4 Triangles In A Rectangle Puzzle

Comments · 159

• GI

  @gigigiogio3 · 1 year ago

  Ok, this one was amazing

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• JO

  @johnklinger2868 · 1 year ago

  It is useful to realize that if you squeeze the rectangle horizontally by a factor and stretch it out by the same factor vertically, then all the areas remain the same. So given the areas, you can't find the lengths of the sides. But you can solve the problem by assuming the rectangle is a square, since it can be squeezed and stretched into one without changing the areas.

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• VI

  @vishalmishra3046 · 1 year ago

  Let A+B = long side and C + D = short side of the rectangle. Now all 4 triangle areas are computable from their orthogonal sides, creating 4 equations that use all the information in the problem, so solving them should answer what is S.<br>BC = 2S and D (A+B) = 2S + 2 and A (C+D) = 2S + 4 and (A+B)(C+D) = Area of the Rectangle = 4S + 6 = (2S + 2) + (2S + 4) = D (A+B) + A (C+D)<br>So, AC + AD + BC + BD = Area of the Rectangle = AD + BD + AC + AD. Cancelling 3 common terms AC AD and BD, we get AD = BC = 2S, so BD = 2 (from 2nd equation) and AC = 4 (from 3rd equation)<br>So, 2 x 4 = BD x AC = ABCD = AD x BC = 2S x 2S = 4 S^2, so S^2 = 2, so S = √2

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• NA

  @navghtivs · 1 year ago (edited)

  I call BC=x1, AB=x2, ED=x3, FD=x4<br><br>write down the areas of the rectangle and three corner triangles:<br><br>a: x1x2=4S+6<br>b: x1(x2-x3)=2S+2<br>c: x2(x1-x4)=2S+4<br>d: x3x4=2S<br><br>e: from a-b you get x1x3 = 2S+4<br>f: from a-c you get x2x4 = 2S+2<br><br>put a*d on one side and e*f on the other you will get a simple equation of S to solve.

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• ES

  @essemque · 1 year ago

  That is so much cleaner than my method, which involved using Heron&apos;s Formula for the area of the central triangle; well done!

  9

• TR

  @treyg5265 · 1 year ago

  This method is quite good not gonna lie.

  18

• PU

  @putyrak · 1 year ago

  The formula is interesting, but not necessary to solve this problem. If you draw a horizontal line through E and vertical through F, then you have small rectangles of areas (starting bottom right ang going counter-clockwise): 2S, 2, 2S, 4. If a &amp; b are the vertical intervals and c &amp; d horizontal, then ac = bd = 2S, bc = 2, ad = 4. Now this is an easy manipulation of equation: by multiplying the first one by b, you get 2a = abc = b^2d = 2Sb and likewise 2b = Sa. From that 2a = 2Sb = aS^2, or S^2 = 2.

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• 飞来

  @飞来峰-x6r · 1 year ago

  Solution: Based on the point markings at the 28th second of the video, draw EG∥BC through point E, intersecting segment AB at point G. Also, draw FH∥AB through point F, intersecting segment BC at point H. Let segments EG and FH intersect at point I. Let the area of ​​□BGIH be x. Then the area of ​​□HIEC S1 = 2 × the area of ​​△BCE - x = 2 × (S + 1) - x = 2S + 2 - x, and the area of ​​□AFIG S2 = 2 × the area of ​​△ABF - x = 2 × (S + 2) - x = 2S + 4 - x. Since the area of ​​□ABCD = S + 1 + S + 3 + S + 2 + S = 4S + 6 = S1 + S2 + 2S + x = 6S - x + 6, we can calculate: x = 2S. Therefore, we can find: S1 = 2S + 2 - x = 2, and S2 = 2S + 4 - x = 4. Therefore, 2S x = S1 × S2 = 8. After substituting x=2S, the above expression can be solved to: S=√2.

  1

• CO

  @coraleap · 1 year ago

  Here’s my solution:<br><br>The total area of the rectangle is 4s + 6. This means that the right edge of the top-right triangle takes up (s+1)/(2s+3) of the length of the right side of the triangle, and the bottom-left edge of the bottom-left triangle takes up (s+2)/(2s+3) of the bottom length of the rectangle. This means the bottom-right triangle’s right and bottom edges take up (s+2)/(2s+3) and (s+1)/(2s+3) of the right and bottom lengths of the rectangles respectively, so the area of that triangle should be (s+1)(s+2)/(2s+3). But it’s also s, so s(2s+3) = (s+1)(s+2). Solving this equation gives us s=sqrt(2).<br><br>It’s a really nice problem!

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• U2

  @U2FwcGhpcmUg · 1 year ago

  I used the coordinate method and found S = sqrt(2) pretty quickly.

  1

• DA

  @damn_daniel511 · 1 year ago

  Wow, I loved the way you solved it, man 😮. Out of my mind, good job 😁👍

• MI

  @milos_radovanovic · 1 year ago (edited)

  Proving complex and obscure theorems for simple geometry problems always seems like the wrong approach to me!<br>Simply convert triangles to rectangles and you can pretty quickly, in a few steps, deduce that the original rectangle is split into four smaller ones of the following areas: [2s, 2; 4, 2s] and using side proportions get the equation (2s)/2=4/(2s) witch imedietly simplifies and solves to s=sqrt(2) asumming positive s.<br>And if you want, for pedagogical purposes, you can then proceed to generalise the solution and, using the same logic, prove your theorem.<br><br>(Two steps I skipped mentioning are the centre dart shape of area 3 and the L shape of area 2s+6)

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