Equilateral triangles

• Jan 17th 2010, 03:04 PM
bloo.tomarto
Equilateral triangles
hi guys

I need some help in solving this problem and another problem both extremely similar and both involving the use of the fraction sqrt3/2. I do not undeerstand this concept.

btw here is my question

In An equilateral triangle ABC of side x, a inscibed circle is drawn and then a cyclic square is drawn circumscribed by the circle. Find the area of the square

my other question is similar

in an equilateral triangle ABC of sides 2x, a rectangle is constructed with the midpoints AB and AC as two vertices and the other two lying on bc. I understand one side is x but how do you find the breadth

• Jan 17th 2010, 05:31 PM
HallsofIvy
Quote:

Originally Posted by bloo.tomarto
hi guys

I need some help in solving this problem and another problem both extremely similar and both involving the use of the fraction sqrt3/2. I do not undeerstand this concept.

btw here is my question

In An equilateral triangle ABC of side x, a inscibed circle is drawn and then a cyclic square is drawn circumscribed by the circle. Find the area of the square

Dropping a perpendicular from the vertex of an equilateral triangle, with side length x, to the opposite side gives you two right triangles, each having hypotenuse of length x and one leg of length x/2. By the Pythagorean theorem, the other leg, the altitude of the equilateral triangle, is $\displaystyle \sqrt{x^2- (x/2)^2}= x\frac{\sqrt{3}}{2}$. Taking the distance from the vertex to the center of the circle, a radius of the circle and a diagonal of the square, to be y, the distance from the center of the circle to the opposite side is $\displaystyle \frac{\sqrt{3}}{2}-y$ and now we have a right triangle with one leg of length $\displaystyle \frac{\sqrt{3}}{2}x-y$, one leg of length x/2, and hypotenuse of length y. Set up the Pythagorean theorem for that triangle, solve for y. Now use the fact that is a square has diagonal of length y, then its side has length $\displaystyle \frac{y}{\sqrt{2}}$ to find the area of the square.

Quote:

my other question is similar

in an equilateral triangle ABC of sides 2x, a rectangle is constructed with the midpoints AB and AC as two vertices and the other two lying on bc. I understand one side is x but how do you find the breadth
Unfortunately, what you "understand" is not true. The angles in an equilateral triangle are 60 degrees. The legs of a right triangle with one angle 60 degrees and hypotenuse of length x/2 are x/4 and $\displaystyle x\frac{\sqrt{3}}{2}$. The sides of the rectangle formed are x/2 and $\displaystyle x\frac{\sqrt{3}}{2}$.

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• Jan 18th 2010, 12:52 PM
bjhopper
equilateral triangle
posted by bloo,tomarto

Looking at this problem I am struck by the fact that many similar replies on this site use the Pythagorean Theorem instead of using the properties of standard triangles.
This one is a good example