Square In A 3-4-5 Triangle Puzzle
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- เผยแพร่เมื่อ 20 ก.ย. 2024
- What is the area of the square? Thanks to Papa in India for the suggestion!
This puzzle is half of of problem 21 in the 2017 AMC 10A.
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Generic solution for sides a, b, c, where c is the hypothenusa: s = abc / (ab + c^2)
When I work it out, it looks like another "general formula" derivation! I love this channel! Thanks, Presh!
Every literate 12 th pass student knows this solution ....it is very basic question for indians
General Formula : s=[AB(A^2+B^2)^0.5/(A^2+B^2+AB)]
A=short leg B=long leg s=side along hypotenuse
Breaking news: a right angle triangle made an appearance on "Mind your decision" and there was NO mention of Gougu anywhere in the video.
Right
@@MasterCivilEngineering Triangle
amazing!
Back to Pythagoras, then ? ;-)
What is gogou?
You could do also a general case for the right triangle with sides a, b, c (such that a
I thought I was going to be hearing "gougu's theorem" a lot on this video!
MATH QUESTION
Numerical / algebra QUESTION
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One time see..
sorry but gougu isn't a person, Phythagoras is a person, gougu is just a name of a theorem
There is a quote on Mathematics given by William Paul Thurston,
"MATHEMATICS is not about numbers, equations, computations or algorithms: it is about UNDERSTANDING"
#PuzzleAdda
Same here
13² likes on this comment...
Are u sure that his name is Papa?
It actually means Father in Hindi.
Log Maze le rhe hai
And actually, papa means father in plenty languages, even in English)
@@IoT_ yes. But Papa was never a Hindi word for father. It has come from Latin word Papa (father) which again has come from the Greek word Papas with same meaning.
When you knew only one language
Indians and Greeks have been bed buddies long before...en.m.wikipedia.org/wiki/Ancient_Greece%E2%80%93Ancient_India_relations ....most records of both civilizations have been destroyed...
Just use this , a ,b c sides of right angle triangle c is hypotenuse of triangle , side of square = (abc)/(a²+b²+ab)
a= 3,b= 4 c = 5
Side of square =
3×4×5/(3²+4²+3×4)
= 60/37
Area of square = 3600/1369
I took a slightly longer route. I figured it would involve similar triangles but I defined x and y as the legs of the bottom left triangle. That gave me some equations x/s=3/5, y/s=4/5, s/(3-x) = 4/5, s/(4-y)=3/5. Doing some algebra to rearrange the equations got me 5x=12-4x, 5s=12-3y, x=(3/5)s, y=(4/5)s. Substituting for either x or y in the first two equations gets the same result 5s=12-(3*4/5)s --> 25s=60-12s --> s = 60/37. The area is (60/37)^2.
""PaPa from India" trolled you real bad.
Next question will be from Mumma.
Papa is also used in english its spanish. Johnny johnny yes papa poem is not a hindi poem if i can remeber
@@AbirInsights Indeed. In fact Johny Johny yes papa is an Italian poem. Right?
🤣🤣🤣
@@pranjalpathak4498 yes it is
@@pranjalpathak4498
But papa is still an english word
Please solve (in your video) this problem: ABC is a triangle, AB=5, AC=4, BC=3. Point K is on segment AB, point L is on segment AC, point M is on segment BC. AB and LM are parallel. KL=LM=MK=x. Find x. Greetings from Poland!
For a general right triangle where [(3)(4)(5)] / [(3)(4) + (5^2}] = (60) / (12 + 25) = 60/37.
From working out this problem, it came to light that 0. + 0. + 0. = 1,
which is one of the reasons why math is so cool.
I solved this one by trigonometry as well as by coordinate geometry... Coordinate Geometry took more time...
I find coordinate geometry to be the cheap way of solving it. Everything can be solved that way, but it takes the fun out of it. Could you tell me the trig way because I could only figure out the way shown in the video.
