Can you find the length AB? | (Fun Geometry) |

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Ай бұрын

Second method (Trigonometry) link:
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Learn how to find the area of the length AB. Important Geometry skills are also explained: Similar triangles; Equilateral triangles. Step-by-step tutorial by PreMath.com
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Пікірлер: 56

@PreMath Ай бұрын

Second method (Trigonometry) link: kzfaq.info/get/bejne/js6leJB0kreZp6c.html

@assyrianatheist3966 Ай бұрын

Area of ABC + Area of ABD = Area of ACD --> 1/2 * 70 * AB * sin (60) + 1/2 * 28 *sin (60) = 1/2 * 70 * 28 * sin (120) , sin (120) = sin (60) 70 * AB + 28 * AB = 70 * 28 98 * AB = 1960 AB = 1960/98 = 20 I think this is much easier

@RAG981 Ай бұрын

Agreed. Brilliant!

@bryanfluhrer1306 Ай бұрын

this is how I did it also. I also factored it to 49*2*AB = 7*10*7*2*2 then left with AB = 10*2 = 20

@marcgriselhubert3915 Ай бұрын

@@bryanfluhrer1306 Very good.

@VerdantGMD Ай бұрын

Yep, thats the best method

@jamestalbott4499 Ай бұрын

Thank you!

@lukeheatley4148 Ай бұрын

am not a fan when you got to k (constant). i find it easier to say that if AB/BP = 5/2, then AB/AP = 5/7. Therefore AB = 5/7 x 28 = 20

@devupmanyu8471 Ай бұрын

I also do😊

@quigonkenny Ай бұрын

AB/(28-AB) = 5/2 would work as well, since we know that AP = 28.

@hungnguyenphu1356 Ай бұрын

Good

@richardleveson6467 Ай бұрын

Clever construction! Thanks.

@thinker821 Ай бұрын

Factor out 14 so AC is 5 units and AD is 2 units, we will multiply it to the result later. Use cosine rule + angle bisector theorem, letting x = AB (BC/BD)^2 = (AC/AD)^2 = (5/2)^2 = 25/4 (x^2 + 25 - 2*5*x*cos(60°))/(x^2 + 4 - 2*2*cos(60°)) = 25/4 Since cos(60°) = 1/2, we have: (x^2 + 25 - 5x)/(x^2 + 4 - 2x) = 25/4 4x^2 + 100 - 20x = 25x^2 + 100 - 50x 30x = 21x^2 => Either x = 0 (impossible) or x = 10/7 Hence x = 10/7 Multiplying the factor of 14: AB = x = 10/7*14 = 20

@laxmikantbondre338 Ай бұрын

Good Solution. But like to add one thing just for more clarification that angle CAD is 120. So as sum of all Angles of traingle is 180. Sum of Angles ACD and ADC is 60. So Angle ADC is less than 60. So the constructed point P will lie outside the traingle on the extension of Segment AB and not on Segment AB inside the traingle.

@phungpham1725 Ай бұрын

1/ By using the bisector theorem, we have: BC/BD=AC/AD= 70/28= 5/2 (1) 2/ Drop two perperdicular DM and CN to AB. Because AMD and CNA are 60-90-30 special triangles, we have: AM= AD/ 2= 14 and AN= 35-> MN=21 The 2 triangles NBC and MBD are similar so, BN/BM= BC/BD=5/2--> BN/5=BM/2=(BN+BM)/(5+2)=MN/7=21/7=3 --> BM= 6 and AB= 14+6= 20

@user-zy1rw6yb9p Ай бұрын

Thanks

@user-yx9kr8ur5q Ай бұрын

Area of triangle ACD = Area of triangle ABC + Area of triangle ABD; let AB = X, then substituting: (1/2) *70*28*Sin(120) = (1/2)*(70)*X*Sin(60)+(1/2)*(28)*X*Sin(60), but Sin(120) = Sin(60) so X = (70*28)/(70+28), X =20.

@thinker821 Ай бұрын

Very neat use of trig!

@CloudBushyMath Ай бұрын

Thinking outside the box👍

@PreMath Ай бұрын

Yes! Thanks for the feedback ❤️

@CloudBushyMath Ай бұрын

@@PreMath You are welcome

@anshumanmohanty979 Ай бұрын

Just use the sine rule and internal angle bisector theorem

@prossvay8744 Ай бұрын

AC/AD=BC/BD(ADis bisector) BC/BD=70/28=5/2 BC=5x ; BD=2x Cos(120)=(70^2+28^2-(7x)^2)/2(70)(28) So x=2√39 BC=10√39 , BD=4√39 AB^2=AC.AD-BC.BD AB^2=(70)(28)-(10√39)(4√39) Hence AB=20 units.❤❤❤ Thanks sir.

@uwelinzbauer3973 Ай бұрын

Hello! As several others I used the angle bisector theorem and the law of cosine, then obtained a quadratic equation. One correct solution 20, the second solution 50, not working, to be rejected. Thanks again for the interesting 👍 video, You are a master of geometry of triangles, 🙏 respect! Best wishes and greetings!

