How To Solve The Rectangle Length Puzzle - Think INSIDE The Box!

[@MindYourDecisions]

I also learned the exam has 60 problems to solve in 2 hours (an average of 2 minutes/problem). Could you have figured it out so quickly? I could not! (You may notice this comment is also 3 weeks old, even though I just published the video. Lately I have been sharing sneak peaks to people who pledge on Patreon--it's your chance to see the video before anyone else and share a comment before there are too many for me to respond to.)

[@piyushpathak1186]

Hello sir I have a math problem which I could not able to solve There is 40 kg stone that broke into 4 pieces such that you can measure any weight between 1kg to 40kg using a balance what are the weight of broken pieces. Please give mathematical explanation to this math puzzle I'm waiting for your answer

[@yurenchu]

Hi Presh, This puzzle is rather easy, especially when you recognize the 7:24:25 and 3:4:5 Pythagoras triples; that gives you |AD|, |BC| and |CF| instantly. Then draw a line from F that's perpendicular to CD, call the intersection J. Due to Z-angles, CFJ is a 3:4:5 triangle (and you already know the length of hypotenuse CF, so you can easily determine |CJ| and |FJ|) and DFJ is a 7:24:25 triangle (and you just determined |FJ|, so you can easily determine |DJ| ). |AB| = |CJ| + |DJ| = 12 + 14/3 = 16 2/3 .

[@abd.137]

Piyush Pathak The answer is: 27, 9, 3, 1

[@abd.137]

Piyush Pathak There were a couple of things not obviously defined in the question: 1. When measuring the weight, the broken pieces could be placed on either side of the balance to get a larger range of weight. More simply; for example if we want to measure 2 kg. But in my answer we have 3 kg and 1kg, regardless of that we can measure 2 kg by placing 3kg on one side and 1 kg on the other side of the balance so the overall weight is 2kg.

[@Fauzan_Botakman]

MindYourDecisions its for junior high school lol

[@araptuga]

I really appreciate that Presh doesn't just show "here's how to solve it". Rather, when he struggles with a problem, he tells us, explains ways he tried that did NOT work, and then how he worked through it from those failures to a workable solution. He also sometimes discusses alternative approaches. All of those things model how math is REALLY done. It's also an inspiration to others to see that it's OK, indeed a good thing, to struggle when figuring things out.

[@ransiri6595]

I solved it differently with trigonometry! Just find the angels of triangle fcd by using shift-sin(or shift-cos) on both triangles, and after you found fc with pythagoras theorom, use the sine law to find dc!

[@andreasmichaelides5565]

ran siri yeah, same

[@serkankoldere]

ran siri Yeah I did it the same way. I'm surprised this problem took him so long tbh.

[@GR0907]

Same. I found the way in about 5 minutes.

[@myutwocar]

This problem is for kids didn't learn sin cos other than Pythagoras. Presh just follow the game goes, he might stumble this one but you might stumble his specialize field of maths be humble ☺️

[@jessstuart7495]

I went about this a different way... 1. Use pythagora's theorem to solve for the side lengths of the upper and lower right triangles. 2. Draw a horizontal line from point F to the edge CD. 3. Label angle ADE as α (alpha) and angle BCF as β. 4. Draw a vertical line from point F to edge AC. 5. Draw the "other" diagonal of the rectangle made from steps 2 and 4. You now have two right triangles with sides that add up to the edge CD and can write expressions for the lengths in terms of angles α and β. (Alternate interior angles for α and the new right triangle drawn) L = 20*sin β + 20*(cos β)*(tan α) Trig functions of the angles can be calculated from the lengths of the known triangles. tan α = 7/24 sin β = 15/25 = 3/5 cos β = 20/25 = 4/5 Substitute back into expression for L... L = 20*3/5 + 20*(4/5)*(7/24) = 12 + 14/3 = 50/3 = 16.667

[@allenz0rzer]

I found AB like you, then found angle ADE with sin^-1 ( 7 / 25 ). 90 - angle ADE to find angle CDF. Same procedure at the bottom to end up with FC = 20 and get the measure of angle FCD. Law of Sines got me 50/3 for CD, which has the same length as AB.

[@memy02]

this was my process as well (minus actually plugging in values because I'm lazy and would need to look up which trig equations to use)

[@MSReidpr]

I did it your way too

[@Gamesmarts194]

Yeah this was my thought process as well. (Probably not the “intended” method since they want to to take note of point F being in between ED, but I’m not going to try to sound like the current US education system.)

[@crazym108]

This is the way I did it as well. I'm a big fan of trigonometry!

[@schmebulock4567]

I thought of this too but was too lazy to look up the equations. Then I watched the video and when he said he had so much trouble with the problem and showed that complicated way I only halfway understand, I thought my method must be wrong because it's so easy. ^^'

[@elleryrajagopal1734]

Hey Presh, I've been watching your videos for a few years now. I love your work, keep doing what you're doing!