Can you show us the trig way to solve this problem
use sin 37 and sin 53 ..... that method is lightning fast
Let that square be PQRS with side "x" taking in the clockwise manner where P point lies on the side AB... And ABC be the triangle(same as in this solution)...in ΔABC, Angle BCA would be 37° as tan37°= 3/4... in ΔPSB, angle PSB would also be 37° because lines AC and PS are parallel... So BP=xSin37°... in ΔAPQ, AP= 3-xSin37° and PQ would be [(3-xSin37°)Cos37°] as angle APQ is 37° as well...which is also equal to x as PQ is a side of the Square PQRS... And now you have got the equation "[ (3-xSin37°)Cos37°= x]... By solving this equation you will get x=60/37 which is the side of the square PQRS... Now we all know that the area of the square= (side)^2= (60/37)^2... I hope you guys got it... Let me know if you guys got it...If you didn't get any step.. You can ask me...
@@MrHehe-qn7zw thanks but how did you know that angel BCA is 37
I did it, yay! At first I didn't believe my answer because the final number is hella ugly, but it was right.
I used pretty much the same idea, but applied over different segment lengths (the longer leg of the upper triangle = the hypothenuse of the tiny one).
To find the side of a square we use base *height (on which side of the triangle the base is touching)/base +height
(Base *height )/base +height
I kind of suspected that all those small triangles might be of the same proportion, but I was not sure. How can you know or prove that they are of the same proportion? At 01:00, he just says "divides the entire triangle into similar triangles" but without an explanation why.
i'm probably a little late, but he knows that because one of the angles is 90 because the square touches the hypotenuse and goes straight out, and another angle is the same as the main triangle
english is not my main language but i hope it helped
Another way to solve this problem is to take x as the length of the side of the square. From there you can derive the base and height of all the other triangles in terms of x. Then you could do sum of the areas of the smaller triangles + square = area of the large triangle
Thats what I would have done
@@samuelwillowcreek8764 i too did same. Little longer method.
Anyway, different from preshji.
Finally, I can solve a quesiton in your videos.
How did you get 3/4s and 4/3s ?
similar triangles, but he did not prove it, im doing the same problem but i cant figure out how to show they are similar. :(.
@@rohangeorge712 .
The two smaller triangles both have an external angle from the square which is 90 degrees. The two smaller traingles also share another angle with the larger triangle (angle BAC for one, angle ACB for the other). Since they both have two same angles (90 degrees and the other angle), they must also have the same third angle as all triangles have 180 degrees in total. Therefore, since they all have the same angles, they must be similar.
The third smaller triangle that was not used is also similar, and this can also be proved but is not necessary.
Good and easy question
Loved that he didn't use that Gogy-Pogy Theorem
Indeed, a delightful problem.
You also can make it for the "legs", with the same result:
Short leg: 5•S/4 + 3•S/5 = 3 ; S = 60/37
Long leg: 4•S/5 + 5•S/3 = 4 ; S = 60/37
Luis Miguel Uribe hi. Can I ask? Since I don’t ever understand y that side is 4/3s and why that other side is 3/4s, can you explain please? Many thx
@@richworld5106 Hi rich world
All 3 triangles are proporcional to the 3-4-5 triangulo! Can you see it?
The let's say the triangle in the right angle will be 3x-4x-5x, being x the proportion ratio, ok?
5x would be the hypotenuse, and turns out to be equal to S (side of the square).
If S = 5x then x = S/5
Now substitute in the legs of the triangle:
Lower leg is 4x, then lower leg is 4•(S/5), that's 4/5•S
Left leg os 3x, then left leg is 3•(S/5), tha's 3/5•S
Can you work with the other triangles?
@@richworld5106 Sorry..., my telephone is based in Spanish language and auto-corrects to spanish words!
I got it. Thx very much. 😊
I used this approach for the long leg.
I just love how he added at the end "what a delighful problem". For whatever reason, it felt so passive aggressive..
