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Prove that if triangle $\triangle RST$ is equilateral, then the area of $\triangle RST$ is $\sqrt{\frac34}$ times the square of the length of a side.

My thoughts:

Let $s$ be the length of $RT$. Then $\frac s2$ is half the length of $\overline{RT}$. Construct the altitude from the $S$ to side $\overline{RT}$. Call the intersection point $P$. Now, you have a right triangle whose sides are $|\overline{RP}| = \frac s2$ and $|\overline{RS}| = s$. By the Pythagorean Theorem, $|\overline{SP}| = \sqrt{s^2 - \frac14 s^2} = \sqrt{ \frac34 s^2} = \frac{\sqrt{3}}{2} s$. The area of the triangle is $$\frac12 \left(|\overline{SP}|\right)\left(|\overline{RT}|\right) = \left(\frac12 s\right) \left(\frac{\sqrt{3}}{2} s\right) = \frac{\sqrt{3}}{4}s^2,$$ as suggested.

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  • $\begingroup$ Hello. Sorry that you immediately received multiple downvotes and close votes without anyone bothering to tell you what you can do to improve. The reason you've been downvoted and voted towards closure is that you haven't really shared your thoughts on the problem at hand, or where you are having difficulty. I urge you to edit the question ASAP adding your ideas before the question is closed. $\endgroup$ – layman Apr 26 '15 at 16:12
  • $\begingroup$ What I can think is the triangle is divide into two, become isosceles triangle. $\endgroup$ – user231076 Apr 26 '15 at 16:17
  • $\begingroup$ I thought the previous comment you made was your attempt at the solution. Was I wrong? I edited it into your question. $\endgroup$ – layman Apr 26 '15 at 16:26
  • $\begingroup$ Good move, now you have shown exactly how to prove this statement. (Notice the typo in the statement.) $\endgroup$ – MonkeyKing Apr 26 '15 at 16:51
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Let the lenght of the side be $s.$

In an equilateral triangle, the lengths of the sides are equal.

So, $RS=ST=TR=s.$

All angles are $\frac{π}{3}$ $radians.$

Area of a triangle as we know, is $$\frac{(RS)(ST)\sin(S)}{2}.$$

$$=\frac{s^2\sin(\frac{π}{3})}{2}$$

The area of the $ΔRST$ is thus, $$\frac{s^2\sqrt3}{4}$$

$Quod $ $erat $ $demonstrandum$

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  • $\begingroup$ Besides, your thoughts are perfect. $\endgroup$ – Of course it's not me Feb 28 '18 at 8:13

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