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Stabiliti daca triunghiul ABC estedreptunghic cu m unghiului BAC =90 grade stiind ca AB=4/3 cm, AC=3/3 cm si BC=5/3 cm

Răspuns :

Triunghiu
Reciproca Teoremei lui Pitagora C₁²+C₂²=ip²
AB²+AC²=BC²
(4/3)²+(3/3)²=(5/3)²
(16+9)/3=25/3
25/3=25/3    ⇒   ΔABC dreptunghic(∡A=90°)
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Abscisa5

[tex]\it AC = 3\cdot\dfrac{1}{3},\ \ AB= 4\cdot\dfrac{1}{3},\ \ BC = 5\cdot\dfrac{1}{3} \\ \\ \\ AC^2+AB^2 = 3^2\cdot\left(\dfrac{1}{3}\right)^2 + 4^2\cdot\left(\dfrac{1}{3}\right)^2 = \left(\dfrac{1}{3}\right)^2 (3^2+4^2) = \\ \\ \\ = \left(\dfrac{1}{3}\right)^2 (9+16) = \left(\dfrac{1}{3}\right)^2 \cdot25 \ \ \ \ (1) [/tex]

[tex]\it BC^2 = 5^2\cdot\left(\dfrac{1}{3}\right)^2 = 25\cdot \left(\dfrac{1}{3}\right)^2 = \left(\dfrac{1}{3}\right)^2 \cdot25 \ \ \ \ (2) \\ \\ \\ (1),\(2) \Rightarrow AC^2+AB^2=BC^2\ \stackrel{RTP}{\Longrightarrow}ABC-dreptunghic,\ m(\hat{A})=90^0[/tex]


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