Can You Find "x"? Geometrical/Mathematical Challenge !

Comments · 82

• MA

  @MataniMath · 1 month ago

  Nice problem and solution

  3

• AN

  @andersonantunesdeleu8591 · 1 month ago

  Beautiful resolution! I wonder if it’s the only one possible though…

• IS

  @isidu4244 · 2 months ago

  Thank you for your efforts!

  1

• IA

  @ianhinson2829 · 2 months ago (edited)

  At the <a href="https://www.youtube.com/watch?v=1H49EmSoig8&amp;t=300">5:00</a> mark, once all lines of length &apos;a&apos; had been set up, then the value of &apos;a&apos; can be directly calculated using the similarity of the trapezoids on each side of the vertical line ie. by comparing the ratio of their left and right sides.<br>Left trapezoid: lhs = 9; rhs = a<br>Right trapezoid: lhs = a; rhs = 3<br>By similarity their rhs/lhs ratio is the same.<br>ie. a/9 = 3/a =&gt; a² = 27<br>=&gt; a = 3√3<br>Soln: x = 2a = 6√3

  4

• BA

  @bartoszczajkowski5765 · 1 month ago

  Bardzo ciekawe zadanie. 👌👍

• MA

  @marcdeangelis3497 · 1 month ago

  I liked it.

• SA

  @SafaldeenAlbayati · 1 month ago

  لو لديكم أسئلة اثرائية حول موضوع مجموعة الاعداد التخيلية أو المركبةa+bi

  1

• KO

  @Kotaro_Mario · 1 month ago (edited)

  If you draw the two trapezoids &nbsp;ABCD and EFCD, then the trapezoids are similar, that is, 3/a=a/9. &nbsp;Solving this equation, you directly obtain a^2=27, and the solution: x=2a=6*sqr(3).

  1

• AB

  @abcMATEMATICA · 1 month ago

  En este tipo de problemas siempre se cumple que: las distancias perpendiculares H y h están en relación a R y r ---&gt; H.h=R.r &nbsp;, y que x = 2(√Rr).

• JV

  @jv_fnaf435 · 1 month ago

  👏👏👏👏👏👏👏👏

• BE

  @benazouzlemouchi85 · 1 month ago

  Nice &quot;half circles&quot; youhave drawn there. Sincerely i thought it was parabolasand triedto solve the problem without any success. Iwascon the wrong path.😅😅😅

• SR

  @srinivasch-re2oq · 3 weeks ago

  We can easily find x.