Răspuns:
c) AB=5√6
BC=25
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Aplici Pitagora si aflli AC
AC²=BC²-AB²
AC²=25²-(5√6)²=
625-150=475
AC=√475=√25*19=5√19m
Aplici teorema catetei si AFlii BD
AB²=BD*BC
(5√6)²=BD*25
150=25BD
BD=150/25=6m
CD=BC-BD=25-6=19m
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d) AB=2x BD=x
Aplici teorema catetei
AB²=BC*BD
(2x)²=BC*x
4x²=BC*x=>
BC=4x²/x=4x
Aflii AC cu Pitagora
AC²=BC²-AB²
AC²=(4x)²-(2x)²=16x²-4x²
AC²=12x²
AC=√12x²=√3*4x²=2x√3cm
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Varianta 2
c)AB=5√6
BC=25
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Aplici teorema catetei
AB²=BC*BD
(5√6)²=25*BD
150=25 *BD
BD=150/25=6m
CD=BC-BD=25-6=19m
Aplici teorema catetei si aflii AC
AC²=CD*BC
AC²=19*25=475
AC=√475=√25*19=5√19
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d)AB=2x
BD=x
Aplici teorema catetei si aflii BC
AB²=BD*BC
(2x)²=x*BC
4x²=xBC
BC=4x²/x=4x
=> CD=BC-BD=4x-x=3x
Aplici teorema catetei si afli AC
AC²=CD*BC=3x*4x=12x²
AC=√12x²=2x√3cm
Explicație pas cu pas:
