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Problem

The following figure is made up of equilaterals triangles and squares of side 24. What's the area of the square that forms connecting the centers of the four squares?The image isn't accurate at all

What i've done so far

I've tried many things and i just got this: -Two of the sides of the red square are paralell to the shared side of the two equilateral triangles, that makes the two small-triangles (up and down) equilaterals and similars to the 2 original ones.

-The other 2 small-triangles (left and right) are congruent to the others (they're equilaterals, and we can prove it playing with the angles).

-We can replace the small triangles (left and right) with the other two (up and down), so the area will be the area of the 2 original triangles and the other four regions, that equals 1 small square. By pithagoras, the height of the triangle of side 24 is

$\sqrt{24^2 - 12^2}$ = $\sqrt{576 - 144}$ = $\sqrt{432}$

And the area of both triangles is

$(\sqrt{432} x 24 )/2 x 2 )$ = $24\sqrt{432}$

Adding the area of a square:

$24\sqrt{432}$ + 24^2 = $24\sqrt{432} + 576$

But i'm not really sure about it, because i made it with a compass and i got that the diagonal of the red square is 48 ( two times the sides of the triangles and the small squares) and, using pithagoras you get that the area is 1152

Is it possible to solve the problem without using trigonometrical functions or any "complicated" methods?

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  • $\begingroup$ I would put the $x$ axis along the line between the triangles with the origin at the center and find the coordinates of the center of one square. You might argue that involves trig. $\endgroup$ – Ross Millikan Feb 12 '17 at 5:41
  • $\begingroup$ Your hand measurement with the compass is deceiving you. I get about 46.36 for the diagonal, which is pretty close to 48 for a hand-drawn figure. I think you did very well with the calculations you made; I got the same result through a somewhat different method. $\endgroup$ – David K Feb 12 '17 at 5:55
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A right triangle is formed, with an angle of 60° and greater leg (blue in the diagram) of length 12. Shorter leg is then $12/\sqrt3=4\sqrt3$ and the hypotenuse is $8\sqrt3$.

The red square side is then $L=2\cdot8\sqrt3+(12-4\sqrt3)=12(1+\sqrt3)$.

enter image description here

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It is just a matter of using pythagorean theorem twice. enter image description here

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Hint: each red box contains a small box and two triangles.

Cut and glue.

To make it more clear, you should try to expand the pattern and add more red boxes.

Edit: Observe the area of a equilateral triangle is given by \begin{align} \text{height}\times\text{base} \times \frac{1}{2}= 12\sqrt{3}\times 24 \times \frac{1}{2} = 12\sqrt{432} \end{align}

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Let the line between the triangles be the $x$ axis with the origin at the center. The corners common to the triangles are $(-6,0)$ and $(6,0)$. Using Pythagoras the top of the triangle is $(0, 6\sqrt 3)$. The center of the right side is then $(3, 3\sqrt 3)$. Now draw a line from $(-6,0)$ through $(3, 3\sqrt 3)$ which will pass through the center of the upper right square. Extend the line $12$ units past $(3,3 \sqrt 3)$ and you will be at the center. The $y$ coordinate is half the side of the red square.

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enter image description here Required area = [⊿ABX] + 2[⊿ADX] + [⊿XCD]. The area of each of them can be found using the formula $Area = \dfrac {1}{2}ab \sin C$. In particular, $AX = AD$ and $DX = \dfrac {12}{\cos 15^0}$.

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Here's a cut and paste solution suggested by @JackyChong 's answer.

The red square is the union of four quadrilaterals and two equilateral triangles. (The equilateral triangles are composed of two pieces each.)

The four quadrilaterals assemble into a square of side $24$. (That's a well known proof of the Pythagorean Theorem - maybe I can find an image). The two triangles have side length $24$. So the area of the red square is $$ 24^2 + 2 \times 24^2\left(\frac{\sqrt{3}}{2}\right) = 24^2(1 + \sqrt{3}). $$

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