Răspuns:
3. Inaltimea AB a triunghiului ABC este si bisectoare si mediatoare si mediana si inaltime .
In triunghiul ABD -dr. cu AB=25cm si AD=20 cm => T.P.=>
[tex]AD^{2}=AB^{2} +BD^{2} \\AD^{2} = 25^{2} +20^{2} \\AD^{2} =5^{2} *5^{2}+5^{2}*4^{2} \\AD^{2}=5^{2} (5^{2}+4^{2})\\AD^{2} =5^{2} (25+16)\\ AD^{2} = 5^{2} *41\\AD=5\sqrt{41} = h_{1}[/tex]
[tex]A=\frac{BC*AD}{2} =\frac{40*5\sqrt{41} }{2} =100\sqrt{41} \\A=\frac{AB*CE}{2} ; 100\sqrt{41}=\frac{25*h_{2} }{2} ; h_{2} =\frac{100\sqrt{41}*2 }{25} =8\sqrt{41} \\h_{1}+h_{2}+h_{3}=5\sqrt{41} +8\sqrt{41} +8\sqrt{41}=21\sqrt{41}[/tex]
4. a)Cum AB=AC => ΔABC este isoscel =>∡B=∡C
Cum ∡A=120 => ∡A+∡B+∡C=180 =>∡A+2∡B =180 => 120 +2∡B=180=> 2∡B=180-120=> ∡B=60/2=> ∡B=30 si ∡C=30
b) In ΔACD -dr, cu AC=8√3 si ∡C=30 => T∡30 AD=1/2 DC
notam cu x latura AD si aplicam teorema lui Pitagora in ΔADC
[tex]DC^{2} =AD^{2} +AC^{2} \\(2x)^{2} =x^{2} +(8\sqrt{3})^{2} \\4x^{2} =x^{2} +8^{2} *3\\4x^{2}-x^{2}=8^{2} *3\\3x^{2}=8^{2} *3\\x^{2} =\frac{8^{2} *3}{3} \\x=8[/tex]
Cum AD=1/2 DC => DC=2AD=> DC=16 cm
∡A=∡BAD+∡DAC =>120=∡BAD+90=> ∡BAD=30
Cum ∡BAD=30 si ∡B=30 => ΔDAB este isoscel => AD=DB=8
Explicație pas cu pas:
