Question

se considera triunghiul abc in care bp perpendicular pe ac
ac=13
bp=6
2 radical ap= 3 radical pc
CERINTA: stabiliti natura triunghiului ABC

Answer 1

2√AP=3√PC
4AP=9PC
AP=9PC/4
AP+PC=AC
9PC/4+PC=13
9PC+4PC=52
13PC=52
PC=4
AP=13-4
AP=9
BP=√AP×PC =√ 9×4=√36=6 cm

Answer 2

"2 radical ap= 3 radical pc"

[tex]\it 2\sqrt{AP} =3\sqrt{PC} \Rightarrow \sqrt{\dfrac{AP}{PC}} =\dfrac{3}{2} \Rightarrow \dfrac{AP}{PC} = \dfrac{9}{4} \stackrel{derivare}{\Longrightarrow } \\\;\\ \\\;\\ \dfrac{AP+PC}{PC} =\dfrac{9+4}{4} \Rightarrow \dfrac{AC}{PC} = \dfrac{13}{4} \Rightarrow \dfrac{13}{PC} =\dfrac{13}{4} \Rightarrow PC = 4 cm[/tex]


$\it AP = AC-PC = 13-4=9cm$

$\it AP \cdot PC = 9\cdot4 = 36 =6^2= BP^2$

Din reciproca teoremei înălțimii ⇒ ABC - dreptunghic în B