[@JLvatron]

I calc'd the lengths like you until 3:31 Then I used trig to define all the angles, by dividing Opposite or Adjacent by Hypotenuse (plus subtraction for complimentary angles). This led to triangle FCD having the same angles at F & C, so lengths FD & DC were identical. So I used Law of Cosines to solve for just the 1 variable, which was AB what we're looking fo!

[@danwiththe8647]

I did the same thing

[@NestorAbad]

I solved it just like you! A pretty combination of Pythagoras and similarity. But I find it really hard if there are 60 like this one to be solved in only 2 h... ¬_¬u I don't like time constraints... You need enough time to enjoy math problems. Doing math is an art, and any pressure goes against creativity and elegance. Thanks for sharing!

[@AnTran-ie7vq]

Nestor Abad I think, logically, this is one of a few hardest questions in that exam, so it supposed to be 5 to 7 mins

[@omariomariomario]

Nestor Abad same here

[@triton62674]

What? Restraints in any problem leads to an innovative, elegant solution which isn't the same as enjoying the problem.

[@kenhaley4]

Agreed!

[@muditmittal4511]

How are the last 2 triangles similar? I can't figure it out By A.A.S?? But how

[@Eltrotraw]

I looked at the thumbnail and immediately tried it, not realizing the length of ED is 24 - I had assumed DF was 24 and tried a lot more convoluted methods. Oops. DF being 24 makes the solution more complicated

[@Eltrotraw]

Guess I was over thinking it in the wrong ways, especially after watching the solution...

[@derschriev5771]

Geigenzahler could you please elaborate a bit. I initially thought DF was 24 and saw no easy derivations

[@micahm.4713]

Geigenzähler I think you used an assumption that ED is 24 to get to that point before you can make the FHD triangle. In other words If you assume FD is 24, you can't use the length of ED to solve AD via Pythagoras theorem and therefore cannot solve BF or GF.

[@Starling2100]

I did the exact same thing

[@omariomariomario]

Eltrotraw haste to a mathematician is like curiosity to a cat

[@jundachen9518]

The location of the 24 in the diagram was quite misleading, I thought it was FD = 24

[@jandemeyer105]

The text was clear but indeed, a figure is supposed to give a clear overview and in this case, for that aspect, it was all but clear, in contrast, as you said, misleading.

[@kavynizde]

I think we can do this problem if FD=24. But it is harder then if ED=24

[@OceaneKudo]

I did the same mistake. So I found a result and realize it was fault -_-'

[@spiderjerusalem4009]

@@kavynizde LOL SERIOUSLY? How??

[@manjutr7691]

Same here

[@austinpowers9740]

I really love this channel, even though I'm not good at mathematics. All these exercises and their solutions make my brain and my mind work together and that's a very good thing. So thank you Presh!

[@Mike-rw1jw]

this is elementary mathematics that a sixth grader would solve easily

[@austinpowers9740]

@@Mike-rw1jw Good for that sixth grader.

[@kennethhicks2113]

Much easier solution IMO. Once you get FC you can use law of cosines to get DC (angles ADE and BCF are simple trig thus angle CFD, FCD and CDF are easily found) Good stuff, Thanks : )

[@H0n3yMonstah]

Yeah this was the way I thought he would have done it, seemed much simpler in my mind than his method.

[@Blue-cm6kt]

how do you get the value of the cosines?

[@kennethhicks2113]

@@Blue-cm6kt Hi, long time ago... will have to review content... surely your not asking for the simple cosine formula.... I'll review in a bit.

[@supercool1312]

ky thinking: figure out ADE, which then allows you to figure out the reflex of BFD, as you know ABF is 45 as it is half of 90, and BAD=90, then when you figure that out, add 90 to it then subtract it from 360, then use trigonometry (along with pythagoras) to find the remaining side

[@lyslol7690]

I solved it by using trigonometry, which I thought that was kind of a tedious solution.

[@beaming_sparkling_trash261]

Trigo >>>>

[@dreadpiratetjoris]

I did this all in my head and got as far as creating line HG, but then I got stuck. As soon as you drew the actual line, I figured it out. Shows that sometimes you just need to draw it out and really visualize it 🙂

[@rhonanbrownfield816]

