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

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

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[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

[vaibhavgupta]

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

[misterenter-izrz]

Let w be the middle width, l be the side of square, d be the diagonal, w=sqrt(l^2-144)-sqrt(l^2-225), d^2=2l^2=729+w^2=729+(sqrt(l^2)-144)-sqrt(l^2)-225))^2, equation of l, solving it gives the area is l^2=369.
But your method is simpler and clever than me.

[HappyFamilyOnline]

Very well explained👍
Thanks for sharing🎉

[misterenter-izrz]

Rather difficult, this puzzle， I give up.😅

[dimitriosgrivas]

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.

[jilalimohad]

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

[kelstra]

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?

[murdock]

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

[santiagoarosam]

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.

[ファミパンaka剛腕]

Was it already known, this shape is square?

[vara]

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

[vidyadharjoshi]

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

[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.