A Very Nice Geometry Problem | You should be able to solve this!

A Very Nice Geometry Problem | You should be able to solve this! | 2 Methods

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Күн бұрын

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Пікірлер: 30

@johnbrennan3372 Ай бұрын

Triangles bqf and cqe are similar so 4/2= cq/bq. Then cq= 2x/3. The triangles qce and pde are similar so 4/1= (x+ed)/ed. so ed=x/3. Using triangle qec 4^2= (x+x/3)^2+ (2x\3)^2. Then 16= 20(x)^2 divided by 9. X=6/ sqroot 5.

@giuseppemalaguti435 Ай бұрын

ED=a ..arccos(a+x)/4=arcsin((x-√(16-(a+x))^2)/2)=arccos(a/1)...x=6/√5...a=x/3=2/√5

@jairoeveliogordillomarin5780 17 күн бұрын

En la figura corro el lado AD hasta que D coincida con E. Se me forma un triángulo rectángulo ECM, con ángulo recto en M y una hipotenusa igual a 6 y catetos: EM= 2X y CM= X.. Aplico teorema de Pitágoras: 6²= X²+(2X)² Entonces: X= 2.683 u, o X= 6÷√5 u.

@frankxu9675 Ай бұрын

x^2 + (2x)^2 = (1+3+2)^2 5 x^2 = 6^2 x = 6 / sqrt(5)

@davidseed2939 Ай бұрын

but you don’t say how you measure to get those equations

@harikatragadda Ай бұрын

Imagine a rectangle with a diagonal 6 and a Square of side X in it. Because of Similarity, the Left box and the Right box bases also add upto X. Hence the rectangle can be divided into two Square boxes of side X. Applying Pythagoras theorem, (2X)² +X²=6² X =6/√5

@Zollaho Ай бұрын

What I did. 15 seconds mentally

@bpark10001 Ай бұрын

There is no need for line PM. Extend line EC. Drop perpendicular from F to intersect EC extended at N. Triangles EDP, ECQ, & ENF are all similar, so ED/EC = 1/ 3. Given DC = X, ED = X/3. ED/EG = 1/2. CG = (2)(X/3) = (2/3)X. EG = x/3 + x + (2/3)X = 2X GF = BC = X. 6² = X² + (2X)². 5X² = 36. X² = 36/5. X = 6/√5.

@Irtsak Ай бұрын

ED/EC=1/4. where is point G ?

@bpark10001 Ай бұрын

@@Irtsak Point G is out to the right (where point N is in the video). I wrote up & made point G, but determined later he had already had named the point N. So all the G's need to be edited to N's.

@Irtsak Ай бұрын

@@bpark10001Your solution is very clever. Allow me to rephrase your solution in order everyone to understand it. OP // QC // FN => ED/EP=DC/PQ=CN/QF (Thales ) => ED/1=x/3=CN/2 => ED=x/3 and CN=2x/3 EN=x/3+x+2x/3=2x and FN=x . Apply Pythagoras theorem in right triangle ENF => 4x²+x²=6² => x=(6√5)/5 Have a nice day.

@santiagoarosam430 Ай бұрын

La pendiente de EF es constante→ PD=a→ distancia vertical entre Q y P =3a→ BQ=2a→ BC=a+3a+2a=6a→ 3²=(3a)²+(6a)²→ a²=1/5→ a=1/√5→ BC=6/√5 =X. Gracias y un saludo cordial.

@davidseed2939 Ай бұрын

let s =sinθ c=cosθ measuring vertically 6s=x, horizontally 3c=x, 6c=2x 36ss=xx 36cc=4xx 36=5xx 6=¥5 * x ¥5 * 6=5x x=1.2¥5

@michaeldoerr5810 Ай бұрын

The answer was def 6[sqrt(5)/5]. Apparently the two methods involved two substitutions: a Pythagorean identity as well as substitution via HL poatulate and parrallel postulate. I actually calculated this quantity using my identical version. I hope that this means that I kind of got it!!! Also do you think it possible to COMBINE those two methods? Im planning doing that when I can!!!

@quigonkenny Ай бұрын

Extend EC to M and draw MF perpendicular to EM so that ∆EMF is a right triangle. Draw QN, where N is the point on AD where QN is perpendicular to AD. As ∠PED = ∠QFB as alternate interior angles, and ∠EDP = ∠FBQ = 90°, ∆EDP and ∆FBQ are similar triangles. As QM is parallel to EC, ∆EDP and ∆QNP are also similar triangles for the same reason. And finally, as ∠EDP = ∠EMF = 90° and ∠E is common, ∆EDP and EMF are similar triangles. As EP + QF = PQ = 3 and ∆EDP, ∆QNP, and ∆FBQ are similar triangles, then ED+BF = QN = DC = x. As CM = BF, EM = ED+DC+CM = 2x. As ABCD is a square and MF is parallel to BC, MF = x as well. Triangle ∆EMF: MF² + EM² = FE² x² + (2x)² = 6² x² + 4x² = 36 x² = 36/5 x = 6/√5 = (6√5)/5 ≈ 2.683 units

@michallesz2 24 күн бұрын

ED + BF = x => x^2 + ( 2x )^2 = (1+2+3)^2 => x^2 + 4x^2 = 36 => 5x^2 =36 => x^2=36/5 = 7,2 x=V7,2 = 2,68328....

@SidneiMV Ай бұрын

ED = y => BF = 2y (x + y)/4 = (x + 2y)/5 5(x + y) = 4(x + 2y) 5x + 5y = 4x + 8y x = 3y => y = x/3 EC = 4x/3 AF = 5x/3 AP² = 25 - 25x²/9 => AP = 5√(1 - x²/9) PD² = 1 - x²/9 => PD = √(1 - x²/9) AP + PD = 6√(1 - x²/9) = x 36(1 - x²/9) = x² 36 - 4x² = x² 5x² = 36 => *x = (6√5)/5*

@Irtsak Ай бұрын

Let PD=y, BQ=ω So QC=BC-BQ => *QC=x-ω* Now triangles EPD,BQF are similar => y/1=ω/2 => *ω=2y* (1) Also AQC,EPD are similar => QC/y=4/1 => QC=4y => x-ω=4y => x=4y+ω => x=4y+2y cause (1) => x=6y (2) Apply Pythagoras theorem in EPD => ED=√(1-y²) AD//BC => x/3=√(1-y²) / 1 => x=3·√(1-y²) (3) (2),(3) => 6y=3√(1-y²)=> 2y=√(1-y²) => 4y²=1-y² => y=1/√5 (2) => x=6/√5

@nexen1041 Ай бұрын

Move the entire diagonal section horizontally to the right until ED is ZERO. QF will be 3 and BF will be X. The new right triangle height will be X, base is 2X and hyp is 6 Finally, Solve for X using Pythagorean theorem

@nexen1041 Ай бұрын

Thumbs up to those who solved it visually, without even the need to use calculator. Thiat is what i did 👍