A Very Nice Geometry Problem | You should be able to solve this! | 2 Methods
Вставка
- Опубліковано 5 вер 2024
- GET MY EBOOKS
•••••••••••••••••••••••
Differentiation : payhip.com/b/W...
Indefinite Integration : payhip.com/b/8...
Definite Integration + Area under the Curve : payhip.com/b/1...
Trigonometry : payhip.com/b/8...
OTHER CHAPTERS : COMING SOON.....
--------------------------------------------------------------------------------
Join the channel to become a member
/ @mathbooster
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.
ED=a ..arccos(a+x)/4=arcsin((x-√(16-(a+x))^2)/2)=arccos(a/1)...x=6/√5...a=x/3=2/√5
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.
x^2 + (2x)^2 = (1+3+2)^2
5 x^2 = 6^2
x = 6 / sqrt(5)
but you don’t say how you measure to get those equations
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
What I did. 15 seconds mentally
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.
ED/EC=1/4. where is point G ?
@@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.
@@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.
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.
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
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!!!
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
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....
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*
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
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
Thumbs up to those who solved it visually, without even the need to use calculator. Thiat is what i did 👍
6/5×root(5)
(1)^2 =1 (3)^2=9,(2)^2 = 4 {1+9+4}= 14 360°ABCD/14=2.164ABCD 2^1.10^10^8^8 2^1.2^52^52^32^3 1^1.1^12^11^11^1 21 (ABCD ➖ 2ABCD+1 )
3/(1+3+2)=3/6=1/2
x²+(x/2)²=3² 5x²/4=9 x²=36/5 x=6/√5
sinα=х/6 ; cosα=x/3
(sinα)^2+ (cosα)^2=1
x=6/√5
6√5/5≈2,36
no 2,36 , but cca 2,683
30/sqrt(61) is it also the answere
Happy Guru Purnima, Sir🫡😇 Thank you for your informative videos😊