how can I find the Side length Two squares inside an equilateral Triangle?

Question

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.

Answer 1

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

Answer 2

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

Answer 3

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 $x$:

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

Simplify:

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

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