Răspuns:
BD, CD, AD, AB, AC, BC
Explicație pas cu pas:
a) 9; 16; 12; 15; 20; 25
Teorema lui Pitagora
[tex]BC = \sqrt {AB^2+AC^2} = \sqrt{ {15}^{2} + {20}^{2} } = 25 \ cm \\ [/tex]
Teorema catetei:
AB²=BD×BC, 225=BD×25, BD=9 cm
CD=BC-BD=25-9=16 cm
Teorema înălțimii:
[tex]AD = \sqrt {BD \cdot CD} = \sqrt{9 \times 16} = 12 \: cm \\ [/tex]
b) 14,4; 25,6; 19,2; 24; 32; 40
[tex]AC = \sqrt {BC^2 - AB^2} = \sqrt {40^2 - 24^2} = 32 \ cm \\ [/tex]
AB²=BD×BC, 24²=BD×40, BD=14,4 cm
CD=BC-BD=40-14,4=25,6 cm
[tex]AD = \sqrt {BD \cdot CD} = \sqrt {14,4 \cdot 25,6} = 19,2 \ cm \\ [/tex]
c) 7,2; 12,8; 9,6; 12; 16; 20
[tex]AB = \sqrt {BC^2 - AC^2} = \sqrt {20^2 - 16^2} = 12 \ cm \\ [/tex]
AB²=BD×BC, 12²=BD×20, BD=7,2 cm
CD=BC-BD=20-7,2=12,8 cm
[tex]AD = \sqrt {BD \cdot CD} = \sqrt {7,2 \cdot 12,8} = 9,6 \ cm \\ [/tex]
d) 18; 32; 24; 30; 40; 50
[tex]AD = \sqrt {BD \cdot CD} = \sqrt {18 \cdot 32} = 24 \ cm \\[/tex]
BC=BD+CD=18+32=50 cm
[tex]AB = \sqrt {BD \cdot BC} = \sqrt {18 \cdot 50} = 30 \ cm \\[/tex]
[tex]AC = \sqrt {CD \cdot BC} = \sqrt {32 \cdot 50} = 40 \ cm \\[/tex]
