New KCC's Quiz AQQ298 about an interlaced square and triangle puzzle

The Area of the Triangle is by Factor 1.07735 greater than the Area of the Square

https://filedn.com/lS8iJvuAuSbL4HBliSQrMrL/adi/aqq298.pdf

Triangle side = A = C(1 + 1/sqrt(3)) (i.e. C + C tan(30))
Triangle Area = sqrt(3)/4 * A^2 = C^2 * (1/2 + 1/sqrt(3)) ~1.07735 C^2

1. Triangle will have the higher area (even without computation we can see that top triangle above square is extra area)
2. Area is (0.5 + 1/sqrt(3) ~ 1.077

YES! You get it steve12 , congratulations!

Well done Sriramg , you get it, bravo!

1.   By inspection the triangle clearly has a larger area than the square by the area of the small equilateral triangle above the square.

2.   We can derive the area of an equilateral triangle (half the base, b, times the height) as b^2 * 3^0.5 / 4 (where ^ is exponent) [1].

The area of the square is c^2.  So all we need is the area of the little triangle.

We can use the trig identity Tan(30) = 3^-0.5 to get the length of the base of the little triangle:

b = c (1- 3^-0.5).  Insert this in [1] and add c^2.

Expand and collect terms:

Area of large triangle = c^2 (2+ 3^0.5) 3^-0.5 / 2

~ c^2 * 1.0774 

Q1 :The area of the red triangle ABC is always greater than that of the square adBe. The difference in area is equal to that of the small triangle abC because the two right triangles Aea and bdB are equal. Indeed, their angles are paired, and one side of the right angle is equal to C.

1. The triangle has the largest area
2. Area Triangle = A = c^2 (3+2sqrt(3))/6

Answer 1 comes from geometrical consideration around the two right triangles determinded by the portion L' of the triangle's sides (hypotenuse) and the sides of the square (cathetus). These two right triangles (first one internal to the equilateral triangle, second one external) are equal because they have two sides equal. So the area of the square fill the equilateral triangle except the area of the smaller equilateral triangle on top.

Answer 2. Let's call A' the area of the small equilateral triangle and L the side of the big equilateral triangle. By geometrical consideration you see the side of the small triangle is c-a, where a= L-c. Since the triangle is equilateral, we know all the angle are 60°. So we know that a=c/sqrt(3)
You can then calculate A'=sqrt(3) (c-a)²/4. Then A=c^2 + A' = c^2 (3+2sqrt(3))/6

I agree entirely @gpairino.  Interestingly this problem is less about logic or analogue circuits, and more about basic trigonometry and algebra.  But still an interesting mental exercise.          

1. The triangle has the larger area

2. Base of triangle = b

c/(b-c) = tan(60) = 1.732

c = (b-c)*1.732

b = 1.5773672*c

Triangle Area = {sqrt(3)/4}*b²

Triangle Area = 1.07737*c²

The triangle area is about 7.7% larger than the square

After reviewing my work, I realize I didn’t simplify my expression properly, so I’m adding this clarification to my answer:

It’s the same result, but it’s simpler…

I’d also like to add that the square is denoted by adBe and not by aDbe, as I wrote too hastily. I apologize for that.