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Question: Figure shows an equilateral triangle with side length equal to $1$ . Two squares of side length a and $2a$ placed side by side just fit inside the triangle as shown.

Find the exact value of $a$.

Its an Assessment question from edX course "A-Level Mathematics Course 1" and I am supposed to use skills that I learnt in Indices and surds,Inequalities and The Factor Theorem.

I have tried finding the height of triangle and then use similar triangles to find the right triangle length still No luck.

I am just looking for food for thought or very small hints thats all.


Hint: You have everything you need along the base. Also, note the two flanking right triangles.

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    $\begingroup$ i think i got it. Thanks $\endgroup$ – fpsshubham Sep 18 '19 at 14:12

From the leftmost right triangle, $$ \frac{a}{x} = \tan(60°) \implies a = \sqrt{3}x $$ From the rightmost right triangle $$ \frac{2a}{1-3a-x} = \frac{a}{x} \\ a = \frac{3-\sqrt{3}}{6} \approx 0.211 $$

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    $\begingroup$ I think you are on the right path to an answer, but there is not enough detail. First, you should typeset your answer with MathJax. Second, give more details. E.g. what is $x$? $\endgroup$ – Physical Mathematics Apr 21 '20 at 21:18

I struggled with this too. But the info is along the bottom. The triangle is equilateral so all angles are $60°$. On the left there is a right-angled triangle - let's call its base $x$.

Triangle 1: Angle = $60°$, opposite = $a$, and adjacent = $x$

On the right there is another right-angled triangle and its base is $1-3a-x$.

Triangle 2: Angle = $60°$, opposite = $2a$, and adjacent = $1 - 3a - x$

From triangle 1: $$\tan60° = \frac{a}{x} \implies x = \frac{a}{\sqrt{3}}$$

From triangle 2: $$\tan60° = \frac{2a}{1-3a-x}$$

Equalising $\tan60°$ and substituting for $x$:

$$\frac{2a}{1-3a-x} = \frac{a}{x} \implies \frac{2a}{1-3a-\frac{a}{\sqrt{3}}} = \frac{a}{\frac{a}{\sqrt{3}}}$$


$$\frac{2}{1-3a-\frac{a}{\sqrt{3}}} = \frac{\sqrt{3}}{a}$$
$$\frac{2a}{1-3a-\frac{a}{\sqrt{3}}} = \sqrt{3}$$
$$2a = \sqrt{3}(1-3a-\frac{a}{\sqrt{3}})$$

$$3a +3a\sqrt{3}= \sqrt{3}$$
$$a= \frac{\sqrt{3}}{(3 +3\sqrt{3})}$$ Rationalise the Denominator
$$a= \frac{\sqrt{3}(3 -3\sqrt{3})}{(3 +3\sqrt{3})(3 -3\sqrt{3})}$$
$$a= \frac{3 -\sqrt{3}}{6}$$


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