Funny, the solution that I used to solve this problem ended up being completely different! I followed this technique up until the point at which you solve for the length of line CF. After that, I calculated the measure of angle ADE by taking the tan-1 (7/24), which equaled roughly 16.26 degrees. Knowing this, I was then able to solve for angle CDF by subtracting the measure of angle ADE from 90 degres (90 - 16.26 = 73.74). After this, I returned to triangle BFC, where I calculated the angle BCF by taking the tan-1 (15/20), which was about 36.87 degrees. Then, as I did before, subtracted that value from 90 to get the measure of angle FCD, which is approximately 53.15 degrees. At this point, I was then able to solve for the final angle in triangle FDC by subtracting the sum of angles FDC and FCD from 180 degrees, which just so happened to be exactly 53.13 degrees. This, of course, told me that triangle FDC was isosceles, as it has two equivalent base angles off of line FC. What I then did was draw in the altitude of triangle FDC from point D to the center of line FC (call it "point G), separating the triangle into two smaller, congruent triangles. Given that this altitude line was perpendicular to line FC, as well as bisected it, I shifted my focus to the right triangle that shared its hypotenuse with line CD -- triangle CDG. From here it was simple algebra. I already knew that angle DCG was 53.13 degrees and that line CG had a length of 10 (given that line CF had a length of 20 and was bisected by line DG), so in order to find the length of line CD I set up the equation cos(53.13) = 10/x. Of course, this meant that x (the length of line CD) was equal to 10/cos(53.13), which was equal to 50/3. Finally, because we're dealing with a rectangle, line CD is congruent to line AB. Therefore, line AB = 50/3. This was a really fun problem to solve regardless! Thanks!

[@elinaevans5803]

Rhonan Brownfield wow))

[@Akun-u2m]

i got tan-1 (7/24) equal to 0.2838 in calculator, how to know the degree ?

[@anandk9220]

Solved this persistently in 3 attempts. The height of triangle FBC helps to find its horizontal distance till C and that can be used to find remaining vertical distance from F to AD using similarity property in triangle ADE by first drawing altitude from E. Checked the answer for first attempt but was little disappointed to be incorrect. Thankfully I didn't give up, just like this great mind Presh !!! 😊

[@GretgorPooper]

I got this solution right as well, but through a very bizarre roundabout method that I'm almost too embarrassed to show. In fact, I double-checked my calculations a few times before finally accepting that the solution was not going to be a cute little integer. It frustrated me that the solution to this problem wasn't neat and rounded. I started by doing pretty much everything just like you did up until the point you find the length of FG. Then I got the length of FG by considering the rectangle with same base and altitude as BFG (whose area is then 15*20) and dividing its area by 25 to get the length of 12. Then I calculated that GC=16 by Pythagoras. I then calculated the height of ADE to be 6.72 (again by thinking of the rectangle with same base and height as ADE), and with it, the distance between E and AB to be 1.96 by Pythagoras. Then I rotated the whole thing 180 degrees, and treated edge DE as a line on the cartesian plane with the origin at D. The function representation of DE was such that f(0)=0 and f(25-1.96)=6.72, which gave me f(x)=7/24x. Then, since the coordinate x of F is 16 in this new diagram, I only needed to calculate f(16)=14/3 to know the distance between F and AD (which in this crazy diagram coincides with the x axis) . So yeah, I made a mess, but it is a mathematically correct mess. Yay me...?

[@theSoberSobber]

Yay!!!

[@GretgorPooper]

Tech Master Pavit yay!

[@lancedejesus6402]

It's not a good solution, I think you can't get perfect marks but at least it's right.

[@derpdee63]

Dude, these are exactly the kinds of problems us “math meanies” want to see! The labeling of ED caused some confusion, but other than that the instructions were clear and the explanations never drew on assumptions. To top it all off, the “think inside the box” slogan for this video really drives home the point that we should keep our minds open to new perspectives (or, as you might say, Mind Your Decisions). These are the kinds of videos people can really learn from. Keep it up Presh! 👍

[@javieralfonsomartin]

I solve it by another way. (sorry for my english, it's not my 1st language) First of all I find the value of AD by Pythagoras. It's 25. Then, I dinf the value of FC, by Pytgagoras also. It's 20. Now I have 2 sides of the triangle FCD, and I want to find the another DC, which is the same of AB. So I need the angles. We have 3 D angles, for the rectangle, for the AED triangle and the FDC triangle. I want the last one. For that I find the angle for AED using the formula cos= contigous leg / hypotenuse (cos D=24/25). It´s aprox. 16.26º. For find the D angle of the another triangle I rest 16.26º to 90º. It´s 73.74º. For the same way we find the C angle. It's 53.13º For find F angle, we plus the other angles and rest it for 180º. It 53.13º too. Now I only have to do the cosine theorem. DC/senF = FC/senD; DC=FC senF /senD; DC = 20 sen53.13 / 73.74. The solution was 16.666627. We can induit that whithout the aproximation the decimal 6 would be periodic. So the final answer is 16.6666666..., which in fraction is 50/3, your answer. So it's right.

[@TheMartinaheimo]

I did the same way, just what I noticed that this is sinus theorem not cosinus theorem😁

[@xz1891]

Not elegantly, in any case, if u get an angle value such as 34.56°, it's a failure. Try only use trig, not the angle itself.

