Find the area of a triangle with the radius 9

Answer 1

Here is my answer for an equilateral triangle is inscribed in a circle of radius 9.

The inscribed triangle can be cut into 6 equal triangles.

One side each of these triangles (the hypotenuse) is equal to the length of the radius.

Each triangle is a rt. triangle. Each triangle is a 30 - 60 - 90 triangle. 

Angle ADO is a right angle. The measure of angle DBO is 30^o. 

cos 30 = BD/BO = x / 9

√3/2 = x / 9

x = (9√3)/2

Now find the height of the triangle ABC using Pythagorean theorem

CD = sqrt (x² + 2x²) = √3x 

CD = √3 *  (9√3)/2 = 27/2

Area of one triangle: A = (1/2)bh

A = 1/2 * AB * CD

A = 1/2 * 2x * 

A = √3x²

A = √3 ((9√3)/2)²

A = 243√3 /4

Comments

Sorry... typo mistake.. corrected here

Now find the height of the triangle BDC using Pythagorean theorem

CD² + BD² = BC²

CD = sqrt( BC² - BD²)

CD = sqrt ((2x)² - x² )= √3x

Area of one triangle ABC: A = (1/2)* base * height

A = 1/2 * AB * CD

A = 1/2 * 2x * √3x

A = √3x²

A = √3 ((9√3)/2)²

A = 243√3 /4

Answer 2

Here is my answer for an equilateral triangle is circumscribed in a circle of radius 9 and length of the side of an equilateral triangle is 2a.

Angle BDO is a right angle. The measure of angle DBO is 30^o. 

tan 30 = OD/DB = 9 / a

1/√3 = 9 / a

a = 9√3.

The formula for area of an equilateral triangle is A = √3/4 side*side.

A = √3/4 (2a)²

A = √3/4 (4a²)

A = √3a² = √3(9√3)² = √3(9√3)² = √3(81*3) = 243√3 square units.