Question

Величина центрального угла развертки боковой поверхности конуса равна 90. Найдите площадь осевого сечения конуса, если радиус основания равна 6см. Срочнооо! Сделайте пожалуйста

Answer 1

Пусть образующая конуса l
Тогда длина дуги развёртки боковой поверхности составит
z = 2*π*l*(90/360) = πl/2
Но эту же самую длину дуги имеет и основание известного радиуса
z = 2*π*r = 2*π*6 = 12π
πl/2 = 12π
l = 24 см
Т.е. осевое сечение конуса представляет из себя равнобедренный треугольник со сторонами 12, 24, 24 см
Его площадь по формуле Герона
p = (12+24+24)/2 = 30 см
S = √(30*(30-12)(30-24)(30-24)) = 6√(30*18) = 6*3√60 = 36√15 см²

Answer 2

Problem Analysis

We are given that the central angle of the lateral surface of a cone is 90 degrees, and the radius of the base is 6 cm. We need to find the area of the axial section of the cone.

Solution

To find the area of the axial section of the cone, we need to determine the radius of the axial section.

The central angle of the lateral surface of the cone is 90 degrees. The lateral surface of a cone is a sector of a circle, and the central angle of the sector is equal to the central angle of the lateral surface.

Since the central angle of the lateral surface is 90 degrees, the central angle of the sector is also 90 degrees.

The radius of the sector is equal to the radius of the base of the cone, which is given as 6 cm.

To find the area of the axial section, we need to find the area of the sector. The formula for the area of a sector is:

Area of sector = (central angle / 360) * π * r^2

Substituting the given values, we have:

Area of sector = (90 / 360) * π * 6^2

Simplifying the expression, we get:

Area of sector = (1/4) * π * 36

Calculating the value, we find:

Area of sector = 9π

Therefore, the area of the axial section of the cone is 9π square units.

Answer

The area of the axial section of the cone is 9π square units.