[@katlegomotlana4544]

Finally!! This is the first math problem I've solved ever since I started watching this channel😊💪🎉 I used the Pythagorean theorem to find the last 2 sides of the 2 given triangles. Then found the angles

[@colinvandusen2072]

Pythagorean theorem and then 2 sine laws to get angles ADE and BCF, subtract both from 90 to get two interior angles of triangle CDF, subtract those from 180 to get the third interior angle angle, use that angle in sine law to get DC = AB

[@davidhernandez7032]

I was thinking useing trig.

[@ReenaKumari-vq2rk]

I did it by trignometry.. 1st i found all the dimensions of BFC. then draw a perpendicular from F on DC and name it FM. since FM//BC//AD, angle CFM=angle BCF. and by applying sin for both angles in thier respective triangles we get 15/25=CM/20 AND WE GET CM=12.then we calculate FM by pytho theo. FM=16. Again since angle DFM= angle ADE. We apply tan in both angles and get 7/24=DM/16 SO WE GET DM=14/3=4.67. Therefore DC=DM+CM=12+4.67=16.67 AND, AB=DC=16.67 ans.

[@winter_c]

I also figure out that with using sine formula and sine and pyth thm

[@mohammedzaheen4675]

Another way of solving this is use pythagorian theoroem and find AD=25 therefore BC=25 now find FC=20 now let angle ADE=theta so angle DFC=90-theta now let angle FCB=alpha so angle FCD=90-alpha and angle DFC will be theta +alpha now aplly sine rule in triangle DFC thus you get DC=AB=50÷3 but before you doing this you must know the formula sin(A+B)=sinAcosB+cosAsinB

[@Bibibosh]

Now we need another video for the answer of 50/3

[@bobjordan5231]

Nicely done. I especially appreciated how you have to labor over some of these problems. The answers sometimes come so fast that I think "how'd he do that!" One doesn't have to be a genius to solve these, but they do have to think and work, and that's the key to life as well as math. Thanks Presh, this engineer is not a math meanie, he's on your side!

[@JBaker452]

I reached the same solution, although I did it without extending any lines. If computed length AD, BC and FC the way you did. But next I began computing angles: Ang BCF = BC*sin(th1) = BF, so BCF = 36.87 deg Ang FCD = 90 - BCF = 53.1 deg Only a few more angles to go.. Ang ADE = AD*sin(th2) = 7, ADE = 16.3 deg Ang FDC = 90 - ADE = 73.7 Now, Ang DFC = 180 - FCD - FDC = 53.1 deg Law of triangles (or sins) means sin(DFC) / CD = SIN(FDC) / FC So length CD = FC SIN(DFC) / SIN(FDC) = 16 + 2/3

[@rcb3921]

Using trigonometry is, in essence, scaling to the unit circle, operating, and scaling back out again. I'd say that extends/constructs quite a lot.

[@Kasei87]

Interesting solution! I did Pythagoras to find FC, then did pi-tan^-1(7/24)-tan^-1(15/20), then used the sine law on CDF and CFD, using CF=20.

[@mohdhumza2071]

We could have drawn a line parallel to BC through point F and could have obtained rectangle FGCX where X lies on DC ( FX=16) and then used Pythagorean theorem in Triangle FXD to obtain the value of DX. The side AB=DX+XC (XC=12, since FGCX is rectangle)

[@elinaevans5803]

mohd humza you are genius, really)) cool method

[@kenhaley4]

I did figure it out! My thought processes were almost identical to yours, Presh. I would say that it took me about 30 minutes. No way could I do 60 problems like this in 2 hours! I think this number of problems is designed so that not even the quickest can finish. Some think this is a good way to rank the students' math abilities. I'm not sure I agree. Being quick and thinking deep do not necessarily go hand in hand.

[@Sam_on_YouTube]

I actually did figure it out. I rarely get the ones that you have this much trouble with. I got the method in a few minutes. Basically, I realized that if you set it on an axis, origin at the lower left, and look at it as vector addition, you could get the vector from the lower left to virtex of the lower triangle. Then if you take the angle of that line going the rest of the way to the corner and allow it to extend until ot hits the width of the box, we can get the sum of the two vectors. The y component is the answer. Once I had the idea and started working through finding the vector components, the math was very similar to what you did. I wasn't going to go all the way to a solution because I didn't have a calculator handy, but once you pointed out the 7/24/25 triangle, I paused again and did the rest in my head, getting the correct answer.

[@GretgorPooper]

Very pragmatic, but I like it.

[@tapasmondal9466]

Phew!

[@MawsSemin]

Sam g

[@TheRandomYoYo]

I love your method. I just did it with the sines rule, but yours is more professional. Cheers for the insight!