@wilfredchan3607 Ай бұрын

Construct a line from B to AC with equal length of AB forming a equilateral triangle, the problem is solved.

@ludmilaivanova1603 Ай бұрын

sorry, can you be more specific?

@wilfredchan3607 Ай бұрын

@@ludmilaivanova1603 Let the line meet AC at E then AEB is an equilateral 🔺️ & EB is parallel to AD, so triangle 🔺️CEB is similar to triangle 🔺️CAD & CA:AD=CE:EB. Let AB=EB=EA=a, then 70:28=(70-a):a.

@ludmilaivanova1603 Ай бұрын

@@wilfredchan3607 Thanks for the idea, I've solved it myself and think it is a better way than shown).

@JLvatron 12 күн бұрын

I made a different equilateral triangle, with a segment from B to line AC. All its sides (including AB) are X. This new triangle (CBE) was congruent to CDA, with its corresponding sides as X and 70 - X. So X/(70-X) = 28/70. This solves X = 20.

@ROCCOANDROXY Ай бұрын

In General, letting AC = a and AD = b implies AB = (a * b)/(a + b).

@nilsalmgren4492 Ай бұрын

I like this because it only requires a basic geometry understanding. No trig, so an approach anyone should be able to follow but few could get on their own. The culmination could be the ratio of the two sides is 5:2 and the total must equal 28...then resolved from there. A drawing of just the equilateral triangle alone at that point might help those who have difficulty seeing it.

@jimlocke9320 Ай бұрын

Extend AD upward and drop a perpendicular to it from C, labelling the intersection as point E. Note that ΔACE is a 30°-60°-90° right triangle with hypotenuse AC having length 70, so AE, opposite the 30° angle, has length 35 and CE has length 35√3. DE = AD + AE = 28 + 35 = 63. ΔCDE is a right triangle and we have the lengths of its sides CE and DE, so we use the Pythagorean theorem to find length CD = √(7644). We use the angle bisector theorem to find BC = (5/7)(√(7644)) and BD = (2/7)(√(7644)). We use the theorem for the length of the angle bisector, in this case (AB)² = (AC)(AD) - (BC)(BD) to find that AB = 20, as PreMath also found.

@marcgriselhubert3915 Ай бұрын

As (AB) is the bissector of angle CAD we know that CB/CA = DB/DA, so CB/70 = DB/28, we have then CB = 5.k and DB = 2.k, with k a certain positive real. The law of cosines in triangle CAB gives: CB^2 = CA^2 + AB^2 -2.CB.AB.cos(60°), so 25.k^2 = 4900 + AB^2 -70.AB The law of cosines in triangle DAB gives in the same way that: 4.k^2 = 784 + AB^2 -28.AB We multiply the first equation by 4, the second by 25, we have: 100.k^2 = 19600 + 4.AB^2 -280.AB and 100.k^2 = 19600 +25.AB^2 - 700.AB Now, by difference we get: 4.AB^2 -280.AB = 25.AB^2 -700.AB, or 21.AB^2 -420.AB = 0. That gives that AB = 420/21 = 20.

@User-jr7vf Ай бұрын

I also did it this way, but his way is more beautiful as it uses a geometrical eye to use less algebra.

@unknownidentity2846 Ай бұрын

I agree to the opinion of @User-jr7vf about the elegance of the solution shown in the video. My approach is also based on the angle bisector theorem, but I determined the length of AB at the end in a different way.

@User-jr7vf Ай бұрын

@@unknownidentity2846 yea, I also did it exactly the same way (before checking the solution or reading the comments). It is nice to see someone has gone the same route.

@johnbrennan3372 Ай бұрын

Very good method

@LuisdeBritoCamacho Ай бұрын

To solve this Problem I only needed a few Equations : 1) Being Point A the Center of Coordinates (0 ; 0), and Line AB belonging to Axis yy. Point C belonging to Axis xx with Length = 70 Then : 2) Being the sin(60º) = sqrt(3) : y = - sqrt(3)*x. This Line makes an Angle of 120º with the Axis xx 3) x^2 + y^2 = 784 4) With these two Equations I found the Intercepting Point D (-14 ; 14*sqrt(3)) 5) Drawing a Line Between Point C (70 ; 0) and Point D (-14 ; 14*sqrt(3)) ; with Slope = - 7*sqrt(3) 6) The distance from Point A to Point B is the interception of Line CD with Axis yy. 7) The Coordinates of Point B are (35*sqrt(3)/3 ; 0) 8) I must conclude that Length AB = [(35*sqrt(3))/3] lin un ~ 20,21 lin un 9) Answer : Length of Line AB is approx. equal to 20,21 Linear Units.

@SanketMarathe-wi6rx Ай бұрын

Better way you draw parallel line to ad from b to ac....

@CJGlobius Ай бұрын

Why not to use just one triangle bisector formula: Lc=(2*a*b*cos(phi/2))/(a+b) and get the same 20 as an answer?