Alright a new video from Presh Talwalkar😁👏👏👏👏👏👏✌🤙
Yup
a and b are the smaller upper and lower measurement of the lower left side triangle, the line shared with the square is z, then a is the longer line, and b is the shorter line. a/b=4/3
Denote sidelength of the square s. The small triangle with a vertex at A is similar to triangle ABC, and the side collinear with AC is in a 3:4 length ratio with the side that is also a side of the square. The small triangle with a vertex at C is also similar to triangle ABC, and the side collinear with AC is in a 4:3 length ratio with the side that is also a side of the square. The length of AC can thus be expressed as the sum of the lengths of the three line segments in the following way:
(3/4)*s + s + (4/3)*s = 5
(37/12)*s = 5
s=60/37
The area of the square equals s^2=3600/1369, or approximately 2.630 units^2.
Time to find out if this is the way Presh Talwalkar will solve it.
Quite easy problem...
Step 1. Find the value of perpendicular from angle B to AC(base)
Step 2. Now value of side of triangle will be = (Base×perpendicular)/(base + perpendicular)
Take S as the side of the square and consider point K is the intersection point of the square with side AB and point M is the intersection of the square with side BC,then
Ak +KB =3------------------(1)
Angle BAC =Angle BKM =€
substituting the values of AK & KB in terms of € yields:
S/sin€+Scos€=3
S(1/Sin€+cos€)=3
S(5/4+3/5)=3
S=1.621
Area of the square =2.63 unit square
a nice follow up question would be: Is this square the largest square that fits in this triangle? If so, why? If not, how can a larger square be constructed?
One of the rare occasions where I was able to answer a question from this channel!
Ikr? Me too!
The only question I was able to do by myself
Well Indians know....
*Papa* .....
🤣
😎😎😎
😜
I took some time to derive a general formula for the side length of a square with one side tangent to a right triangle's hypotenuse and opposite corners touching the right triangle's legs, where the hypotenuse is c and the legs are a and b, exactly like the one in the video, and I came up with:
s=c/(((a^2+b^2)/ab)+1).
Using the data given in the video, a=3, b=4, c=5, plugging these numbers in gives:
s=5/((25/12)+1), which is equal to 1.621621..., exactly what the result was in the video! Testing with other Pythagorean (Gougu?) Triples, I believe this formula should work for scaled versions of this problem. Good stuff MYD!
"We also have that this triangle will be similar to the entire triangle." isn't obvious to me. I guess I need to relearn my geometry.
Yea, that seemed like a big fact to gloss over.
@Hamza Zaidi Thanks for clearing that up for me buddy.
@Hamza Zaidi Thanks Hamza. Your short clear answer was exactly what I was looking for regarding the those last two triangles.
EXACTLY!!! I was contemplating on that assumption but I wanted to prove it, and indeed I can't! Unlike PAPA who confidently stated that the 3 smaller triangles are similar triangles!🤔😟
AA similarity theorem, taught in high school geometry
I didn't got it :|
How can we say those side to be 4/3 of S and 3/4 of S? Can someone explain?
Tonmoy Das me too
Hi, Presh, I would be thankful if you explain or give me the link to the video where you talk about time: 1:24 until 1:30 (porportions on pythagorean triangles for perimeters)
Exactly what I was wondering too. I see where he gets the values from but is this only applicable to right triangles? or Pythagorean triples? or any triangle with a square w/ one side and 2 corners tangent to the interior of triangle? I'll test it out on my own eventually but an explanation would have been great.
@@HeckaS When you have two triangles with the same angles, if you divide the length of two sides of one triangle, you get the same number that if you divide the lenghts of the equivalent sides of the other triangle.
In this case, all the triangles, the biggest, and the 3 which sides with the square, have the same angles: all of them have a 90° angle, two of the4m share one angle with the biggest (therefore, the angles which doesn't share are the same, because in a triangle the three angles must sum up 180°), and the one which shares de 90° angle with the biggest, has a hypotenus paralel to the biggest.
In the biggest triangle, if you divide the catetus, you get 4/3 (or 3/4, if you change the order).
In the triangle to the left, the smallest catetus has a length of S (the same than side of the square). If x is then lenght of its biggest catetus, you get x / S = 4 / 3, therefore x = (4 / 3) S.
Thanks
it is a revise on simular triangles that by ratio to get an area. so from a square we can explan a triangle like this and a rectangle of the second half orientated orthotically.
