If the given equilateral triangle has a side measure of 4. What is the sum of the areas of the stacked squares inside it?
+328 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.
+1641 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
+129899 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
+26393 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
\(\begin{array}{|rcll|} \hline sum_{\infty} &=& s_0^2\dfrac{q}{1-q} \quad | \quad q = 0.2153903092 \\ sum_{\infty} &=& 4^2\times 0.2745190528 \\ \mathbf{sum_{\infty}} &=& \mathbf{4.3923048454} \\ \hline \end{array}\)