[@SuperAbciximab]

Hi Presh, thanks for the detailed explanation. But I was still not clear in the last part where you said AE/ED =7/24 = HF/ HD ? Why couldn't take HF/ FD to keep it consistent with AE/ ED? The ratio is not same as shorter leg/longer leg as applicable for similar triangles ? Or is it? Am I missing something here. Why were you taking different sides ? I will be grateful for your advice 🙏

[@Jukka70]

I wanted to thank you for your channel. I have just started teaching math, but have tutored math for over 20 years. My hope is to be teaching high school Math next year. One of the things that I would love to do, is to give student one of these types of problems while they work in groups to come up with plans of attack on how to solve it. I do this now when tutoring students that are going for their high school equivalency test. I'll teach students at the same level a few things, then we increase the difficulty and their job is to form what I call the plan of attack.

[@clotz1820]

FD and ED are confusing on picture

[@alex44119]

I haven't solve it like you at all !!! (Except the Pythagorean Theorem to find AD and then CF). I start calculating AD and CF an then with the trigonometry I prouved that

[@rickielukas4558]

The pinned comment by MindYourDecisions was 3 weeks ago... Time travel?

[@danwiththe8647]

I used to just watch your videos, now I pause and really try to solve them first. excellent work

[@clubstepdj]

I spent my time being confused because i thought the length '24' is for FD, not ED

[@sakshamtyagi6527]

Same

[@godzillaninja1644]

I didn't get it, but my favorite part of the video was when you said that this one stumped you and that they aren't always easy even for you. I thought that was encouraging for people like myself who like math but arent as good at it as others. Thanks.

[@jawaharprakash3486]

Why can't you use cosine formula in triangle CFD. You can easily find angle ADE & BCF

[@Chewychewychoochoo]

Irk?! That's how I solved it

[@cj_xx2001]

Find angle ADE then find angle CDE. Then find angle GCF and then find angle DCF. Hence find angle DFC. Use the sine rule for triangle CDF to find DC :)

[@GeoDetective]

Ah, man! I had 16 2/3...

[@sw3aterCS_]

50/3 does equal 16 and 2/3

[@GeoDetective]

sw3aterCS I know 😁

[@jetjl412]

sw3aterCS wooosh

[@sw3aterCS_]

JuJuJetastic r/

[@skoockum]

:P

[@Chewychewychoochoo]

I went all out, using law of cosine, law of sine, Pythagorean theorem, and so forth. For me, I found all the angles of triangles: AED, BFC AND FDC using Law of Cosine. And using Pythagorean theorem to find side FC equal to 20, I used Law of Sine to find DC. I never liked drawing/cutting more lines/triangles into the picture, thus my method involved finding the angles of Given triangles and sides. But maybe your method could be more time efficient, who knows. Trig was always my top favorite math, above calculus and algebra.

[@roshandon3157]

And I was watching tonight how to guess MCQs...😂

[@twwc960]

I did it slightly differently. I used Pythagoras like you to get the three sides of triangle BCF, then I calculated its area, multiplying 1/2 by the two legs (BF and FC) to get an area of 150. I then expressed the area differently, as 1/2 base times height, the base being BC=25. This gives the altitude of point F as 12. I then drew a horizontal from F to intersect AB at G and used Pythagoras on triangle BGF to get GF=9 units. I then set up a Cartesian coordinate system with the origin at B, so that point F has coordinates (9,12). The point D has x coordinate 25, and the y coordinate is unknown. But from the upper triangle, the tangent of angle ADE is 7/24, so that is the slope of the line FD. Knowing the slope, as well as the coordinates of a point on the line, enabled me to calculate the equation of the line as y=7/24 x + 75/8. This allowed a calculation of the y coordinate of D to be 50/3 which is also the length of CD and hence AB.

[@maxsimes]

Wait what?? I thought that DF = 24, not DE = 24...my bad..

[@manikaninika5881]

Same

[@alexsong9211]

Took me about 5 minutes looking at the thumbnail then I realized the title was think inside instead of outside then I solved it 2 minutes later. After realizing similarity on your supposed FHD triangle. Pretty solid problem.

[@flashbond]

I think DF is 24

[@mustafaocak6522]

Me too, unfortunately ı tried to solve on thumbnail it take my 10 min

[@sophiacristina]

1:16 - I solved using this... Make two lines until it meet the other lines, call those '"I" (from 'F' to bottom) and "H" (from 'F' to top), let's call 'h' the altitude from 'E' to top. Then we do this with the upper triangle: 7^2 - h^2 = x^2. 25 - x = y; The triangle BCF, we can figure out the other side is '20'. Now with 'I' we just multiply it (using _soh_cahtoa) by the sine value, 'I = (15 / 25) * 20'. And now we know CG is 16. That means what "16" is to "i" is congruent with what "h" is to "H". 'H + I = AB', our answer! h = 6.72; CG = 16; y = 23.04; I = 12; H = 14 / 3;

[@tawhidandtahmid1469]

I have question for you which I couldn't ans. Please answer it. :-( . ❄ Two tangent PQ and PR are drawn from external point P to a circle with center O; where Q, R are not the point of tangency. Q, R are you two points such that PQ=PR and O is the midpoint of line QR. X, Y are two points situated on PQ and PR respectively in such a way so that XY is a tangent to the circle. If QR=10 then find the value of QX. RY

[@yurenchu]

Not solvable without further details. However, if XY is parallel to QR, then |QX| = |RY| = 5 .

