Point O is the midpoint of side AB of square ABCD. The radius of the circle

Point O is the midpoint of side AB of square ABCD. The radius of the circle around triangle AOC is √10. Calculate the perimeter of the square.

1. Since ABCD is a square, then AB = BC = CD = AD = a.

By hypothesis, point O is the midpoint of side AB ⇒ OA = OB = AB / 2 = a / 2.

From the rectangular △ OBC, we find the hypotenuse OC by the Pythagorean theorem:

OC = √ (OB² + BC²) = √ ((a / 2) ² + a²) = √ (a² / 4 + a²) = √ ((a² + 4 * a²) / 4) = √ ((5 * a²) / 4) = a√5 / 2.

From the rectangular △ ABC we find the hypotenuse AC by the Pythagorean theorem:

AC = √ (AB² + BC²) = √ (a² + a²) = √ (2 * a²) = a√2.

1. The area of ​​a triangle can be found by the formula:

S = (a * b * c) / (4 * R),

where a, b and c are the lengths of the sides of the triangle, R is the length of the radius of the circumscribed circle.

Find the area △ AOC:

S = (OA * OC * AC) / (4 * R) = (a / 2 * a√5 / 2 * a√2) / (4 * √10) = (a³√10) / 4: (4 * √10) = (a³√10) / 4 * 1 / (4 * √10) = a³ / 16.

1. The area of ​​a triangle can be found by the formula:

S = (a * h) / 2,

where a is the length of the side of the triangle, h is the length of the height drawn to side a.

Find the area △ AOC:

S = (OA * BC) / 2 = (a / 2 * a) / 2 = a² / 4.

1. Thus:

a³ / 16 = a² / 4;

a³ / 16 – a² / 4 = 0;

a³ / 16 – (4 * a²) / 16 = 0;

(a³ – 4 * a²) / 16 = 0;

a³ – 4 * a² = 0;

a² * (a – 4) = 0;

a – 4 = 0;

a = 4.

1. The perimeter of the square is:

P = a + a + a + a = 4 * a = 4 * 4 = 16.

Answer: P = 16.