Triangle ABC is regular, its side is 18cm. Find the radius OB of the circle around it

According to the definition, the center of the circumscribed circle is equidistant from the vertices of the triangle, therefore, the radius of the circumscribed circle can be found from the triangle OBC.

Consider it and see that it is isosceles, and its base is 18 cm, and the sides are equal to the length of the radius of the circumscribed circle R.

The height of an isosceles triangle OH will be both the bisector and the median (by the property of an isosceles triangle). We apply the Pythagorean theorem to the resulting right-angled triangle:

OB ^ 2 = R ^ 2 = OH ^ 2 + (18/2) ^ 2

The angles of an equilateral triangle are 60 °.

And the angle OBH = 60 ° / 2 = 30 °.

OH – leg against an angle of 30 ° is equal to half of the hypotenuse = R / 2.

R ^ 2 = (R / 2) ^ 2 + 9 * 9;

R ^ 2 = R ^ 2/4 + 81;

4R ^ 2 = R ^ 2 + 81 * 4;

4R ^ 2 – R ^ 2 = 324;

3R ^ 2 = 324;

R ^ 2 = 108;

R = 6√3.