MHB Problem about equilateral triangles

[Nich6ls]

Hi! I need help with this exercise: A side of one equilateral triangle equals the height of a second equilateral triangle. Find the ratio of the perimeter of the larger triangle to that of the smaller. A "detailed solution" to be analysed by me then.

Answer: \[ 2/3 sqrt3 \]

Thanks

 

[Euge]

Use the fact that triangles are similar. If $x$ represents the side of first triangle, show that the triangle's height is $x\frac{\sqrt{3}}{2}$. The second triangle has height $x$, and since the triangles are similar, the ratio of their perimeters is equal to the ratio of their heights. Conclude.

 

[HOI]

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.)