Find area of the Green Square inscribed in the right triangle | Important Geometry skills explained

[@prbmax]

Can't say enough good things about your content and technique.

[@PreMath]

Greatly appreciated😎 Glad you think so! Thanks for your continued love and support! You are awesome, Paul. Keep it up 👍 Love and prayers from the USA! 😀 Stay blessed 😀

[@williamwingo4740]

As usual, here's a different approach: The big triangle ABC is 3-4-5, and all the smaller triangles are similar to the big one and to each other by angle-angle-angle; so triangles: AGF, EBF, and DCE are all also 3-4-5. Let EB = x; BF = y; and EF = z. z is also the side of the square we want. Because triangle EBF is 3-4-5, z = 5x/3 = 5y/4. Let's concentrate on y. Now consider triangle AGF. Here, AF = 4 -- y; so z = 3(4 -- y)/5. We now have two independent expressions for y. Equating them, we get: 5y/4 = 3(4 -- y)/5 = (12 -- 3y)/5; cross-multiplying: 25y = 48 -- 12y; collecting terms: 37y = 48; solving for y: y = 48/37. So far. so good; but what we really want is z. Remembering that z = 5y/4: z = (5/4)(48/37) = (5*48)(4*37) = (5*12)/37 = 60/37; and z^2 = (60/37)^2 for the area of the square. We could have done the same thing by figuring out x instead of y; and we could use any two of the three smaller triangles to get two independent expressions for either x or y. Cheers. 🤠

[@murphygreen8484]

I'm a bit confused. Just because you have right angle triangles, can you assume all the other angles always equal beta or alph? How do you know the triangles are similar before you know their angles?

[@douglasfeather3745]

He starts off by looking at triangle ABC. B is 90degrees so he labels angles A alpha and C beta. Then he looks at AGF, G is 90, A is alpha so F has be beta. Then he looks at CDE, D is 90deg, C is beta so E must be alpha. Then he looks at point F, this has 3 angles on a straight line or 180deg, two of them are know to be 90deg and beta so the 3rd must be alpha. Finally he looks at triangle BEF, B is 90deg, F is alpha and so E must be beta. So he has established that all 4 triangles have angles 90deg, alpha and beta.

[@MrPaulc222]

I've saved this one to watch again because I didn't understand beyond a certain point. I understand the the triangles are similar but figuring out what was happening with the 'a' variable looked confusing. I will get it, but not today :)

[@AnonimityAssured]

A general formula might come in handy one day. For a right triangle, a-b-c, for which c is the hypotenuse, the side-length, s, of the largest square that can fit inside it with one side along the hypotenuse, is given by a fairly simple formula. Spoiler alert. s = a ∙ b ∙ c / (a² + b² + ab). In our case: s = 3 ∙ 4 ∙ 5 / (3² + 4² + 3 ∙ 4); = 60 / (9 + 16 + 12); = 60 / 37. So: s² = 60² / 37²; = 3600 / 1369; ≈ 2.63. Interestingly, in any right triangle, the largest square that can fit inside it with its sides parallel to the triangle's orthogonal sides is always larger than the largest square with one side along the hypotenuse. There is an even simpler formula for the side-length, s, of the largest square that will fit inside a right triangle a-b-c: s = a ∙ b / (a + b). This is essentially the same as the simplest formula for the Crossed Ladder Theorem, but that's another story.

[@alster724]

This Math problem is tricky yet nostalgic.

[@PreMath]

Thanks for your feedback! Cheers! You are awesome. Keep it up 👍 Love and prayers from the USA! 😀

[@mffthefrog1307]

Thank you SO MUCH dude. I was struggling with a problem because I missed the class and the transcript was too confusing, and I swear I was in a monstrous amount of stress. This saved me. Thx bro 🔥🔥🔥❤❤❤

[@quickshortsmathapproach7669]

(3,4,5) are pythagorian triplets so , hypotenuse is 5. And length of perpendicular on a hypotenuse is =(p×b)/h= 4×3/5=2.4 And side of any square its a=( Height×base)/(height+base) Square lay on hypotenuse ,so hypo its base a= 2.4×5/(2.4+5) a=( 60/37) A²= (60/37)²

[@RahulGupta-uk7xy]

Hell sir i have a question that how can make our geometry good means the mindset you have to solve such problems how can we develop that , when question comes from geometry i am not able to think that so your tips pls sir

[@PreMath]

Dear Rahul, I call these math puzzles "gymnastics for the mind!" They make us think and improve mental agility! We are all lifelong learners! Just keep persevering. Thanks for asking. You are awesome. Keep it up 👍 Love and prayers from the USA! 😀

[@nandisaand5287]

I see a right triangle, I can't not think in trigonometric terms. SinA=S/AF=3/5; AF=5S/3 SinB=FB/S=4/5; FB=4S/5 AF + FB = 4 5S/3+4S/5=4 Multuply by 15: 25S + 12S = 60 37S = 60 S = 60/37 S^2 = (60/37)^2 = 2.63 There's likely several different variations of this calculation (using different sides) in this problem.

