Explicație pas cu pas:
1) BC = 6
[tex]AB = AC = \sqrt{ {3}^{2} + {4}^{2} } = \sqrt{9 + 16} \\ = \sqrt{25} = 5[/tex]
perimetrul ΔABC = 2×5 + 6 = 10 + 6 = 16
AO = 4
aria ΔABC =
[tex]= \frac{AO \times BC}{2} = \frac{4 \times 6}{2} = 12 \\ [/tex]
2) BC = 4
[tex]AB = AC = \sqrt{ {2}^{2} + {2}^{2} } = \sqrt{4 + 4} \\ = \sqrt{8} = 2 \sqrt{2} [/tex]
perimetrul ΔABC =
[tex]= 2 \times 2 \sqrt{2} + 4 = 4 \sqrt{2} + 4 = 4( \sqrt{2} + 1) \\ [/tex]
AO = 2
aria ΔABC =
[tex]= \frac{AO \times BC}{2} = \frac{2 \times 4}{2} = 4 \\ [/tex]
3) BC = 3
[tex]AB = \sqrt{ {2}^{2} + {2}^{2} } = \sqrt{8} = 2 \sqrt{2} \\ [/tex]
[tex]AC = \sqrt{ {5}^{2} + {2}^{2} } = \sqrt{25 + 4} = \sqrt{29} \\ [/tex]
perimetrul ΔABC =
[tex] = 3 + 2 \sqrt{2} + \sqrt{29}[/tex]
d(A, BC) = 2
aria ΔABC =
[tex]= \frac{d(A, BC) \times BC}{2} = \frac{2 \times 3}{2} = 3 \\ [/tex]