I do it as area.
6=(x^2)(1+6/9 + 6/16 + 6/25)
6=(x^2) 8214/3600
x^2=3600/1369
We have 3 345 triangles of scale a b and c (clockwise from bottom left).
Also the sides of the square are 5a 4b or 3c. So b=5/4 a and c=5/3 a
Also the left side of the big triangle is 3a+5b=3 So 3a+25/4 a=3 So 37/4 a = 3 So a=12/37
The square is 25.a^2 = 3600/1369
PS: we can verify this with the 2 other sides of the big triangle...
As mentioned by other there are three ways to find the solution,
by considering two little triangles (all similar to the main one) and one of the side of the main triangle (5, 4 or 3).
Here you are :
For the side of length 5 :
(3/4 + 1 + 4/3)x=5 and so ((3x3 + 4x3 + 4x4)/4x3)) x = 5 then x = (12x5)/(9 + 12 + 16) = 60/37 , as shown in the video
OR For the side of length 4 :
(4/5 + 5/3)x=4 and so ((4x3 + 5x5) / 5x3) x = 4 then x = (15 x 4)/ (12 + 25) = 60/37
OR For the side of length 3 :
(5/4 + 3/5)x=3 and so ((5x5 + 3x4)/5x4) x = 3 then x = (20 x 3) /(25 + 12) = 60/37
Side of square= abc/(a^2+b^2+ab)
where c is the length of the hypotaneous.
Alternate formula which I derived from the way I solved it: s=abc/(c^2+ab). It's basically the same since Pythagougu says a^2+b^2=c^2.
Could I know the software being used to create these diagrams.
Photoshop ... isn't it?
@@supratipsantra365 Thanks. How could I know for sure.
Yes
Power of Belief, Dare to believe 👇
th-cam.com/video/ygc-B_OHz9w/w-d-xo.html
May be
Me: doing mental calculation for 30 min
Getting area 2+πi 😁
MATH QUESTION
Numerical / algebra QUESTION
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One time see.
wut...
This was quite easy. Took me 5 minutes.
Just had to figure out similar triangles
I'm annoyed that no proof is given for the similar triangles. I knew the bottom left triangle was similar but how can I know the other two are?
@@VinTheFox at first I thought the same, but I will help you, there are two triangle that make you ask why it could be similar, the triangle on the left top and on the right. why the right triangle is similar? Because the triangle ABC and the triangle on the right have the same corner, which is C, and we know that it's a right triangle (the corner that stick on the square is 90°), so the other corner would be the same as A (because the total of each corner for every triangle would always be 180°), which mean they're the same triangle, this technique will also be used for the left top triangle
@@VinTheFox For the other two triangles one of the angles of them is in the big triangle . A square has 90° and the entire angle of the internal angle of square + internal angle of triangle has to be 180°. Therefore internal angle of the both the triangles has to be 90°. So they are similar because two angles are same which means third angle also must be the same. Ok now coming to the small triangle the square lengths are always parallel , so hypotenuse should be parallel to the hypotenuse of smaller triangle so as corresponding angles are equal and the fact that small triangle also has 90° makes it similar to the big triangle. So that's all the proof you need. And the ratios of the similar triangles have to be equal using which problem can be solved
freedom of speech thank you for that. A good explanation. I think it would’ve been good if he could have shown that in the video, (maybe with a short animation) but you made it clear. Nice one.
@@georgecaplin9075 your welcome, happy to hear that my explanation could help others understand. I believe he should have explained it in the video too because at first I also confuse about it, but then after thinking about it for some time, I finally get the answer to why it's similar
Maybe the easiest calculation here: let h be the height of the triangle with 5 as the base: 5*h/2=3*4/2 => h=2.4. Then calculate the side length of the square x using the intercept theorem 5/x = h/(h-x).
Ooooh! With a more general right triangle of side lengths a, b and c, it is a method for proving the Pythagorean theorem!
