Side of a equilateral chord triangle , which is drawn into a circle is 9 cm . Find the area of a square , drawn into the same circle .

Question

phi · Answer

Find the radius of the circle using the fact that the side of the triangle is 9.

anonymous · Answer

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so, for a triangle inscribed in a circle, there are 2(easy) ways to find the radius of the circle.
Method 1: by drawing the 3 altitudes of the triangle, which intersect at the center of the circle. This point, since it is equilateral, is the centroid, circumcenter, incenter, & orthocenter. Because it is the centroid, the intersection point is 2/3 the length of the altitude from the vertex.
The altitude length can be found using pythagorean theorem, or trig, by looking at half the equilateral triangle, such that our hypotenuse is 9cm and our small leg is 9/2 cm. This gives us an altitude length of$$9\sqrt{3}/2$$, which means our radius, or the distance from the vertex to the intersection, is $$2/3 * 9\sqrt{3}/2=3\sqrt{3}$$
Method 2: (magically) know that the radius you are looking for is called the circumradius, and that the area of a triangle is equal to $$a*b*c/(4R)$$ where a,b, and c are your side lengths (in this case all are equal to s) and R is your circumradius. the area of an equilateral triangle is $$\sqrt{3}/4 *s^2$$ setting those two areas equal to each other:$$s^3/4R=\sqrt{3}/4s^2$$$$s=\sqrt{3}R$$$$9=\sqrt{3}R$$$$R=3\sqrt{3}$$


so now that you have the radius of the circle, the inscribed square is going to have a diagonal equal to the diameter of the circle. The area of a square in terms of the diagonal, d, is: $$A=d^2/2$$, where d is twice what we solved for above. substituting in:$$A=(6\sqrt{3})^2/2=(\sqrt{108})^2/2=54$$

hope that helps/makes sense. if you need clarification go ahead and ask :)

phi · Answer

area of the square is 2 *r*r

anonymous · Answer

Thank you very much , p-4!