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Math Help - Equilateral triangles

  1. #1
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    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

    thnx in advance
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  2. #2
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    Quote Originally Posted by bloo.tomarto View Post
    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 \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 \frac{\sqrt{3}}{2}-y and now we have a right triangle with one leg of length \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 \frac{y}{\sqrt{2}} to 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
    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 x\frac{\sqrt{3}}{2}. The sides of the rectangle formed are x/2 and x\frac{\sqrt{3}}{2}.

    thnx in advance
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  3. #3
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    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

    radius of circle =x/2divided by radical 3
    diameter of circle = x/radical 3 =diagonal of square
    side of square =x/radical3radicical2
    area= side squares=x^2/6


    bjh
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