[@TheAnica51]

simple and nice

[@douglasfeather3745]

At 4:31 you side EF having length 5a in the end the side length turns out to be 60/37 so here a = 12/37. At 4:56 you have side DE with length 4a and here a must be 15/37. Then latter you change both these sides to 20a with a = 3/37. You don't clearly point out that you are changing the a as you do this.

[@WaiWai-qv4wv]

Ok Very thanks

[@HappyFamilyOnline]

Great explanation👍 Thanks for sharing😊

[@KAvi_YA666]

Thanks for video.Good luck sir!!!!!!!!!!

[@prithvisinghpanwar007]

but wait at 5:00 1 side of square is 5a and other is 4a so 5a=4a =>5=4🤔🤔

[@beverleygomesz]

A is constant.. omg thankss

[@calspace]

I called line segment EB x. Then the side of the square is 5 * (x/3). Line segment CE is therefore 3-x, which means that the side of the square is 4 * (3-x)/5. Therefore 5x/3 = (3-x) * 4/5 25x = 3 * (3-x) * 4 25x = 36 - 12x 37x = 36. x = 36/37. Therefore the side of the square is 5 * (36/37) / 3 ≈ 1.62. And the area of the square ≈ 1.62^2 or 2.63. It's exactly equal to (180/111)^2. As an aside, I think it's improper to label one side of the square 5a and another 4a. I get that you're trying to say that one is the 5 side of a 3-4-5 triangle and the other is the 4 side of a 3-4-5 triangle. But I think it would be better to label ∆CDE as having sides 3x, 4x, and 5x (with the side of the square being 4x) and ∆BEF as having sides 3y, 4y, and 5y (with the side of the square being 5y).

[@ybodoN]

Of course, we can apply the same procedure as in the video but using the triangles AGF and BEF instead of the triangles CDE and BEF. In this case, it is AB that will be 37a and EF (a side of the square) will be 15a. So we will obtain a = 4 / 37 and EF = 4 (15) / 37 = 60 / 37.

[@devondevon4366]

I did something quite different. Since it is a 3-4-5 triangle with angles 90, 53.13 and 36.87 Let the sides of the square = 120 x. I use 120 because it is easier divide it by 3, 4 and 5 So all four sides were labeled as 120x; hence AG=160x AF= 200x since A = 36.87 degrees, G=90 degrees FB=96x since E is 53.13 degrees EB=72x CE= 150x since CE is the hypotenuse (or 5 in the 345 triangles) DC=90x The sum of AB= 296x (200x + 96x) The sum of CB=222x ( 150x + 72x) The area of the triangle hence is [296][222]/2 = 32856x The area of the square is 120x^2 or 14,400. Recall side of the square was 120x The area of the square hence is 0.43287 ( 14,400/32856) of the triangle or 43.287% of it So what is the area of the triangle? Since it is a 3-4-5 triangle, the area is 12/2 = 6 square units But since the square is 43.287% of the triangle , and 43.287% of 6 is 2.6296. Then the area of the square is 2.6296 round to 2.63 Answer /Would have been better to use 60x instead

[@devondevon4366]

Could have used any number for the side of the square. It would be easier if I used 60 x instead of 120x. Note that all three triangles are 3-4-5 formed due to the inscribed square. The goal is to know if 120 x is the 3, 4, or 5. Take triangle AGF; 120 corresponds to 3. Hence AG is 160x and AF= 200x. Take triangle BGF; 120x corresponds to the 5; 120/5= 24. Hence the other two sides are 72x (3*24) and 96x ( 24*4) Take triangle CDE; the 120x corresponds to the 4. 120 divided by 4= 30; hence the other two sides are 150x ( 30*5) and 90x ( 30*3).

[@theoyanto]

Brilliant stuff, I didn't know where to begin 🤔. I love the way you just know where to stick variables and manipulate them, thats some real skill you've got there. Thanks again 👍🏻

[@PreMath]

Glad you enjoyed it! Thanks for your continued love and support! You are awesome, Ian. Keep it up 👍 Love and prayers from the USA! 😀

[@shadabshaikh92264]

Is this mathematically right?

[@hongningsuen1348]

A method not using similar triangles: 1. Let H be height of triangle ABC perpendicular to AC. By equal area equation 1/2 x 5 x H = 1/2 x 3 x 4. Hence H = 12/5. 2. Let A be side of square. Divide triangle ABC into trapezium ACEF and triangle BEF. Area of trapezium ACEF = 1/2 x (5 + A) x A = 1/2 x (5A + A^2) Area of triangle BEF = 1/2 x A x (H - A) = 1/2 x A x (12/5 - A) = 1/10 x (12A - 5A^2) 3. Area of triangle ABC = area of trapezium ACEF + area of triangle BEF Hence 6 = 1/2 x (5A + A^2) + 1/10 x (12A - 5A^2) = 25A + 12A = 3.7A. Hence A = 60/37 and area of sqaure = A^2 = (60/37)^2.

[@bhavyajindal7136]

Hey, I have solved also by using similar triangles property, but my answer is exact 3 sq. mts. for area of square. So, i am not getting why it has been solved as deriving area which is equal to 2.63 approx.