Indeed, with (a/b).s + s + (b/a).s = c, we find s = abc/(ab+a²+b²)
but with (c/b).s + (a/c).s = a, or (b/c).s + (c/a).s = b, we find s = abc/(ab+c²)
Then ab+a²+b² = ab+c², and eventually a²+b² = c²
Kindly explain the split-up value ofhypotnuse as 3/4S ,S and 4/3S .We are interested by similar triangle ratios.
as we know that the big triangle is similar to the small one we can equal their ratios and get the value of the side which is 4s\3.
How can you assume all triangles are similar?
All three of the right triangles created by inserting the square are similar to the original triangle. This lets us write the hypotenuse, working from A to C, as (3/4)*S + S + (4/3)*S = 5. This is trivial to solve and yields S = 60/37; squaring that yields the area as right at 2.63.
To see this, note that in the top left small triangle, the side with length S plays the role of the "4 long side." The side that's part of the hypotenuse fills the role of the "3 long side," so it has to have length (3/4)*S. Similarly in the bottom right triangle, the roles are inverted, so the hypotenuse section has to have length (4/3)*S. The rest is just arithmetic.
Note you could have done this for any of the sides of the original triangle. For example, the vertical side yields, from top to bottom, (5/4)*S + (3/5)*S = 3. For the bottom edge, left to right, you get (4/5)*S + (5/3)*S = 4. These both lead to the same result, of course.
The 3 small triangles all have sides with lengths 3:4:5.
The 3 sides of the square inside the big triangle represent the 3-side, the 4-side and the 5-side of their small triangle. So, say the side of the square is 3*4*5s = 60s.
The sides of the top triangle are 45s, 60s and 75s.
The sides of the bottom left triangle are 36s, 48s and 60s.
The sides of the bottom right triangle are 60s, 80s and 100s.
The left side of the big triangle is 3, is 75s+36s. s = 3/111 = 1/37.
The side of the square is 60/37. Etc.
It is still possible to solve this problem without using information about similar triangles. I used the Tg (Tangent) of the BCA and CAB angles to calculate the length of the divided hypotenuse. The Tg values are possible to calculate from the lengths of the sides of the triangle ABC and then they should be compared to the Tg of the sides of the small triangles. We have 3 unknowns - s, x, y, where x and y are the two parts of the hypotenuse of the triangle. The third equation is s + x + y = 5. From these three equations we compute s and get the same result.
Nice solution.I have the same solution but longer Sum of area interior triangles=area exterior triangle .Ex xy/2+(5-z)2z+z²=6 ,we have x=3z/5 , y=4z/5 , a+b+5=z etc......I get 37z²+125z-300 =0 ,z=1.621 z²=2.62
A proposed generalisation of 1:42
For any legs a and b of the right triangle: s^2=(a^2+b^2)(ab)^2/(a^2+ab+b^2)
height of the triangle = 12/5 (with respect to the base(5cm ))
side of square = base of triangle* height of the triangle/ base of triangle + height of the triangle
At 1:23 why is it (4/3)s?? Someone please explain
Because similar triangles, the small triangle’s sides have the same ratio as the big triangle (3,4,5). If one side is 3, the next side is 4/3 x 3 = 4. The other side has the ratio 5 /3 x 3 = 5. The same ratios apply to the smaller triangles.
Nice, I got exactly the same answer using trigonometry. It took me approximately 5 minutes and it makes me more confident in math.
Please show your process.