[@tawhidandtahmid1469]

( according to question QR=10) QR lines middle point is O but the diameter of O circle is not given . Plz give me your gmail address I will sent you the question and the geometric shape

[@tawhidandtahmid1469]

( according to question QR=10) QR lines middle point is O but the diameter of O circle is not given . Plz give me your gmail address I will sent you the question and the geometric shape

[@nikolamarijanovic6261]

abdus salam B [e-mail]

[@nikolamarijanovic6261]

Not enough information

[@TheAveryFilms]

solved using internal angles... finding internal angles of DFC... found out angles DFC and CFD are both 53.13 degrees...isosceles triangle... heght bisects base at 90 degrees in half... leg is 10 ... cos(53.13) = 10/x ... x= 10/cos(53.13) ... x= 10/.6 x= 16.67 = 50/3

[@ColeJT]

I thought FD=24 looking at the picture :/

[@fridhmzyev6477]

me too when i knew the right i did 5 minutes

[@K3K900]

Yes, the problem can also be resolve with the sines law, if you get the angles of all the sides and then use the sines law you get the side of the rectangle which is also the side of a triangle made with the intersection of the two triangles, Then using the sines law you get the side missing which is also the side from the triangle

[@J0EB0B555]

I'm surprised you had trouble with this. Its basic trig. Took me like a minute.

[@habeebhabib3160]

Ikr

[@lifeofphyraprun7601]

I gave an examination today,that had exactly this problem,and we had to figure out the value of 3AB.Unfortunately,I was unable to remember how to solve this problem,so I'm watching this video again.

[@coop7894]

Why don't you formally prove the right-angled triangles are similar rather than just saying "By similar triangles, the ratios of the matching sides are constant"?

[@MichaelOnines]

Cooper, any two right trianglea that share an acute angle are similar by the angle-angle-angle definition of similar triangles. QED.

[@coop7894]

Yes, so why not say that... Lol. I never said there was anything to be proven, but if you're formally proving something in mathematics you should at least show how you got from a to b

[@MichaelOnines]

Cooper, where do you draw the line, though? Does he need to recite the definition of a rectangle and the theorem that opposite sides are equal to transfer AB to the opposite side? Drawing similar right triangles using altitudes is the bread and butter of almost every right triangle geometry puzzle.

[@coop7894]

Draw the line? I'm not suggesting he should prove that a square has 4 sides or anything, just to state in less than a sentence why 2 triangles are similar.

[@MichaelOnines]

Cooper, :)

[@rinkeborg]

I solved it in a different way, a bit tedious, but it worked, and took about 10 minutes. From a quick scan over the comments, it seems this type of solution was not yet mentioned! Just put the big rectangle in a grid, where B is at (0,0), A has coordinates (0,q), and D has coordinates (25,q) - the solution starts in the same way by finding this out via Pythagoras. Then you can work with vectors, and calculate the length of AE (known to be 7) from the coordinates of vector A and vector E, and the length of ED in the same way. This gives that E is at (1.96, q - 6.72). Same trick for BF and CF, which gives that F is at (9,12). Then it is given that F is on line ED. Create a formula for line ED: (25,q) + lambda (25-1.96, 6.72), and fill in F. This will give q = 50/3. Solved.

[@MrRetiJunior]

Not gonna lie I just use my CAD program to solve these

[@patrick2231]

so, before watching, i was able to come to conclusion of using half circles with diameter equal to the long side of the retangle. doing this, solve for the heights of each, and you add them to get the length of AB. this was really quite roundabout but i was very happy i got the answer. When looking at it in terms of circles. the radius of the circle is half your longer side, which you find right away to be 25. so 25/2 is half of that. you then know that the area of a right triangle in a semicircle is largest when it's height is equal to r. you also know that the height of any isosceles triangle will bisect the base of that triangle. using what you know, you can determine the area of each right triangle above, and from there, determine the height using the equation A=1/2b*h, where 1/2b is known to be the radius=25/2. doing this, you can solve the bottom to have a height of 12. doing the same above you will get 14/3. I would really love it if you showed a visual of what i was tried to explain here. i think it'd be neat.

[@hybridz1237]

i cracked it in seconds lol

[@mohdjuddin]

boaster. dont discourage others who took time to solve it.

[@golfchan3613]

Oh, cool. I saw right angles and turned immediately to Pythagoras, and solved all the sides but did not think to think enough inside the box. I eventually came to the conclusion of doing trigo identities to obtain angle BFC, 90-(the answer), repeating for ADF and using triangle angle sum to obtain DFC and FINALLY use sine rule to get DC I'm not sure if this is correct but great job presh!