[@nandisaand5287]

Without showing your calculations, we can't see where you went wrong. This video has only been up a few hours, and already there's multiple comments showing different approaches to get to the same answer.

[@santiagoarosam430]

ABC es triángulo rectángulo tipo 3,4,5 y AGF, EBF y CDE son semejantes→ AC=5 → DE=EF=FG=GD=a → AF+FB=AB→(5a/3)+(4a/5)=4 → a=60/37 → Área verde =a²=60²/37² =2.6296 Un saludo cordial.

[@calvinham1]

Maybe he did it that way so that people who are not your level can benefit

[@murdock5537]

Nice and awesome, many thanks, Sir! 4 = k + n sin⁡(φ) = 3/5 = a/k → k = 5a/3 sin⁡(φ) = 3/5 → cos⁡(φ) = √(1 - sin^2(φ)) = 4/5 = n/a → n = 4a/5 → 4 = 5a/3 + 4a/5 → a = 60/37 or: ∆ ABC → AB = AF + BF = k + (4 - k) = 4; BC = BE + CE = 3 → AC = AG + DG + CD = 5 DE = EF = FG = DG = a sin⁡(φ) = 3/5 = a/k → k = 5a/3 = BE/(4 - k) = (3/4)(4 - k) = CD/CE = 3/5 = CD/(5a/4) → CD = 3a/4 cos⁡(φ) = 4/5 = AG/k → AG = 4a/3 → AC = 5 = 4a/3 + a + 3a/4 = 37a/12 → a = 60/37

[@aryan-ayandaredevils9780]

AB =4, BC =3 it is already given Let understand AC =5 (how)

[@pramodsingh7569]

Good problem

[@PreMath]

Glad you think so! Thanks for your feedback! Cheers! You are awesome, Pramod. Keep it up 👍 Love and prayers from the USA! 😀

[@phungpham1725]

Let a be the square side length. We notice that all the right triangles are 3-4-5 triples so the ratio of the relevant side lengths are equal. Consider the two right triangles AGF and FBE, we have GF/AF = short leg/hypotenuse =3/5-------> a/AF = 3/5 -------> AF= 5a/3 (1) and EF/FB = hypotenuse/long leg =5/4--------> a/FB= 5/4--------> FB= 4a/5 (2) (1) + (2)-----> AF+ FB = 4= 5a/3 + 4a/5--------> a= 60/37 Area of the green square = sq(60/37) = 2.63 sq units

[@vidyadharjoshi5714]

Let FE = 3a. All triangles are 3:4:5 triangles. So AF = 5a. FB = 3a*4/5. AB = 4 = 5a + 12a/5. a = 20/37. Area = 2.63

[@mohamedb6645]

Why fb=3a*4/5

[@vidyadharjoshi5714]

@@mohamedb6645 FEB is 3:4:5 triangle with EB = 3, FB = 4 & FE = 5. Since FE = 3a = 5, So FB = 4/5*FE = 3a*4/5.

[@msafasharhan]

diffcult to understand sir there is no another method to solve

[@giuseppemalaguti435]

A=(60/37)^2...tutti i triangoli rettangoli bianchi sono simili e i loro cateti stanno in proporzione 3:4

[@montynorth3009]

Let side length of square = X units. Triangles ABC and AFG are similar. So X/AF = 3/5 = 0.6. X = 0.6AF...........(1) Triangles ABC and FBE are also similar. So X/FB = 5/4 = 1.25. X = 1.25FB. Equating the two values of X. 0.6AF = 1.25FB. AF + FB = 4. FB = 4 -AF. Substituting for FB. 0.6AF = 1.25 (4 - AF). 0.6AF = 5 -1.25AF. 1.85AF = 5. AF = 5/1.85. From (1). X = 0.6AF. X = 0.6 x 5 /1.85. X = 3/1.85. Area (X^2) = 3^2/1.85^2. 9/1.85^2. 2.63.

[@devondevon4366]

11

[@devondevon4366]

.x

[@devondevon4366]

2.63

[@yakupbuyankara5903]

3600/1369.

[@divyabarad4795]

I did it in some other way and got the ans approx 2.78😢

[@hassegreiner9675]

You're too elaborate on the most basic things after my taste, but I like to learn the applied strategies to solve the problem.

[@PreMath]

Glad you think so! Thanks for your feedback! Cheers! You are awesome, Hasse. Keep it up 👍 Love and prayers from the USA! 😀

[@mariopopesco]

Usually your explanation are very clear. Not this time.

[@subratamalakar5472]

According to the queston, we can write that in between triangle CDE and Tri EBF angCDE =angEBF=90degree again, angDEC=angEFB And side DE=sideEF Therefore, Both the triangle are equal. It means sideFE =sideCE Then DE=CE Therefore angCDE=angDCE, which is absurd. The problem itself wrong.

[@WernHerr]

Warning, misconception: CDE and FBE are only similar! In CDE you have CE as hypotenuse, in FBE it is FE, with not the same length!

[@DevilMayCry01]

calculate height assuming ac as base which is (4×3)/5=12/5. then the side of square will be (5×12÷5)/(5+12÷5)=60/37 now square this side you'll get the area.. you don't need paper pen to solve this question