Side of the square be 'a'
Inner triangles being similar to (3,4,5) triangle, their sides are
(3a/4, a, 5a/4), (a, 4a/3, 5a/3),
(3a/5, 4a/5, a),
Now a(4/5 +5/3) = 4
a = 3*4*5/(5*5+3*4)
this is consistent with other two results
a(3/5 +5/4) = 3
a = 3*4*5/(5*5+3*4)
Now a(1+4/3 +3/4) = 5
a = 3*4*5/(3*3+4*4+3*4)
If we write your equation 3/4 s + s + 4/3 s = 5 in the form
9s + 12s + 16s = 60, we can see a pattern
3^2 s + 12s + 4^2 s = 5(3)(4) , also notice that 12 is the geometric mean for 9 and 16
So if we were to find the area of the square of side s in the right triangle with sides 5, 12, 13 ,
our equation would be 25s + 60s + 144s = 780 , then solve for s
After watching solution
Others=oh my god
Indians=oh bhaisahab
Brief sketch of a co-geom. sol'n:
[1] put the entire figure on the Cartesian plane where
~ B is at (0,0);
~ BC is on +ve x-axis; &
~ BA is on +ve y-axis;
call the square S
[2] eqn. of L₁ (line containing AC):
y = -3/4 x + 3 => 3x + 4y - 12 = 0
[3] eqn. of L₂ (line containing the side of S // to L₁, but not in AC):
y = -3/4 x + c => 3x + 4y - 4c = 0 (for some c>0)
[4] eqn. of L₃ (line containing the side of S ⊥ to L₁, & intersects AB (at (0,c)):
y = 4/3 x + c => 4x - 3y + 3c = 0
[5] eqn. of L₄ (line containing the side of S ⊥ to L₁, & intersects BC (at (4c/3,0), where L₂ meets x-axis):
y = 4/3 x + k where 0 = 4/3 (4c/3) + k
=> 4x - 3y - 16c/3 = 0
[6] let s be the length of a side of S,
then dist(L₁,L₂) = s = dist(L₃,L₄) ...(*)
=> dist(O,L₁)-dist(O,L₂) = dist(O,L₃)+dist(O,L₄)
=> |-12/√(3²+4²)| - |-4c/√(3²+4²)| = |3c/√(4²+3²)| + |(-16c/3)/√(4²+3²)|
=> c = 36/37 ...(#) (recall: c>0)
[7] (*) & (#) => s = 60/37
=> area of S = s² = 3600/1369
As simple as that. U always amaze me with the extraordinary math tricks...... this is very delightful.
Where the similarity of triangles came from?
I want to know that too!!!
Since they're all right triangles you can derive that their corresponding angles are all equal. If angle A = x, the other acute angle in its smaller triangle is 90-x, which means the angle touching it on the other small triangle will also be x, and so on
I have another solution.
assume each side of square = 5x.
cut the square, throw away. cut the smallest triangle opposite to the slope = 5 and throw away too.
combine the two pieces of triangles with common line of 5x.
this is the new triangle with 3 sides: 3(1-x), 4(1-x) and 5(1-x) with a height of 5x.
Area of triangle = 0.5 * (3(1-x) ) * (4*(1-x)) = 0.5 * 5x * 5(1-x)
Tally, 12(1-x) = 25x
12-12x = 25x
37x = 12
x= 12/37
So, side of 5x = 5* 12/37 = 60/37
area = (60/37) square = 3600/1369
I dont wanna solve it rn or watch the solution either but one side of the square is paralel to the hypotenuse so theyre similiar triangles and u can solve it from there
My variation on this puzzle: same 3-4-5 triangle but now inscribed with a CIRCLE touching the three sides of the triangle... what is the radius of the circle?
In an acute-angled triangle ABC the interior bisector of the angle A intersects BC at L and intersects the circumcircle of ABC again at N. From point L perpendiculars are drawn to AB and AC, the feet of these perpendiculars being K and M respectively. Prove that the quadrilateral AKNM and the triangle ABC have equal areas.
Hi Vijay. In case you are interested in math competitions, please consider
th-cam.com/video/l5ef8BNduDs/w-d-xo.html and other videos in the Olympiad playlist. Hope you enjoy.
My approach:
Area of triangle = 6
Area of trapezoid containing the square is obviously: 1/2*(S+5)*S
Area of Triangle in the south-west corner is obviously: 1/2*(3/5)S*(4/5)S
Then: 6 = 1/2*(S+5)*S + 1/2*(3/5)S*(4/5)S
And solve for S and then S-squared
You want to thank YOUR PAPA who Lives in India for this math problem. YOUR PAPA would be proud of you as you thanked PAPA.
I used trigonometry.
