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May 17th, 2011, 09:45 PM   #1
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Height equilateral triangle - formula

Hi, why is the height of an equilateral triangle equal to (s*sqrt3)/2?
I understand you could break the triangle in two, leaving a 30-60-90 triangle. Then the relative amounts are 1:sqrt3:2, and thus the attached formula results. The height becomes: h= 2^2 - (sqrt3)^2, but how does the above formula result?

Thanks for explaining in steps!
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 May 17th, 2011, 09:55 PM #2 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,898 Thanks: 1093 Math Focus: Elementary mathematics and beyond Re: Height equilateral triangle - formula sin(60) = h/s, where h is height and s is side length. Since sin(60) = ?(3)/2, h = ?(3)s/2. $h^2\,=\,\frac{3}{4}s^2,\,h\,=\,\frac{\sqrt{3}}{2}s$ (you want the positive root).
 May 17th, 2011, 10:55 PM #3 Guest   Joined: Posts: n/a Thanks: Using Pythagorean theorem, h² + (s/2)² = s² i.e. h = ?(s² - (s/2)²) = ?(s² - s²/4) = ?(3s²/4) = ?(3s²)/2 = ?3s/2.
 May 18th, 2011, 09:30 PM #4 Newbie   Joined: May 2011 Posts: 22 Thanks: 0 Re: Height equilateral triangle - formula Hi, how do you come up with h^2+(s/2)^2 = s^2? The 30-60-90 triangle has thee relative amounts are 1:sqrt3:2. The height becomes: h= 2^2 - (sqrt3)^2, but how does the above formula result? How do you come from s^2-(s^2/4) to S^3/4? Thanks
 May 18th, 2011, 09:41 PM #5 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,209 Thanks: 517 Math Focus: Calculus/ODEs Re: Height equilateral triangle - formula Take an equilateral triangle having side lengths s and orient it such that the bottom edge is horizontal. Now, from the top vertex drop a vertical line down to the bottom edge, bisecting the triangle. You now have two 30°-60°-90° triangles. Take one of these triangles and observe that the side opposite the 30° angle is s/2, the side opposite the 60° angle call h and the side opposite the 90° angle is the hypotenuse s. Now apply the Pythagorean theorem: $$$\frac{s}{2}$$^2+h^2=s^2$ $h^2=s^2-\frac{s^2}{4}=s^2$$1-\frac{1}{4}$$=\frac{3}{4}s^2$ Taking the positive root, we have: $h=\frac{\sqrt{3}}{2}s$ The method of greg1313 is easier though. Taking the sine of 60°, we have: $\sin$$60^{\circ}$$=\frac{h}{s}$ $h=\frac{\sqrt{3}}{2}s$

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