Can you find area of the red Triangle ABC? | Square and Equilateral triangle | [Math Olympiad]

[@bigm383]

Thanks Professor!❤🥂

[@PreMath]

My pleasure!
You are awesome. Keep it up 👍

[@bighominid]

As another commenter suggested, once you have the triangle side length, finding the area just means applying the formula for the area of an equilateral triangle: (s^2√3)/4. No need to determine h. Easy!

[@jimlocke9320]

Trigonometry can be avoided by making use to the ratios of sides for the 30°-60°-90° special triangle: short side, long side, hypotenuse = 1 : √3 : 2. ΔAED is such a triangle, and its long side is 1, so short side is 1/√3 = (√3)/3 and length AB = 1 + (√3)/3 = (3 + (√3))/3. The formula for the area of an equilateral triangle is A = ((√3)/4)s², where s is the length of a side. So s² = ((3 + (√3))/3)² = (9 + 6√3 + 3)/9 = (9 + 6√3 + 3)/9 = (12 + 6√3)/9. Multiply by (√3)/4) to get A = ((√3)/4)(12 + 6√3)/9 = (12√3 +(6)(3))/((4)(9)) = (12√3 +18)/36 = (3 + 2√3)/6, as PreMath also found.

[@thewolfdoctor761]

Once you get the equilateral triangle side length, then you can use the formula Area = 1/2(ab)Sin(x), in this case ((side length)^2 * sin(60))/2 ~= 1.08

[@PreMath]

Very true! Thanks, dear

[@soli9mana-soli4953]

Triangle AED is a 30°60°90° triangle and DE = 1 (side of the square) that means X√ 3=1 so X = √ 3/3
then the side of equilateral triangle is 1 + √ 3/3... and so on...

[@KAvi_YA666]

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

[@JamesDavy2009]

If a variable ends up as a denominator in a trigonometric ratio, use a reciprocal ratio (csc, sec, cot). The value x is simply cot(60°), equalling sqrt(1/3); and instead of rationalising the denominator, you could have used the laws of radicals: 1/sqrt(3) = sqrt(1/3). Either approach yields the same value.

[@murdock5537]

Nice! BE = DE = 1; EAD = 2φ = 60°; AE ∶= a → DE = BE = a√3 = 1 → a = √3/3 →
AB = a(√3 + 1) = (1/3)(3 + √3) → AB/2 = (1/6)(3 + √3) →
BM = AM = AB/2 → sin⁡(BMC) = 1 → CM = (AB/2)√3 = (1/6)(√3)(3 + √3) →
area ∆ABC = (1/6)(3 + √3)(1/6)(√3)(3 + √3) = (1/6)(√3)(2 + √3) = (1/6)(3 + 2√3)

[@PreMath]

Super!

[@georgebliss964]

I used a Trig.method to calculate the side length of triangle.
Joining D to B forming the diagonal of the square DB.
Since the sides of the square = 1, then DB = Sq.rt 2.
Angle DBA = 45 degrees.
Angle DAB = 60 degrees.
Therefore angle ADB = 180 - 45 - 60 = 75 degrees.
Applying Sine Rule to triangle DBA.
DB / sin 60 = AB / sin 75.
Sq.rt 2 /sin 60 = AB / sin 75.
AB = 1.57735, same as ( 3 + root 3) /3 quoted.

[@mohabatkhanmalak1161]

Thank you Ustad (ie.,Proffessor)!😎

[@PreMath]

So nice of you, dear
You are very welcome!
You are awesome. Keep it up 👍
Stay blessed 😀

[@marcovargasglobant7923]

Once you have the side lenght of triangule, no need for calculate "h", just use the formula for equilateral triange!! (L²√3) / 4

[@PreMath]

True! Thanks

[@raya.pawley3563]

Thank you!

[@svmartins1]

Biutiful, i did solve it.
Biuti problem
Kind regards from Portugal

[@giuseppemalaguti435]

Per la similitudine dei triangoli risulta sqrt3/2*l:l/2=1:(l-1)...l=1/sqrt3+1...calcolato l ,A=sqrt3/4*l^2

[@PreMath]

Thanks for sharing! Cheers!
You are awesome. Keep it up 👍

[@stark2518]

Area of equilateral triangle √3/4 multiplied by a²

[@tedn6855]

I calculated the lengths of blue triangle outside. And added missing bottom length from 1. Added that to diagonal and that gave me equilateral length. I think your way was one less step. Same answer though.

[@marioalb9726]

Sidei 's' of square:
s = √1 = 1 cm
Base or side 'b' of triangle:
tan 60° = s / (b - s)
(b - s) tan 60° = s
√3 b - √3 s = s
√3 b = (1+√3) s
b = (1/√3 + 1) s
b = 1,5774 cm
Area of triangle
A = ½ b² sin 60°
A = 1,077 cm² (Solved √ )

[@misterenter-iz7rz]

AD=2/root 3, FG=1/root 3, so DC=1-1/root 3, then AC=2/root 3+1-1/root 3=1+1/root 3, therefore the answer is (1+1/root 3)^2root 3/4=(4/3+2/root 3 )root 3/4=1/root 3+1/2=1.07735 approximately. 😊

[@PreMath]

Excellent!
Thanks for sharing! Cheers!
You are awesome. Keep it up 👍

[@vidyadharjoshi5714]

Tan 60 = DE/AE = 1.732Let DE = X, X = 0.577. Height of the triangle = sqrt( 1.577sq - 0.25*1.577sq) = 1.366.
Area of triangle = 1.577*1.366*0.5 = 1.077

[@ybodoN]

Generalized: the area of the equilateral triangle ABC is (3√S + √3)² / (12√3) where S is the area of the square BEDF.

[@PreMath]

Excellent!
Thanks for sharing! Cheers!
You are awesome. Keep it up 👍

[@wackojacko3962]

When you can't use Pythagoras....use SOH CAH TOA too find side lengths of right triangles. Good problem! 🙂

[@PreMath]

Excellent!
Thanks for your feedback! Cheers! 😀
You are awesome. Keep it up 👍

[@saumypandey2288]

Directly apply area of equilateral triangle

[@hernan860]

1/3 cm2

[@halitiskender1324]

Pembe alan1/3