Question

В правильной четырёхугольной пирамиде SABCD точка О - центр основания, S - вершина, SO=15,BD=16 Найдите боковое ребро SA

Answer 1

SA^2= SO^2+DB^2/4
SA=17

Answer 2

Problem Analysis

We are given a right quadrilateral pyramid SABCD, where point O is the center of the base, S is the vertex, SO = 15, and BD = 16. We need to find the length of the lateral edge SA.

Solution

To find the length of the lateral edge SA, we can use the Pythagorean theorem. Let's consider the right triangle SOD, where OD is the perpendicular from the vertex S to the base ABCD.

Using the Pythagorean theorem, we have:

SD^2 = SO^2 - OD^2 Since OD is the perpendicular from the vertex S to the base ABCD, it is also the height of the right triangle SOD. The height of a right quadrilateral pyramid is the perpendicular distance from the vertex to the base.

Now, let's find the value of SD. We can use the fact that point O is the center of the base ABCD. Therefore, OD is half the length of BD.

OD = BD/2 Substituting the given value of BD = 16, we have:

OD = 16/2 = 8

Now, we can substitute the values of SO = 15 and OD = 8 into the Pythagorean theorem equation to find SD:

SD^2 = 15^2 - 8^2

Simplifying the equation:

SD^2 = 225 - 64

SD^2 = 161

Taking the square root of both sides:

SD = √161

Now, we have the value of SD. To find the length of the lateral edge SA, we can use the fact that point O is the center of the base ABCD. Therefore, SA is twice the length of SD.

SA = 2 * SD

Substituting the value of SD = √161, we have:

SA = 2 * √161

Therefore, the length of the lateral edge SA is 2√161.

Answer

The length of the lateral edge SA in the right quadrilateral pyramid SABCD is 2√161.