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If the given equilateral triangle has a side measure of 4. What is the sum of the areas of the stacked squares inside it?

 

 Jan 10, 2021
 #1
avatar+122 
0

The entire triangle itself will have an area of 16, because the side length of the entire triangle is 4, so 4x4=16. 

 

Then, we'll have to find the area of the space that is NOT the stacked squares.

 

If you look closely there are two half-triangles to accompany each square so we can put them together and if you look even MORE closely, all of the side triangles equal HALF of all of the stacked squares combined.

 

I hope this gave you a clue, but I am warning you that I am not exactly great at math and let alone geometry.

 Jan 10, 2021
 #3
avatar+861 
+1

Hi, Guest!

The area of an equilateral triangle is  ( √3 / 4) * 42    ==> 6.92820323 square units

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

2 (a/√3) + a = 4        a = 1.856406461          a2 = 3.446244947

 

2 (b/√3) + b = a        b = 0.861561236          b2 = 0.742287764

 

2 (c/√3) + c = b        c = 0.399851961          c2 = 0.159881591

 

2 (d/√3) + d = c        d = 0.18557194            d2 = 0.034436945

 

The sum of the areas of all squares         A = 4.382851247 square units

jugoslav  Jan 10, 2021
edited by Guest  Jan 10, 2021
edited by Guest  Jan 10, 2021
 #2
avatar+114361 
+1

It appears   that  the   side of the  bottom  square, S,  can  be  computed as

 

sin 60    =   S/2

2sin 60 =  S

2 sqrt (3) /2   = S

sqrt (3)  =  S

 

And it appears  that  each  square after  this one has side lengths 1/2 as long as the previous square  =   sqrt (3) / 2 , sqrt (3)/4, sqrt (3) /8

 

So  the combined areas are

 

(sqrt (3))^2  +  ( sqrt (3)/2)^2  + (sqrt (3)/4)^2  + (sqrt (3)/ 8)^2   =

 

3  + 3/4  +   3/16  +  3/64   =

 

3 (  1+  1/4 + 1/16 + 1/64)   =

 

3  (64 + 16 + 4 + 1)  / 64   =

 

255/ 64  units^2

 

 

cool cool cool

 Jan 10, 2021
 #4
avatar+25656 
+1

If the given equilateral triangle has a side measure of 4.
What is the sum of the areas of the stacked squares inside it?

 

My attempt:

 

\(\text{Let}~ \tan(60^\circ) = \sqrt{3} \\ \text{Let the side of the equilateral triangle }~ s_0 = 4 \\ s_1~\text{side of stacked square 1} \\ s_2~\text{side of stacked square 2} \\ s_3~\text{side of stacked square 3} \\ s_4~\text{side of stacked square 4} \\ \ldots \\ s_n~\text{side of stacked square n} \\\)

\(\begin{array}{|rcll|} \hline s_0 &=& s_1 + 2x_1 \quad | \quad \tan(60^\circ) = \dfrac{s_1}{x_1}~ \text{or}~ s_1 = x_1\tan(60^\circ)=x_1\sqrt{3} \\ s_0 &=& x_1\sqrt{3} + 2x_1 \\ s_0 &=& x_1(2+\sqrt{3}) \\ x_1 &=& \dfrac{s_0}{(2+\sqrt{3})}\times \dfrac{(2-\sqrt{3})} {(2-\sqrt{3}) } \\ x_1 &=& \dfrac{s_0(2-\sqrt{3})} { (2+\sqrt{3})(2-\sqrt{3}) } \\ x_1 &=& \dfrac{s_0(2-\sqrt{3})} { 4-3 } \\ \mathbf{x_1} &=& \mathbf{s_0(2-\sqrt{3})} \\ \hline s_0 &=& s_1 + 2x_1 \quad | \quad \mathbf{x_1=s_0(2-\sqrt{3})} \\ s_0 &=& s_1 + 2s_0(2-\sqrt{3}) \\ s_1 &=& s_0 - 2s_0(2-\sqrt{3}) \\ s_1 &=& s_0\Big(1-2(2-\sqrt{3})\Big) \\ s_1 &=& s_0(1-4+2\sqrt{3}) \\ \mathbf{s_1} &=& \mathbf{s_0(2\sqrt{3}-3)} \\ \hline \end{array}\)


In the same way...

\(\begin{array}{|rcll|} \hline s_2 &=& s_1(2\sqrt{3}-3) \quad | \quad \mathbf{s_1=s_0(2\sqrt{3}-3)} \\ s_2 &=& s_0(2\sqrt{3}-3)(2\sqrt{3}-3) \\ s_2 &=& s_0(2\sqrt{3}-3)^2 \\ \hline \ldots \\ s_3 &=& s_0(2\sqrt{3}-3)^3 \\ s_4 &=& s_0(2\sqrt{3}-3)^4 \\ \ldots \\ \mathbf{s_n} &=& \mathbf{s_0(2\sqrt{3}-3)^n} \\ \hline \end{array} \)

 

sum

\(\begin{array}{|rcll|} \hline \mathbf{sum_4} &=& \mathbf{s_1^2+s_2^2+s_3^2+s_4^2} \\ sum_4 &=& s_0^2(2\sqrt{3}-3)^2+s_0^2(2\sqrt{3}-3)^4+s_0^2(2\sqrt{3}-3)^6+s_0^2(2\sqrt{3}-3)^8 \\ sum_4 &=& s_0^2\Big( \underbrace{ (2\sqrt{3}-3)^2+(2\sqrt{3}-3)^4+(2\sqrt{3}-3)^6+(2\sqrt{3}-3)^8\Big) }_{Geometric~Sequence~q=(2\sqrt{3}-3)^2} \\ sum_4 &=& s_0^2(q+q^2+q^3+q^4) \\ sum_4 &=& s_0^2\dfrac{q}{1-q} (1-q^4) \quad | \quad q = 0.2153903092 \\ sum_4 &=& 4^2\times 0.2745190528 \times 0.9978476909 \\ sum_4 &=& 4.3923048454 \times 0.9978476909 \\ \mathbf{sum_4} &=& \mathbf{4.3828512478} \\ \hline \end{array}\)

 

Infinite Geometric Sequence

\(\begin{array}{|rcll|} \hline sum_{\infty} &=& s_0^2\dfrac{q}{1-q} \quad | \quad q = 0.2153903092 \\ sum_{\infty} &=& 4^2\times 0.2745190528 \\ \mathbf{sum_{\infty}} &=& \mathbf{4.3923048454} \\ \hline \end{array}\)

 

laugh

 Jan 10, 2021
edited by heureka  Jan 10, 2021

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