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

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

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[prbmax]

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

[williamwingo]

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. 🤠

[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.

[quickshortsmathapproach]

(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

[douglasfeather]

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.

[nandisaand]

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.

[murphygreen]

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?

[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).

[misterenter-izrz]

Let s be the side of the square, then DC=3s/4, CE=5s/4, EB=3s/5, BF=4s/5, FA=5s/3, AG=4s/3, therefore (1+4/3+3/4)s=5, or s=5/(37/12)=60/37, and thus the area is 60^2/37^2=2.63 approximately, done.🙂

[alster]

This Math problem is tricky yet nostalgic.

[MrPaulc]

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 :)

[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.

[RahulGupta-ukxy]

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

[mffthefrog]

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 🔥🔥🔥❤❤❤

[hongningsuen]

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.

[devondevon]

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

[prithvisinghpanwar]

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

[HappyFamilyOnline]

Great explanation👍
Thanks for sharing😊

[santiagoarosam]

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.

[murdock]

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