Question

Диагональ АС основания правильной четырехугольной пирамиды SABCD равна 6. Боковое ребро SB равно 5. Найдите высоту пирамиды SO

Answer 1

ВО=6/2=3
По теореме Пифагора найдём:
3^2+4^2=9+16=25=5 см

Answer 2

Given Information

We are given the following information about a regular quadrilateral pyramid SABCD: - The diagonal of the base, AC, is equal to 6. - The lateral edge SB is equal to 5.

We need to find the height of the pyramid, SO.

Solution

To find the height of the pyramid, we can use the Pythagorean theorem and the properties of similar triangles.

Let's consider triangle SAB. It is a right triangle with SA as the hypotenuse, SB as one of the legs, and AB as the other leg. We can find the length of AB using the Pythagorean theorem:

AB^2 = SA^2 - SB^2 Substituting the given values, we have:

AB^2 = 6^2 - 5^2 = 36 - 25 = 11

Taking the square root of both sides, we get:

AB = √11

Now, let's consider triangle SBO. It is a right triangle with SO as the height, SB as the base, and OB as the hypotenuse. We can find the length of OB using the Pythagorean theorem:

OB^2 = SO^2 + SB^2 Substituting the given values, we have:

OB^2 = SO^2 + 5^2

Since OB is the slant height of the pyramid, it is equal to the height of the triangular face, which is AB. Therefore, we can substitute OB with AB:

AB^2 = SO^2 + 5^2

Substituting the value of AB from earlier, we have:

(√11)^2 = SO^2 + 5^2

Simplifying, we get:

11 = SO^2 + 25

Rearranging the equation, we have:

SO^2 = 11 - 25 = -14

Since the square of a real number cannot be negative, we can conclude that there is no real solution for the height of the pyramid, SO.

Therefore, the height of the pyramid, SO, is undefined or non-existent.

Please let me know if you need any further clarification.