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A Nice Geometry Problem | Can you find length X ? 

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A Nice Geometry Problem | Can you find length X ?
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27 янв 2024

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Комментарии : 13   
@miguelgnievesl6882
@miguelgnievesl6882 6 месяцев назад
The triangle EFC is notable (3k,4k,5k) so it is obvious that
@kelvin-rj9yr
@kelvin-rj9yr 5 месяцев назад
Similar and then pytha. Good thought!!! I like ED = X/4 the most.
@MarieAnne.
@MarieAnne. 6 месяцев назад
I used a similar approach. Since △FAE ~ △EDC then AF/DE = AE/CD = EF/CE → AF/(x−AE) = AE/x = 6/8 = 3/4 AE/x = 3/4 → AE = 3x/4 AF/(x−AE) = 3/4 → AF = 3x/4 − (3/4)(AE) = 3x/4 − (3/4)(3x/4) = 12x/16 − 9x/16 = 3x/16 Using Pythagorean theorem in △FAE we get: AF² + AE² = EF² (3x/16)² + (3x/4)² = 6² 9x²/256 + 9x²/16 = 36 153x²/256 = 36 153x² = 36 * 256 (3√17)x = 6 * 16 *x = 32/√17*
@manojkantsamal4945
@manojkantsamal4945 5 месяцев назад
Respected Sir 🙏 Really your way of answering goes with the grammar of mathematics....
@plamenpenchev262
@plamenpenchev262 6 месяцев назад
AFE and CDE are similar then AE = 6x/8, i.e. y = AD - ED = x - 6x/8 = 2/8x In CDE (2x/8)^2 + x^2 = 8^2 i.e. x = 32/sqrt(17)
@rick57hart
@rick57hart 6 месяцев назад
It would be more interesting if triangle CEF would not be 90°. 😜
@manojkantsamal4945
@manojkantsamal4945 5 месяцев назад
X=5×root 2, if the quadrilateral is a square...
@yakupbuyankara5903
@yakupbuyankara5903 5 месяцев назад
X=32/(17^(1/2))
@nunoalexandre6408
@nunoalexandre6408 6 месяцев назад
Love it🎉
@skwest
@skwest 6 месяцев назад
Hmmm... (32/17)√17 ?
@comdo777
@comdo777 6 месяцев назад
asnwer=9cm
@MathBooster
@MathBooster 6 месяцев назад
answer is 32/sqrt(17)
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