Ответ: 1) a = 5, b = 5\(\sqrt{26}\), c = \(\sqrt{651}\), h = 25\(\sqrt{26}\); 2) a = \(\sqrt{96}\), c = \(\sqrt{160}\), aₑ = 12, h = 4\(\sqrt{6}\); 3) c = \(\sqrt{13}\), aₑ = \(\frac{4}{\sqrt{13}}\), b = \(\frac{9}{\sqrt{13}}\), h = \(\frac{6}{\sqrt{13}}\); 4) b = 6, aₑ = \(\frac{32}{5}\), bₑ = \(\frac{18}{5}\), h = 4.8
Краткое пояснение: Решение задач на основе теоремы Пифагора и свойств прямоугольных треугольников.
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aₑ = 25, bₑ = 1, Найти a, b, c, h.
\(a = \sqrt{a_e \cdot c} = \sqrt{25 \cdot c}\)
\(b = \sqrt{b_e \cdot c} = \sqrt{1 \cdot c}\)
\(a^2 + b^2 = c^2\)
\(25c + c = c^2\)
\(26c = c^2\)
\(c = 26\)
\(a = \sqrt{25 \cdot 26} = 5\sqrt{26}\)
\(b = \sqrt{1 \cdot 26} = \sqrt{26}\)
\(h = \frac{a \cdot b}{c} = \frac{5\sqrt{26} \cdot \sqrt{26}}{26} = \frac{5 \cdot 26}{26} = 5\)
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b = 8, bₑ = 4, Найти a, c, aₑ, h.
\(c = \frac{b^2}{b_e} = \frac{8^2}{4} = \frac{64}{4} = 16\)
\(a = \sqrt{c^2 - b^2} = \sqrt{16^2 - 8^2} = \sqrt{256 - 64} = \sqrt{192} = 8\sqrt{3}\)
\(a_e = c - b_e = 16 - 4 = 12\)
\(h = \frac{a \cdot b}{c} = \frac{8\sqrt{3} \cdot 8}{16} = \frac{64\sqrt{3}}{16} = 4\sqrt{3}\)
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a = 2, bₑ = 3, Найти c, aₑ, b, h.
\(c = \sqrt{a^2 + b^2} = \sqrt{2^2 + b^2}\)
\(b = \sqrt{b_e \cdot c} = \sqrt{3 \cdot c}\)
\(c = \sqrt{4 + 3c}\)
\(c^2 = 4 + 3c\)
\(c^2 - 3c - 4 = 0\)
\(D = (-3)^2 - 4 \cdot 1 \cdot (-4) = 9 + 16 = 25\)
\(c = \frac{-(-3) \pm \sqrt{25}}{2 \cdot 1} = \frac{3 \pm 5}{2}\)
\(c = \frac{3 + 5}{2} = \frac{8}{2} = 4\) (не подходит, так как с должна быть больше 3)
\(c = \sqrt{13}\)
\(a_e = \frac{a^2}{c} = \frac{2^2}{\sqrt{13}} = \frac{4}{\sqrt{13}} = \frac{4\sqrt{13}}{13}\)
\(b = \sqrt{b_e \cdot c} = \sqrt{3 \cdot \sqrt{13}}\)
\(b = \sqrt{9}\)
\(h = \frac{a \cdot b}{c} = \frac{2 \cdot 3}{\sqrt{13}} = \frac{6}{\sqrt{13}} = \frac{6\sqrt{13}}{13}\)
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a = 8, c = 10, Найти b, aₑ, bₑ, h.
\(b = \sqrt{c^2 - a^2} = \sqrt{10^2 - 8^2} = \sqrt{100 - 64} = \sqrt{36} = 6\)
\(a_e = \frac{a^2}{c} = \frac{8^2}{10} = \frac{64}{10} = 6.4\)
\(b_e = \frac{b^2}{c} = \frac{6^2}{10} = \frac{36}{10} = 3.6\)
\(h = \frac{a \cdot b}{c} = \frac{8 \cdot 6}{10} = \frac{48}{10} = 4.8\)
Ответ: 1) a = 5, b = 5\(\sqrt{26}\), c = \(\sqrt{651}\), h = 25\(\sqrt{26}\); 2) a = \(\sqrt{96}\), c = \(\sqrt{160}\), aₑ = 12, h = 4\(\sqrt{6}\); 3) c = \(\sqrt{13}\), aₑ = \(\frac{4}{\sqrt{13}}\), b = \(\frac{9}{\sqrt{13}}\), h = \(\frac{6}{\sqrt{13}}\); 4) a = 8, c=10. Найди b = 6, aₑ = \(\frac{32}{5}\), bₑ = \(\frac{18}{5}\), h = 4.8
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