Question

9. În triunghiul ABC, m(ZB) = 90, BD LAC, DE (AC). Dacă 5BD=4AD AC = 41cm, aflaţi lungimile segmentelor [BD], [AD] şi [CD].​

Answer 1

Răspuns:

20;25,  respectiv 16cm, in ordinea cerintelor

Explicație pas cu pas:

BD/AD=4/5=4k/5k

Teorema inaltimii in tr.dr ABC

BD^2=AD*DC

16k^2=5k*DCD

DC=16k/5

AD+DC=41

5k+16k/5=41

41k/5=41

k/5=1

k=5 atunci

BD=4k=4*5=20cm

AD=5k=..=25cm

CD=15k/5=16cm

Answer 2

$\it Not\breve am\ BD=5x \Rightarrow AD=4x \Rightarrow CD=41-4x\\ \\ Teorema\ \hat\imath n\breve al\c{\it t}imii \Rightarrow (5x)^2=(41-4x)\cdot4x \Rightarrow 25x^2=41\cdot4x-16x^2\bigg|_{+16x^2} \Rightarrow \\ \\ \\ 41x^2=41\cdot4x\bigg|_{41x} \Rightarrow x=4\\ \\ \\ BD=5\cdot4=20\ cm;\ \ AD=4\cdot4=16\ cm;\ \ CD=41-16=25\ cm$