[@mikgigs]

you made mistake in the plot, according to plot FD is 24...instead ED please, correct

[@dirt3554]

It took me about 20 minutes and 9 different steps of using the sine law and cosine law. And I got the same answer as you!

[@bennyfactr6122]

Got it, but it took 11/3 minutes

[@Jaojao_puzzlesolver]

Same here ; )

[@anthonyeaton9049]

This was a fun problem. Nice chance to flex the right triangle trig identities. Keep 'em coming!

[@yossefbudagov8748]

Are you serious it took you so long to solve it? It took me like 5 mins cause I used trigonometry..so easy

[@bluerizlagirl]

yossef budagov Wherever you see (7, 24, 25) and (3, 4, 5), trig probably is a method of last resort.

[@yossefbudagov8748]

bluerizlagirl In addition to pythagorian theorem I used sinus theorem and got the solution very quickly. I prefer this method over his.

[@bluerizlagirl]

Yeah, but bear in mind that this is the sort of stuff you have to learn before you start with trig. If it was meant for kids who have not yet learned trigonometry, it's a bit like cheating.

[@yossefbudagov8748]

bluerizlagirl The thing is I already know the method that is described in the video because it's simple geometry. In math everything is connected to everything so no matter what method you use you should be able to solve it correctly. Finding the easy way to solve a problem isn't cheating for me.

[@bluerizlagirl]

True ..... there is no cheating in maths really, just different properties you can exploit to get to the same right answer.

[@wasfiiwasfi]

i love your videos ! and your channel topics are unique ...

[@robertjohnson4401]

I solved it mostly like you. I used trigonometry at the end rather than taking ratios of similar triangles. I will learn to recognize similar triangles as it makes it much easier to solve.

[@nachomarimba]

Wow! That last part was amazing! Great video!

[@melkartankido1994]

Keep going Persh.. You are wonderful

[@اممدنحمظ]

تمرين جميل جيد. شرح واضح مرتب. رسم واضح مرتب . شكرا جزيلا لكم والله يحفظكم ويرعاكم ويحميكم . تحياتنا لكم من غزة فلسطين .

[@technoultimategaming2999]

I was estimating, rounding up etc and 50÷3=16+2/3 I got 18. Closest I ever got to solving any of your actual math questions. Thank you

[@technoultimategaming2999]

Also. Was trying to do it in a very similar way but I got stuck on |BC| =25 and got 10.6 which x2 is 21 I split BFC into half so I split 90 into 45 and did 15 sin 45 and 15 cos 45. Than did it with the other side too. So I looked up for a clue and you said FC = (25^2-112^2)^1/2. Flipped the page and face palmed myself. Than split CFD into 2 right angle triangles (Solving °FDC is 74° rounded) and by spliting BFC into 90° and 45° i knew FCD is 45° too Did sohcahtoa added the 2 missing parts together (That would be G+H in your case) and got 18! In other words if I didn't round anything up. I would get the same answer

[@mungodude]

Oh wow! My first instinct was that it would be something to do with Pythagorus, but I dismissed that very quickly because I couldn't figure out where to go next either, then I got completely stumped as well and just unpaused the video. Nice to see I was on the right track, just couldn't see where the similar triangles were - I think I need to make myself a tangram set to play with, so I can practise thinking about turning and flipping triangles :)

[@javharsingh6072]

I solved it using trig and pythagoras wihout calculating the altitude! Great job with the proof, it is very innovative.

[@pipertripp]

Elegant. I did a lot of law of sines work. Got to the same result so it's all good. The similar triangles technique allows for an exact solution without a calculator, so bonus style points.

[@emilioschmidt2106]

I did it. It was One of those problems where you don't need calculus or higher math. It really was doable

[@amvkarthik]

Thanks. I first tried looking at the thumbnail. Did not know similarity concept. Subscribed.

[@arnauab25_03]

I can't understand why you had several problems in this problem. For me, this problem is one of the easiest problems of your channel. I saw easily how to solve it!

[@oposdeo]

Before Watching: I got 50/3. The method I used was first using the Pythagorean theorem to determine the width of the rectangle and the missing side of triangle BCF. Having all three sides of each triangle and one right angle, I use the law of sines to get the angles ADE and BCF. Then by subtracting from 90 degrees I got angles FCD and CDF. By adding those two angles and subtracting from 180 I got angle CFD. I finally used the law of sines again to determine CD, which equals AB. After Watching: Your method was interesting too! It would be easier than mine to do without a calculator.

[@sanahaskuranage8071]

I was close but didn’t think about the altitudes until you mentioned them. Learnt a lot.

