Question

Aflati inaltimile triunghiului isoscel ABC cu AB=BC=13cm si AC=10 cm.
P.S daca se poate ma repede va rog mult...

Answer 1

Fie $p= \frac{AB+BC+AC}{2}= \frac{36}{2}=18$.
$A= \sqrt{p(p-AB)(p-BC)(p-AC)}= \sqrt{18*5*5*8}=$$= \sqrt{2*9*5*5*8}= \sqrt{ 3^{2}* 5^{2}*16 }=3*5*4=60 ( cm^{2} )$

$A= \frac{ h_{AC}*AC }{2}=> h_{AC}= \frac{2A}{AC}= \frac{120}{10}=12(cm).$

$A= \frac{ h_{AB}*AB }{2}=> h_{AB}= \frac{2A}{AB}= \frac{120}{13} (cm)$

$h_{BC}= h_{AC}= \frac{120}{13}$(cm).