Insights Blog
-- Browse All Articles --
Physics Articles
Physics Tutorials
Physics Guides
Physics FAQ
Math Articles
Math Tutorials
Math Guides
Math FAQ
Education Articles
Education Guides
Bio/Chem Articles
Technology Guides
Computer Science Tutorials
Forums
General Math
Calculus
Differential Equations
Topology and Analysis
Linear and Abstract Algebra
Differential Geometry
Set Theory, Logic, Probability, Statistics
MATLAB, Maple, Mathematica, LaTeX
Trending
Featured Threads
Log in
Register
What's new
Search
Search
Search titles only
By:
General Math
Calculus
Differential Equations
Topology and Analysis
Linear and Abstract Algebra
Differential Geometry
Set Theory, Logic, Probability, Statistics
MATLAB, Maple, Mathematica, LaTeX
Menu
Log in
Register
Navigation
More options
Contact us
Close Menu
JavaScript is disabled. For a better experience, please enable JavaScript in your browser before proceeding.
You are using an out of date browser. It may not display this or other websites correctly.
You should upgrade or use an
alternative browser
.
Forums
Mathematics
General Math
Problem about equilateral triangles
Reply to thread
Message
[QUOTE="HOI, post: 6786166, member: 567247"] If you drop a perpendicular from one vertex of an equilateral triangle to the opposite side, you divide the equilateral triangle into two right triangles that have hypotenuse equal to the length of a side of the equilateral triangle, one leg half that length, and the second leg, whose length you can get from the Pythagorean theorem, is the altitude of the equilateral triangle. If the length of a side of the original equilateral triangle is "s" then the altitude is $\sqrt{s^2- (s/2)^2}= \sqrt{s^2- s^2/4}= \sqrt{3s^2/4}= \frac{s\sqrt{3}}{2}$. So the first equilateral triangle has perimeter 3s while the second has perimeter $\frac{3s\sqrt{3}}{2}$. The ratio of those is $\frac{3s}{\frac{3s\sqrt{3}}{2}}= 3s\frac{2}{3s\sqrt{3}}= \frac{2}{\sqrt{3}}= \frac{2\sqrt{3}}{3}$. (What you wrote, "2/3sqrt(3)" would correctly be interpreted as $\frac{2}{3\sqrt{3}}$, which is wrong, but I suspect you meant "(2/3)sqrt(3)"or $\frac{2}{3}\sqrt{3}= \frac{2\sqrt{3}}{3}$, which is correct.) [/QUOTE]
Insert quotes…
Post reply
Forums
Mathematics
General Math
Problem about equilateral triangles
Back
Top