Răspuns :
A Sa zicem ca ΔABC este si isoscel , si dreptunghic...
|\ AC=40 cm
| \ ________________________________________________
| \ A=?
| \ ________________________________________________
| \ B=90°⇒AB²+BC²=AC²
|_____\ BC=AB ⇒ 2BC²=40²
B C 2BC²=1600 ⇒ BC²=800 ⇒ BC=20√2
A=C*C/2=(20√2)²/2
A=800/2=400 cm²
|\ AC=40 cm
| \ ________________________________________________
| \ A=?
| \ ________________________________________________
| \ B=90°⇒AB²+BC²=AC²
|_____\ BC=AB ⇒ 2BC²=40²
B C 2BC²=1600 ⇒ BC²=800 ⇒ BC=20√2
A=C*C/2=(20√2)²/2
A=800/2=400 cm²
Acest triunghi il notam ABC.
A = 90 grade; BC=ipotenuza;
Teorema lui Pitagora spune: [tex] AB^{2} + AC^{2} = BC^{2} [/tex] Triunghil este si isoscel --> AB=AC -- > [tex] 40^{2} = 2* Cateta^{2} [/tex] --> 1600=2*[tex] Cateta^{2} [/tex] --> [tex] Cateta ^{2} = [/tex]800 --> Cateta=[tex] \sqrt{800} [/tex]=[tex] \sqrt{800} =20 \sqrt{2} [/tex]
Aria=[tex] \frac{ (20 \sqrt{2} )^{2} }{2} = \frac{800}{2} = 400 cm^{2} [/tex]
A = 90 grade; BC=ipotenuza;
Teorema lui Pitagora spune: [tex] AB^{2} + AC^{2} = BC^{2} [/tex] Triunghil este si isoscel --> AB=AC -- > [tex] 40^{2} = 2* Cateta^{2} [/tex] --> 1600=2*[tex] Cateta^{2} [/tex] --> [tex] Cateta ^{2} = [/tex]800 --> Cateta=[tex] \sqrt{800} [/tex]=[tex] \sqrt{800} =20 \sqrt{2} [/tex]
Aria=[tex] \frac{ (20 \sqrt{2} )^{2} }{2} = \frac{800}{2} = 400 cm^{2} [/tex]