[@pratapjadhao388]

Great sir Advanced knowledge , Thanks Pratap

[@chinareds54]

Pythagorean Theorem gets us AD (=BC) = 25. Applying it to the bottom triangle gets us CF = 20. Now, draw a horizontal line from F to DC, we'll call this new intersection point G. This splits DCF into two right triangles DFG and CFG. Because opposite interior angles are equal

[@PsyKosh]

Cool. I solved it a bit differently, though. After a bit of poking at it, (also considered extending some of the lines), I worked out AD and FC, then extended an altitude from F to a point G' (primed to differentiate from your G) on DC. Then a did similar triangle jiggerpokery to find FG' which then helped enable further similar triangle jiggerpokery for me to get DG' and CG'. (Which I then added, of course, to get the solution)

[@chinareds54]

I did the same thing. Fewer new lines, fewer new points, simpler solution.

[@delianaw7051]

Thank you very much Mr. Presh Talwalkar for your kindness and your guidance to solve this question. Keep teaching us to understand mathematics, Mr. Presh. 🙏🏻🙏🏻🙏🏻

[@creatureofhabit7049]

Well this was neat. What i did was i matched the triangles at the the vertex E and F, calculated how much offset the vertex A and B will be in this way and then to correct this offset the triangle BFC would have to travel along ED and multiplied my offset with ED's slope with a horizontal (line llr to AD) which gave me the offset in the altitude sum of the triangle, subtract the offset from the sum and you get AB. I thought OUTSIDE the box.

[@lesbloches1142]

How do you deduce the formula u used for similar triangles ? How can u say they are similar ?

[@BlueSwede92]

I solved it immediately when I saw the problem, but with a method that only gives a numerical solution (or at least I don't know how to get an exact one). Angle DCF equals angle CBF, and angle CDF equals angle DAE (the angles in each pair are pi/2 minus the same angle). sin(DCF)=sin(CBF)=4/5, sin(CDF)=sin(DAE)=24/25. CFD=pi-arcsin(24/25)-arcsin(4/5). Then I used the law of sines (which I learned from my dad long before I started learning trigonometry in school!) to get CD/sin(pi-arcsin(24/25)-arcsin(4/5))=20/(24/25) => AB=CD=20*25/24*sin(pi-arcsin(24/25)-arcsin(4/5))=16.67.

[@alexdelargy9904]

I’m a junior in high school and I solved this in 5 minutes, there’s a much easier way than what you did.

[@Shreyansh_1485]

Good one

[@mikem6549]

My solution was to flip the triangle top to bottom at the top giving a gap of 1 at the top left but preserving the intersection as the angle is the same. The intersection with a ratio of 8:16 yielding 7*16/24+12 which gets to 50/3.

[@thetruth5084]

Yup Completed it mate.

[@rcb3921]

I also used Pythagoras to complete the triangles' lengths as a fairly natural step first step. The observation I made was that AB=CD. I constructed the altitude of DFC, placing my G on CD. CGF is similar to BFC; DGF is similar to AED. Scale the values, add CG and GD. Presto.

[@sgifford1000]

I ended up getting the problem correct but used more complexity. I had forgotten the concept with similar triangles (but did relearn, thanks!) Instead I basically proved the concept like so. The angle ADE is constant for both right triangles in the top portions so the tangent (short leg/long leg) is constant. Thus the ratio of short leg to long leg is constant. Fun puzzle!

[@marcooliveira9335]

Thanks for this great math video

[@dushyanthabandarapalipana5492]

Thanks!Your work became honey bees to me in this dangerous period in my area.

[@dwaraganathanrengasamy6169]

Hi presh. I too got answer from your first attempt. Finding altitude lengths. It works. Although quite long. Your solution is cool.😋Thnx....!

[@viniciusfernandes2303]

Thanks for the video!!

[@allenminch2253]

The approach presented in the video, using the Pythagorean Theorem and a lot of similar triangles, is a very cool and ingenious approach to solving the problem, which is certainly a way for people who have studied geometry but not yet trigonometry in depth to solve the problem. If you know some trigonometry, such as sum and difference formulas for sine and cosine and the Law of Sines, there is an alternative interesting approach you can use to solve the problem which does not require the formation of any auxiliary line segments. You can notice with some simple geometry that triangle DFC has two acute angles which are the same as one in right triangle AED and one in right triangle BFC, and you can obviously calculate sine and cosine of those angles using the side lengths in the right triangles. You can then use some trigonometry in triangle DFC to find the length of DC, and consequently the length of AB, exactly the same as it is found to be in the video approach. Isn't it so cool when there is more than one way of solving a given math problem?

[@victoriam6569]

Ooh, I've got the same result, but I overcomplicated it: Instead of simply getting FH from similarity proportion, I first found altitude EK (K is on AD), and KD, and then used the similarity of triangles DHF and DKE... (I also had the same problem and couldn't just get FG, but had to make system of equations) Well, I've learnt something new :) And got the result anyway :) It's a win-win :)