The chord AB is 10 cm, the arc AB of the circle is 90 degrees. Find the length of the perpendicular drawn

The chord AB is 10 cm, the arc AB of the circle is 90 degrees. Find the length of the perpendicular drawn from the center of the circle to the chord AB.

Let’s draw the segments OA and OB, the length of which is equal to the radius of the circle, then OA = OB, and the triangle AOB is isosceles.

Since the degree measure of the arc AB, by condition, is equal to 90, the central angle AOB will also be equal to 90, and the triangle AOB is rectangular and isosceles.

Then, by the Pythagorean theorem, OA ^ 2 + OB ^ 2 = AB ^ 2.

2 * OA ^ 2 = 10 ^ 2 = 100.

ОА = √50 = 5 * √2 cm.

The segment OH is the height and median of the triangle AOB, then AH = BH = AB / 2 = 10/2 = 5 cm.

In a right-angled triangle AOH, by Pifgor’s theorem, we define the leg OH.

OH ^ 2 = OA ^ 2 – AH ^ 2 = 50 – 25 = 25.

OH = 5 cm.

Answer: The length of the perpendicular is 5 cm.