in O, aplicam teorema catetei:
AO²=AM*AB⇒AM=AO²/AB=8²/10=6.4cm
OB²=MB*AB⇒MB=OB²/AB=6²/10=3.6cm
Aplicam teorema inaltimii: OM=√(AM*MB)=√(6.4*3.6)=4.8cm

In ΔCOD, dreptunghic in O, aplicam teorema catetei:
CO²=CN*DC⇒CN=8²/10=6.4cm
OD²=ND*DC⇒ND=6²/10=3.6cm
Aplicam teorema inaltimii: On=√(CN*ND)=√(6.4*3.6)=4.8cm

In final: MN=MO+ON =4.8+4.8=9.6cm