Tan(ACB) = 3/4 = x/EC (I called E the point down)
Tan(BCA) = 4/3 = x/AD (the point above)
x = (4/3)AD = (3/4)EC
5 = AD + 5 + EC = (3/4)x + x + (4/3)x = (9/12)x + (12/12)x + (16/12)x = (37/12)x
x = (5 * 12)/37 = 60/37
A = x² = 3600/1369 ≈ 2,63 u²
😊
Oh man!😂😂 Papa is not a name..it means daddy in hindi and I think in english too..didn't u suspect his name😂
@@crazyfrog1190 🤣🤣🤣🤣🤣🤣🤣🤣👏🏻
If the guy signed "Papa", what else could he call him?... Besides, Papa means daddy in pretty much every language so I'm sure he's aware of it
MATH QUESTION
Numerical / algebra QUESTION
th-cam.com/video/soN5NmkaXeM/w-d-xo.html
One time see..
Lol Americans
@@shashibhushankr16 Presh Talwalkar is an American? Really? The name does not sound like it. The accent is not American either.
It looks so simple but hard to solve, lol
3dplanet100: walking and tours videos not very hard
really? All you need to do are name every angle and do similliar triangle, done.
The hardest part is to see what you have to do if you know what you have to do solving is the easiest part
@@AH-tm2cr nah, if you know every unknown angle, you'll get the answer very easily
Noob
There is a formula
abc/a²+b²+ab = (3×4×5)/(3²+4²+3×4)=60/37
This is the only question you have done that I solved by myself and I am proud
How do you know that the inscribed square divides the triangle into a series of similar triangles?
Side of the square = L ; AB=3=AD+DB
The interior triangles are similar to ∆ABC ⇒ AD=5L/4 and DB=3L/5 ⇒ 3=(5L/4)+(3L/5) → 37L/20=3 →L=60/37 ⇒ Area of the square = L²=3600/1369=2.63
Another way.
Let the side of square be x. (In this type of square inscribed in a triangle wherein one side of the square lies on one sode of the triangle.
Let a,b be base and height; c be hypotenuse.
Then x = abc/a^2+b^2+ab
.
Concept is the same..but a quicker method! 😅
First (and probably last) problem i have solved in this channel i am proud of myself
It's just wonderfully simple.
It is dear
Interesting problem as always. Thanks for your time and compile that kind of problems
Good dear
Cool problem. I wasn't sure of my answer until I watched the rest of the video.
Nice and easy :) There are 3 possible eqations (1 for each side) leading to the same result x=60/37 ->Area=x2
I solved this problem using similarity concept in 10th class. Got correct answer 👍
Can you explain please?
I'm not proud that I failed to see how short the video was before I attempted to solve the problem (and let myself be misled down a much more needlessly complicated path). I'm not proud that I needlessly calculated the lengths of the sides of the smaller right triangles. I am a little proud that, despite all that, I arrived at the same solution.
My solution:
The three smaller triangles are similar to the big one.
So the sides of the square are the 3,4 and 5 sides of the triangle.
3*4*5=60, so I assume square edge is 60.
Then 60*3/4=45 and 60*4/3=80. So the long side would be 80+60+45=185.
But it is only 5. So the side of the square is 60*5/185=1,622
The area is then square of that, so 2,263.
An easier method is side of square = b*h/(b+h). From sinBAC * 3 which gives height = 3*4/5, base = 5. So side of square = 60/37.
For the majority of these geometry problems, I just open up Fusion 360 and sketch some constrained dimensions instead of doing any math. It's lazy, but it works. At least, I realized that the triangles are similar before Fusion finished launching. 🤷🏻♂️
Hi Presh, Can you please help explaining how the longer side of a triangle is (4/3)s and shorter side is (3/4)s? Due to law of similar triangles?
(60/37)^2. If the side of square is x, then (5/4) * x + (3/5) * x = AB = 3, so (37/20)* x = 3, hence x = 60/37. It's possible to derive x from similar equations for AC and BC as well. The areas of all three smaller triangles inside are also trivially computed.
wish you had also shown the generalized formula Side S= a*b*sqrt(a^2+b^2)/(a^2+b^2+ab) where a,b are the sides of the triangle.
