Question

5. In triunghiul isoscel ABC, (AB) congruent (AC), se ştie BC = 24 cm, sin(<B)=4/5. Să se calculeze perimetrul şi aria triunghiului. ​

Answer 1

Construim inaltimea in triunghiul isoscel:

AN⊥BC; N∈BC

⇒ΔANB este dreptunghic in ∡N.

ΔABC isoscel; AN-h ⇒ BN=NC=BC/2=24/2=12cm

sin∡B=4/5

sin∡B=cateta opusa/ipotenuza=AN/AB

⇒AN/AB=4/5 ⇒AB=4AN/5

Aplic T.P. in ΔANB: $AN^{2}$+$NB^{2}$=$AB^{2}$

⇒16$AN^{2}$/25=$AN^{2} +BN^{2}$

$BN^{2}$=16$AN^{2}$/25 -$AN^{2}$

$BN^{2}$=9$AN^{2}$/25

BN=3AN/5

3AN/5=12

AN=12*5/3

AN=20cm

AN/AB=4/5 ⇒ 20/AB=4/5 ⇒ 20*5=AB*4

⇒AB= 25cm=AC

P=AB+AC+BC=25+25+24=74cm

A=b*h/2

AN-inaltimea

BC-baza

A=AN*BC/2=20*24/2=240$cm^{2}$