The diagram shows an equilateral triangle $ADE$ inside a square $ABCD$ . What is the value of $\frac{[\Delta ADE]}{[\Delta DEC]}$ .

Question

The diagram shows an equilateral triangle $ADE$ inside a square $ABCD$ . What is the value of $\frac{[\Delta ADE]}{[\Delta DEC]}$ .

What I Tried: Here is the diagram :-

You can see I marked the angles as required. Now let $AB = x$ . We then have :- $$[\Delta ADE] = \frac{\sqrt{3}}{4}x^2$$ Now from here :- https://www.quora.com/What-is-the-ratio-of-sides-of-a-30-75-75-angle-triangle-without-sine-rule , I could understand and show that :- $$ED : DC : CE = \bigg(\frac{\sqrt{3} + 1}{2} : \frac{\sqrt{3} + 1}{2} : 1\bigg)$$

So let $EC = k$ , $CD = DE = \frac{(\sqrt{3} + 1)k}{2}$ .

From here :- $$x = \frac{(\sqrt{3} + 1)k}{2}$$ $$\rightarrow k = EC = \frac{2x}{(\sqrt{3} + 1)}$$

Now, we can find area by Heron's Formula. We have :- $$s = x + \frac{x}{(\sqrt{3} + 1)}$$ $$\rightarrow s = \frac{x\sqrt{3} + 2x}{(\sqrt{3} + 1)}$$ So :- $[\Delta DEC] = \sqrt{s(s-a)(s-b)(s-c)}$

$$\rightarrow \sqrt{\Bigg(\frac{(x\sqrt{3} + 2x)}{(\sqrt{3} + 1)}\Bigg)\Bigg(\frac{x}{(\sqrt{3} + 1)}\Bigg)\Bigg(\frac{x}{(\sqrt{3} + 1)}\Bigg)\Bigg(\frac{(x\sqrt{3})}{(\sqrt{3} + 1)}\Bigg)}$$

This looks like really a complicated expression, and I really am not going to attempt to simplify this. So can anyone give me a different solution?

Thank You.

Answer 1

If $a,b$ are sides of a triangle and $x$ the mesure of angle between them, then the area of it is $${a\cdot b \cdot \sin x\over 2}$$

We use that formula here.

We have $AD = AE =DE =DC=a $ so$$\frac{[\Delta ADE]}{[\Delta DEC]} = {{a^2\sin 60 \over 2}\over {a^2 \sin 30 \over 2}} = \sqrt{3}$$

Answer 2

By Area of triangle using trigonometry

Area (DEC) $ = \frac12 (CD)(DE) \sin (30^\circ) = \frac 14 x^2$

Answer 3

$DC=x$ and the altitude of $\Delta DEC$ from $E$ is $\frac{x}{2}$. This gives us $$[\Delta DEC] = \frac{x^2}{4}$$ And the result $$\frac{[\Delta ADE]}{[\Delta DEC]}=\sqrt{3}$$