Well ,how to prove this propertionality ?
Think showning proof why square divide triangle in congruent triangles important (//-lines), just floating triangle around won't do in exam, but probably exercise/proof leave to the viewer.
But how? Where did you get the 3/4 and 4/3? I’m still learning so I’ll appreciate any help
similar triagles
since it's a 3-4-5 triangle, any triangle similar to it would have one leg be s and if it's the smaller one the larger is 4/3s, and vice versa if the larger is s then the smaller is 3/4s
Rotate and flip the 3-4-5 triangle until you get a similar shape of those small triangles. For the top left triangle you get 4 / s = 3 / unknown length L => L = (3/4)s. Similarly for bottom right triangle you get 3 / s = 4 / M => M = (4/3)s
Use similarly!
If two triangles have the same angles, then their sides must be in the same proportions. We know that one of the angles of the big triangle is 90°, so the top and bottom angles must also add up to 90°. Each of the little triangles obviously keeps at least one angle from the big triangle (the bottom left one must keep all three, since the sides of the square are parallel). The triangle at the bottom right corner keeps the angle in its shared vertex, and introduces a new 90° angle; this fixes the third angle, and now this triangle has all the same angles as the big triangle. The same is true for the top triangle. So we know that in each of these little triangles, as in the big triangle, the short side must be 3/4 of the long side.
Mummy: Say thanks to him
Children: Thank u papa
Everyone needs a Papa from India.
I solve it in an another way..... I am from India ...Since the smaller triangle(right angel) is similar to the bigger Triangle So, s=a*h/a+h,a=hypothenious ,h=height of the bigger triangle, By calculation, s=60/37 ,And we get Area ....The formula comes from Similarity Property
It was easy
I did it in 10 min
And that to bcz of forgetting geometry theorem I had to recollect all
At last I will say it was easy
I truly enjoyed in tackling this
By the way I did this with different approach
You don't even need geometry to solve this problem. Just make a quadratic equation by equating area of the whole triangle equal to all the other parts. Only the knowledge of area Calculation and quadratic equation requires in this approach.
Question:
How do we know the triangles are similar?
and for the triangle from left-bottom, note that the longest line is parallel to the longest line of the whole triangle, you can get the two same angle using internal stagger
I don't understand 0:47. Those fractions are coming from what theorem? And what leg is he referring to when he says the longer leg and the shorter leg? Is he talking about the sides of triangle ABC or the unknown values of the smaller triangles that run along line AC where AC=5?
A little shorter: AB=5s/4+3s/5 =3, so s=60/37....
Hi Mike. In case you are interested in math competitions, please consider
th-cam.com/video/l5ef8BNduDs/w-d-xo.html and other videos in the Olympiad playlist. Hope you enjoy.
lol by just looking at the length time of this video i knew it was easy so i tried and solved it before watching it. Though it actually took me hours to figure out how to come up with an equation that matters lol. thanks for this!
I’m guessing we will be able to come up with 3 equations from Gogou theorem and solve it to get length of side. Or use similar triangles
Well I was wrong kinda
I have another solution.
Let h be the altitude to the hypotenuse, a = AB, b = BC and c = CA. Then ab/2 = ch/2, since both give us the area of the triangle, this implies that h = ab/c.
Now that we know c and h we can easily calculate the area of the square.
h = 3*4/5 = 12/5
Let x be the side of the square.
x = ch/(c+h) (1)
x^2 = ((ch)^2)/((c+h)^2) = 3600/1396
Proof of (1):
Let ABC be a triangle with sides a, b and c. Let h be the altitude of ABC in respect to a. Let DEFG be a square with sides x, inscribed in this triangle such that the side a contains one side of the square.
Then
ah/2 = (a-x)x/2 + x(h-x)/2 + x²
=> ah/2 = (a-x)x/2 + x(h-x)/2 + 2x²/2
=> ah = (a-x)x + x(h-x) + 2x²
=> ah = ax - x² + xh - x² + 2x²
=> ah = x(a+h)
=> x = ah/(a+h)