Question

Из точки A к окружности с центром O проведены две касательные, В и С точки касания. Известно,что АО=10 см ,а радиус окружности = 6 см. Найдите отрезки АВ и АС.........Спасибо заранее!! :)

Answer 1

Радиус и касательная составляют прямой угол. ТреугольникиАОВ и АОС- прямоугольные. АВ и АС их гипотенузы. =10см, а ВО и ОС-катеты 6 см. По т. Пифагора АВ=АС= корень кв. из 100-36=8 см

Answer 2

Problem Analysis

We are given a circle with center O and radius 6 cm. Two tangents, AB and AC, are drawn from point A to the circle, with points B and C being the points of tangency. We are also given that AO = 10 cm. We need to find the lengths of AB and AC.

Solution

To find the lengths of AB and AC, we can use the properties of tangents to a circle. The tangent to a circle is perpendicular to the radius drawn to the point of tangency. Therefore, triangle AOB is a right triangle, with AB being the hypotenuse and AO being one of the legs.

Using the Pythagorean theorem, we can find the length of AB: AB^2 = AO^2 + OB^2

Since AO = 10 cm and OB = radius = 6 cm, we can substitute these values into the equation: AB^2 = 10^2 + 6^2

Simplifying the equation: AB^2 = 100 + 36 AB^2 = 136

Taking the square root of both sides: AB = sqrt(136) AB ≈ 11.66 cm

Therefore, the length of AB is approximately 11.66 cm.

Similarly, triangle AOC is also a right triangle, with AC being the hypotenuse and AO being one of the legs. Using the same logic, we can find the length of AC: AC^2 = AO^2 + OC^2

Since AO = 10 cm and OC = radius = 6 cm, we can substitute these values into the equation: AC^2 = 10^2 + 6^2

Simplifying the equation: AC^2 = 100 + 36 AC^2 = 136

Taking the square root of both sides: AC = sqrt(136) AC ≈ 11.66 cm

Therefore, the length of AC is approximately 11.66 cm.

Answer

The lengths of AB and AC are approximately 11.66 cm each.