Question

In trapezul dreptunghic ABCD, cu AB || CD, AB < CD, m ( <A)= 90 de grade si DB perpendicular pe BC, se cunosc BC= 2√6 cm  si AD= 2√2 cm.
a) Calculati lungimile bazelor [AB] si [CD].
b) Calculati lungimile diagonalelor trapezului ABCD.
c) Determinati lungimea perpendicularei duse din A pe BD.
Va rog frumos sa ma ajutati!

Answer 1

am pus un calcul mare in paranteza.........nu il lua in seama....am vrut sa incerc alta metoda...dar e mai complicat

Answer 2

VEZI FIGURA
a) <ABD=<BDC (alterne interne)
<DAB=<DBC=90⇒ ΔADB≈ΔDBC⇒AD/BC=AB/DB⇒2√2/2√6=AB/DB⇒
√2/√6=AB/DB ridicam la patrat⇒AB²/DB²=1/3
in tr  dreptunghic ADB⇒cf pitagora; DB²=AD²+AB²=8+AB²⇒
AB²/(AB²+8)=1/3
3AB²=AB²+8⇒2AB²=8⇒AB²=4⇒AB=2
ΔBEC cf pitagora: EC²=BC²-BE²=24-8=16⇒EC=4
AB=DE
DC=DE+EC=2+4=6⇒DC=6
b) ΔADB cf pitagora⇒DB²=AD²+AB²=8+4⇒DB=2√3
ΔADC cf PITAGORA⇒AC²=AD²+DC²=8+36⇒AC=2√11
c) AF perpendicular DB
Aria ΔADB= AD·AB/2=AF·DB/2⇒
2√2·2/2=AF·2√3/2⇒ AF=$\frac{2 \sqrt{6} }{3}$