Question

Дана правильная четырёхугольная пирамида MABCD, все рёбра основания которой равны б. Угол между прямыми DM и AL, L — середина ребра MB, равен 60°. Найдите высоту данной пирамиды.

Answer 1

Вот нашла такое же задание, надеюсь поможет)

Answer 2

Problem Analysis

We are given a regular quadrilateral pyramid MABCD, where all the edges of the base are equal. The angle between the lines DM and AL, where L is the midpoint of edge MB, is 60°. We need to find the height of the pyramid.

Solution

To find the height of the pyramid, we can use the properties of similar triangles. Let's analyze the given information and solve the problem step by step.

1. We are given that the angle between the lines DM and AL is 60°. This means that triangle DML is an equilateral triangle, as all angles in an equilateral triangle are 60°.

2. Since triangle DML is equilateral, the angles at points D, M, and L are all 60°.

3. Let's consider triangle DML. The height of the pyramid is the perpendicular distance from point D to the base ABCD. Let's call this height h.

4. In triangle DML, the side DM is perpendicular to the base ABCD. Therefore, the height h is equal to the length of DM.

5. We are also given that L is the midpoint of edge MB. This means that triangle LMB is a right triangle, with LM as the hypotenuse and angles at points L and M equal to 90°.

6. Since triangle LMB is a right triangle, we can use trigonometric ratios to find the length of LM.

7. Let's consider triangle LMB. The angle at point L is 90°, and the angle at point M is 60°. Therefore, the angle at point B is 30° (since the sum of angles in a triangle is 180°).

8. Using trigonometric ratios, we can find the length of LM. The ratio of the length of the side opposite the angle B (which is MB) to the length of the hypotenuse LM is equal to the sine of angle B. Therefore, we have:

sin(30°) = MB / LM

9. Simplifying the equation, we get:

MB = LM * sin(30°)

10. Now, let's consider triangle DMB. The angle at point M is 60°, and the angle at point B is 30°. Therefore, the angle at point D is 90° (since the sum of angles in a triangle is 180°).

11. In triangle DMB, the side DM is perpendicular to the base MB. Therefore, the length of DM is equal to the height h.

12. Using the Pythagorean theorem in triangle DMB, we can find the length of DM. The sum of the squares of the lengths of the two shorter sides (DM and MB) is equal to the square of the length of the hypotenuse (DB). Therefore, we have:

DM^2 + MB^2 = DB^2

13. Substituting the value of MB from step 9, we get:

DM^2 + (LM * sin(30°))^2 = DB^2

14. Since triangle DML is equilateral, the length of DM is equal to the length of LM. Therefore, we can simplify the equation as:

LM^2 + (LM * sin(30°))^2 = DB^2

15. Simplifying further, we get:

LM^2 + (LM^2 * sin(30°)^2) = DB^2

16. Using the trigonometric identity sin(30°) = 1/2, we can simplify the equation as:

LM^2 + (LM^2 * (1/2)^2) = DB^2

17. Simplifying further, we get:

LM^2 + (LM^2 * 1/4) = DB^2

18. Combining like terms, we get:

(5/4) * LM^2 = DB^2

19. Taking the square root of both sides, we get:

LM = (2/√5) * DB

20. Now, we know that the height h is equal to the length of DM, which is equal to LM. Therefore, we can substitute the value of LM from step 19 to get:

h = (2/√5) * DB

21. Finally, we need to find the length of DB. Since the base of the pyramid is a regular quadrilateral, all the edges of the base are equal. Let's call the length of each