Odpowiedź:
[tex](5\sqrt{3} )^{2}[/tex] + [tex]x^{2}[/tex] = [tex]10^{2}[/tex]
75 + [tex]x^{2}[/tex] = 100
[tex]x^{2}[/tex] = 100 - 75 = 25
x = [tex]\sqrt{25}[/tex] = 5
Pole Δ ABC = [tex]\frac{5\sqrt{3}x5}{2}[/tex] = [tex]\frac{25\sqrt{3} }{2}[/tex]
Pole Δ ABC innym sposobem = [tex]\frac{10h}{2}[/tex]
[tex]\frac{10h}{2}[/tex] = [tex]\frac{25\sqrt{3} }{2}[/tex]
10h = 25[tex]\sqrt{3}[/tex]
h = [tex]\frac{25\sqrt{3} }{10}[/tex] = [tex]\frac{5\sqrt{3} }{2}[/tex]
Trójkąt EBC (Pole):
[tex](\frac{5\sqrt{3} }{2}) ^{2}[/tex] + [tex]y^{2}[/tex] = [tex]5^{2}[/tex]
[tex]\frac{75}{4}[/tex] + [tex]y^{2}[/tex] = 25
[tex]y^{2}[/tex] = 25 - [tex]\frac{75}{4}[/tex] = [tex]\frac{100}{4}[/tex] - [tex]\frac{75}{4}[/tex] = [tex]\frac{25}{4}[/tex]
y = [tex]\sqrt{\frac{25}{4} }[/tex] = [tex]\frac{5}{2}[/tex]
Trójkąt ECD (Pole):
[tex](\frac{5\sqrt{3} }{2}) ^{2}[/tex] + [tex](\frac{15}{2})^{2}[/tex] = [tex]z^{2}[/tex]
[tex]\frac{75}{4}[/tex] + [tex]\frac{225}{4}[/tex] = [tex]z^{2}[/tex]
[tex]z^{2}[/tex] = [tex]\frac{300}{4}[/tex]
z = [tex]\sqrt{\frac{300}{4} }[/tex] = [tex]\frac{\sqrt{300} }{2}[/tex] = [tex]\frac{10\sqrt{3} }{2}[/tex] = 5[tex]\sqrt{3}[/tex]
Trójkąt BCD:
Obwód:
5 + 5 + 5[tex]\sqrt{3}[/tex] = 10 + 5[tex]\sqrt{3}[/tex] = 5(2 + [tex]\sqrt{3}[/tex])
Pole:
P = a × h = 5 × [tex]\frac{5\sqrt{3} }{2}[/tex] = [tex]\frac{25\sqrt{3} }{2}[/tex] = 12,5[tex]\sqrt{3}[/tex]
