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There are $12$ similar triangles with angles $30°, 60°$ and $90°$. In the diagram below you can see how the first three triangles are arranged. They are arranged so all of them have one of their points at the point $O$. One of the sides of the $(n+1)$th triangle is the same length as the $n$th triangle’s hypotenuse, and they are arranged so those two sides line up.

If the side $OX$ has length $1$, then what is the area of all of the triangles added together? Below is the image: http://www.ucl.ac.uk/clie/placement-tests/UPC/Maths-2/images/question21a.jpg

I only labeled the angles, other than that I have no clue as to how to solve this problem.

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    If $OX$ has length $1$, what is the length of the hypotenuse of the first triangle? – 2017-01-19
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    i did that, now how can that help me? – 2017-01-19
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    the area of the figure is the area of a regular $12$-gon with apothem $1$. You can find the perimter using the $30-60-90$ triangle theorem. – 2017-01-20

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Hint

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By the picture:

$$\cos 30º=\frac{l_n}{l_{n+1}}=\frac{\sqrt{3}}{2} \rightarrow l_{n+1}=\frac{2\sqrt{3}\cdot l_n}{3}$$

So the sides are in a geometric sequence.

We also have that the area of the $n-th$ triangle will be, as a function of $l_n$:

$$A_n=\frac{h_n\cdot l_n}{2}=\frac{l_n\cdot \tan 30º\cdot l_n}{2}=\frac{(l_n)^2\cdot \tan 30º}{2}$$

Can you finish?

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    sorry i did not get your explanantion. first what does cathetus mean? – 2017-01-19
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    if possible, may u please provide a drawing? – 2017-01-19
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    i found the area of the first triangle to be 1/sqrt3, how can that be helpful? – 2017-01-19
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    thank u sooo mcuh @arnaldo – 2017-01-19
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    @exchangehelpforuni: you are very welcome! – 2017-01-19