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Геометрия

Геометрия. Автор: Славена Маринова. Дадено:. C. Четириъгълник ABCD AC = 30cm MP = 26cm NQ = 28cm S ABCD = ?. P. D. N. Q. А. B. M. Решение:. C. MN – ср. отсечка в ∆ ABC => MN= ║½ AC PQ – ср. отсечка в ∆ ABC => PQ= ║½ AC => MN= ║ PQ => => MNPQ – успоредник

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Геометрия

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  1. Геометрия Автор: Славена Маринова

  2. Дадено: C • Четириъгълник ABCD • AC = 30cm • MP = 26cm • NQ = 28cm • SABCD = ? P D N Q А B M

  3. Решение: C • MN – ср. отсечка в ∆ABC => MN=║½AC • PQ – ср. отсечка в ∆ABC => PQ=║½AC => MN=║PQ => => MNPQ – успоредник => MN=║PQ = 15 MP² + NQ² = 2(MN² + PQ²) 26²+ 28² = 2(15² + MQ²) MQ = √505 Но MQ ср. oтсечка в ∆ABD => BD = 2√505 P D N Q А B M

  4. C SABCD = ½ d1.d2.sinγ MQ║BD и MN║AC=> <QMN = γ ∆QMN – cosT 28² = =15² + 505 – 2.15.√505 . cosγ 2.15.√505 . cosγ = 730 - 784 P D N О Q γ γ 9 cosγ = γ 5√505 А B M

  5. C sin²γ = 1 – (- 9 )² = 5√505 =1 - 81 = 12544 = 25.505 25.505 => sinγ = 112 5√505 P D N О Q γ γ γ А B M

  6. C S = ½ d1.d2. sinγ S = ½.30.2√505. 112 5√505 => S = 6.56 = 336cm² P D N О Q γ γ γ А B M

  7. Край  Благодаря за внимаието

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