Ülesanne.

On antud punkt A koordinaatidega (0; 1) ja vektorid AB = (3; -1), BC = (-1; 1), CD = (1; -4) ja DE = (-5; -1). Leidke kolmnurga ACE pindala ning kolmnurga ABD ümbermõõt (vastused ümardage kümnendikeni).

Vastus.

Leiame kõik punktid:

$$B = A + AB = (0 + 3; 1 - 1) = (3; 0)$$ $$C = B + BC = (3 - 1; 0 + 1) = (2; 1)$$ $$D = C + CD = (2 + 1; 1 - 4) = (3; -3)$$ $$E = D + DE = (3 - 5; -3 - 1) = (-2; -4)$$

Kolmnurga ACE pindala:

$$\vec{AC} = C - A = (2 - 0; 1 - 1) = (2; 0)$$ $$\vec{CE} = E - C = (-2 - 2; -4 - 1) = (-4; -5)$$ $$\vec{AE} = E - A = (-2 - 0; -4 - 1) = (-2; -5)$$ $$\vert{\vec{AC}}\vert = \sqrt{2^2 + 0^2} = 2$$ $$\vert{\vec{CE}}\vert = \sqrt{(-4)^2 + (-5)^2} = \sqrt{16 + 25} = \sqrt{41}$$ $$\vert{\vec{AE}}\vert = \sqrt{(-2)^2 + (-5)^2} = \sqrt{4 + 25} = \sqrt{29}$$ $$S = \sqrt{p(p-a)(p-b)(p-c)}$$ $$p = \frac{a+b+c}{2}$$ $$p = \frac{2 + \sqrt{41} + \sqrt{29}}{2}$$ $$S = 5$$

Kolmnurga ABD ümbermõõt:

$$\vec{AB} = B - A = (3 - 0; 0 - 1) = (3; -1)$$ $$\vec{BD} = D - B = (3 - 3; -3 - 0) = (0; -3)$$ $$\vec{AD} = D - A = (3 - 0; -3 - 1) = (3; -4)$$ $$\vert{\vec{AB}}\vert = \sqrt{3^2 + (-1)^2} = \sqrt{9 + 1} = \sqrt{10}$$ $$\vert{\vec{BD}}\vert = \sqrt{0^2 + (-3)^2} = 3$$ $$\vert{\vec{AD}}\vert = \sqrt{3^2 + (-4)^2} = \sqrt{9 + 16} = 5$$ $$P = \sqrt{10} + 3 + 5$$ $$P \approx 11.2$$