Calculate area of the Blue Square | Fun Geometry | Important Geometry skills explained

[@spiderjump]

By congruent triangles(ASA) the 2 larger right triangles are congruent and their sides are 12 and 15 Hence by Pythagorean theorem, hypotenuse of triangle = (12^2 + 15^2)^1/2 Area of square = 369

[@HappyFamilyOnline]

Very well explained👍 Thanks for sharing🎉

[@kelstra1997]

When I went to school about 65 years ago I used to hate maths but you make it fun. If you had been my teacher who knows what may have happened?

[@theoyanto]

Another great example 👍🏻

[@bigm383]

Nice work Professor!🥂❤

[@PreMath]

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

[@vaibhavgupta5359]

We can use Pythagoras triplet to find answer (2m)^2+(m^2-1)^2 =(m^2+1)^2 Let (m^2-1)=15 m=4 So substitute value of m in (m^2+1)^2 as this is the hypotenuse So the hypotenuse=17 Area of square=side^2 17^2=289

[@alster724]

Wow!!! Awesome solution

[@KAvi_YA666]

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

[@murdock5537]

Nice and awesome, many thanks, Sir! sin⁡(φ) = 12/a cos⁡(φ) = 15/a → sin^2(φ) + cos^2(φ) = 1 → 144 + 225 = 369 = a^2 = area ∎ btw: a = 3√41 → sin⁡(φ) = 4√41/41 or: (a√2)^2 = (15 + 12)^2 + (15 - 12)^2 = 738 = 2a^2 → a = 3√41

[@garypaulson5202]

Fun to watch and re-learn

[@paramasivamparamasivam525]

Nice things ur posting

[@PreMath]

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

[@santiagoarosam430]

Si desde las cuatro esquinas trazamos segmentos iguales al de 15 de longitud del esquema original, el cuadrado azul queda descompuesto en cuatro triángulos rectángulos de 12x15 y un pequeño cuadrado central de 3x3 》 Área azul =4(12×15/2) +3×3=369 Saludos cordiales.

[@jilalimohad3255]

Let's name y the other side of the blue triangle and y' the other side of the yellow triangle Then 12^2+y^2=x^2 (1) 15^2+y'^2=x^2 (2) By doing (2)-(1) we have y'^2-y^2=81 Which reformulate (y'-y)(y'+y)=81 81 = 1*81 or 3*27 or 9*9 For 81=3*27 y'+y=27 or 3 (but obviously 3 is not possible because y'+y > y'-y) And y'-y=3 or 27 (but obviously 27 is not possible because y'+y> y'-y) Then y'=15 and y=12 by solving the system And x = 12^2+15^2 = 369 For the last two cases 81=9*9 and 81=1*81 we find two different values for x which is absurd

[@vara1499]

You mentioned the angles beta and alpha in the yellow shaded triangle.. How did you arrive at this step?

[@AtomizerX]

Because blue α and β must equal 90°, and the same is true for blue α and the adjacent yellow angle, which must be β, because they form the blue square's right angle.

[@Methodgood]

Tính diện tích hình vuông.❤❤❤

[@AnonimityAssured]

I'm pleased with this. I did it by _ad hoc_ means, thinking inside the box, as it were. Spoiler alert. I imagined the internal line segments extended to span the whole square, and then mentally added another line segment extending down and left from the top right corner, parallel with the ascending diagonal shown. The resulting 'virtual' figure then consisted of four congruent right triangles surrounding a small square. That reduced the calculation to: A = 2 ∙ 12 ∙ 15 + (15 − 12)²; = 360 + 9; = 369 square units. Now to see whether my method was the expected one.

[@vidyadharjoshi5714]

15 and 12 are similar triangles so the third side = sqrt(15sq+12sq) Area = 15sq+12sq = 369

[@ファミパンaka剛腕]

Was it already known , this shape is square?

[@bienvenidos9360]

Yes. It says in the directions to find the area of the blue square.

[@samriddhadutta2104]

369 sq unit

[@dimitriosgrivas6231]

This exercise is not realistic. The blue square, if we considered it as square it should have both 4 sides equal to x, but its 2 diagonals should cross eachother in 90 degrees and meet eachoder in their centers.

[@bienvenidos9360]

The lines are not diagonals. As you can see one of the smaller shapes is not a triangle. There are lines inside the square that meet at perpendicular 90° angles but they all don't meet at the same place. This has nothing to do with being diagonals.