Question

consideram triunghiul ABC cu m unghiului A=135 grade,AB=6√2 cm si AC=AB√2.Calculati lungimile inaltimilor ΔABC

Answer 1

AC=AB√2=6√2*√2=12
fie BFperpendicular pe AC, F∈AC( atentie ca BF cade pe prelungirea lui AC deoarece mA = 135 grade)⇒ΔBAF dreptunghic isoscel cu mBAF=mFBA=45grade
(mFAB+mBAC=180grade⇒mFAB=45grade)⇒AB=l√2(ipotenuza in Δdreptunghic isoscel)⇒6√2=FA√2⇒FA=6⇒FB=6cm
fie CE perp pe AB, E∈AB⇒ΔACE drept isoscel cu mCAE=mECA=45grade⇒AE=EC
AC ipotenuza⇒AC=CE√2⇒12=CE√2⇒CE=12/√2⇒CE=6√2cm⇒AE=6√2 cm
BE=BA+AE=6√2+6√2
CE=12√2

Ne incadram in ΔBEC cu mBEC=90grade si aplicam Pitagora
BC²=BE²+EC²
BC²=(12√2)²+(6√2)²
BC²=288+72
BC²=360
BC=6√10cm
Aria ΔABC=BE*EC/2=12√2*6√2/2=72cm²
fie AG perpendicular pe BC⇒ Aria ΔABC = BC*AG/2⇒72=6√2*AG/2⇒
AG=72*2/(6√2)=24/√2=24√2/2=12√2cm