@johnf.kennedy2879 Ай бұрын

If AB is bigger than 28 ,the equilateral triangle will be inside

@TheIsmat987 Ай бұрын

Another solution Teacher. Calculate CD Lenghth by cosine formula=87.43 Calculate angel acd and angel ADC. Equal 16.101 and 43.89837 Then calculate cb=62.45 BD=(87.43-62.45=24.98) Last by sine formula AB =20

@pralhadraochavan5179 Ай бұрын

Good morning sir

@petrileskinen2988 Ай бұрын

My solution was to add a new point E along the line AC so that |AE] = |AB|, so the triangle ABE is equilateral. In this figure the lines AD and EB are parallel. if |AB| = x, then |AE| = x and EC = 70-x. Triangles ADC and EBC are congruent so I can write an equation |AD|/|AC| = |EB|/|EC| => 28/70 = x/70-x => ... => x = |AB| = 20

@farzad1343 Ай бұрын

There is a much easier solution. Just draw a line from B parallel to AD. We know that CB/BD=AC/AD

@giuseppemalaguti435 Ай бұрын

CBA=α ..70/sinα=AB/sin(60+α)...28/sinα=AB/sin(α-60)...elimino AB, ,risulta tgα=-7√3/3...quindi AB=70sin(60+α)/sinα=70(√3ctgα/2+1/2)=70(-3/14+1/2)=20

@misterenter-iz7rz Ай бұрын

(1/2 70×28×sin 120)×(5/7)=1/2×70×s×sin 60, s=28×5/7=20.😮😅😊

@robertstuart6645 Ай бұрын

I presume the Law of Cosines would work in solving the problem.

@comdo777 Ай бұрын

asnwer=35cm isit

@User-jr7vf Ай бұрын

no lol

@jamesrocket5616 26 күн бұрын

AB=20

@unknownidentity2846 Ай бұрын

Let's face this challenge: . .. ... .... ..... According to the law of cosines we can conclude: CD² = AC² + AD² − 2*AC*AD*cos(∠CAD) CD² = 70² + 28² − 2*70*28*cos(60°+60°) CD² = 7²*[10² + 4² − 2*10*4*cos(120°)] CD² = 7²*[100 + 16 − 80*(−1/2)] CD² = 7²*(116 + 40) = 7²*156 = 7²*2²*39 ⇒ CD = 14√39 Since AB is the angle bisector of the angle ∠CAD, we can conclude: BC/BD = AC/AD = 70/28 = 5/2 CD = BC + BD = (5/2)*BD + BD = (7/2)*BD ⇒ BD = (2/7)*CD = (2/7)*14√39 = 4√39 ⇒ BC = 10√39 Now we apply the law of cosines to calculate the length of AB: BC² = AC² + AB² − 2*AC*AB*cos(∠BAC) (10√39)² = 70² + AB² − 2*70*AB*cos(60°) 3900 = 4900 + AB² − 2*70*AB*(1/2) 0 = AB² − 70*AB + 1000 AB = 35 ± √(35² − 1000) = 35 ± √(1225 − 1000) = 35 ± √225 = 35 ± 15 BD² = AD² + AB² − 2*AD*AB*cos(∠BAD) (4√39)² = 28² + AB² − 2*28*AB*cos(60°) 624 = 784 + AB² − 2*28*AB*(1/2) 0 = AB² − 28*AB + 160 AB = 14 ± √(14² − 160) = 14 ± √(196 − 160) = 14 ± √36 = 14 ± 6 Since both values have to be the same, we can conclude: AB = 35 − 15 = 14 + 6 = 20 Best regards from Germany

@unknownidentity2846 Ай бұрын

Appendix: After the calculation of BC and BD there exists also an easier method: AB² = AC*AD − BC*BD = 70*28 − (10√39)*(4√39) = 1960 − 1560 = 400 ⇒ AB = √400 = 20

@hongningsuen1348 Ай бұрын

@@unknownidentity2846 I solved the problem exactly in this way. Solving problem by construction is an art which I am not particularly good at. I usually go by the standard methods. When I saw triangle ACD had given values for SAS (side-angle-side), I knew I could solve any angle and side of that triangle. That is the gist of congruence test, sine and cosine rules - knowing 3 gives you 6. Coupled with angle bisector theorems, solution is an easy target.

@unknownidentity2846 12 күн бұрын

@@hongningsuen1348 Good to know that I am not the only one. When I see such kind of problems I also think about which methods can be used and then I give them a try. But from time to time it can be really helpful to take a sheet of paper and a pen in order to find useful auxiliary lines or constructions. I remember that I tried to solve a problem from this channel when I was sitting in the train some time ago. At the end I found an expression for the angle that has been asked for. The expression gave the correct result but it was not satisfactory since the angle was a whole number (given in degree) and I was not able to prove that my expression turns out to be exactly that value. Accordingly the solution given on this channel was by far much more elegant. Best regards from Germany