Question

В правильно треугольной пирамиде SABC М - середина ребра АВ,S-вершина.Известно что ВС=4см,а SM=29 см.Найдите площадь боковой поверхности . Помогите пожалуйста

Answer 1

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Answer 2

Problem Analysis

We are given a right triangular pyramid SABC, where M is the midpoint of edge AB and S is the vertex. We are also given that BC = 4 cm and SM = 29 cm. We need to find the lateral surface area of the pyramid.

Solution

To find the lateral surface area of the pyramid, we need to find the area of the four triangular faces.

Let's start by finding the length of AB. Since M is the midpoint of AB, we can use the Pythagorean theorem to find the length of AB. Let's denote the length of AB as x.

Using the Pythagorean theorem, we have:

AB^2 = BC^2 + AC^2

Since it is a right triangle, we know that AC = BC. Substituting the given value of BC = 4 cm, we have:

AB^2 = 4^2 + 4^2

Simplifying, we get:

AB^2 = 16 + 16

AB^2 = 32

Taking the square root of both sides, we get:

AB = sqrt(32) cm

Simplifying further, we have:

AB = 4 * sqrt(2) cm

Now, let's find the area of one triangular face. We can use the formula for the area of a triangle, which is:

Area = (1/2) * base * height

In this case, the base is AB and the height is SM. Substituting the values, we have:

Area = (1/2) * AB * SM

Substituting the values of AB = 4 * sqrt(2) cm and SM = 29 cm, we have:

Area = (1/2) * 4 * sqrt(2) * 29 cm^2

Simplifying, we get:

Area = 58 * sqrt(2) cm^2

Since there are four triangular faces, the total lateral surface area of the pyramid is:

Lateral Surface Area = 4 * Area

Substituting the value of Area = 58 * sqrt(2) cm^2, we have:

Lateral Surface Area = 4 * 58 * sqrt(2) cm^2

Simplifying, we get:

Lateral Surface Area = 232 * sqrt(2) cm^2

Therefore, the lateral surface area of the pyramid is 232 * sqrt(2) cm^2.

Answer

The lateral surface area of the pyramid is 232 * sqrt(2) cm^2.