Question

В пра­виль­ной четырёхугольной пи­ра­ми­де бо­ко­вое ребро равно 22, а тан­генс угла между бо­ко­вой гра­нью и плос­ко­стью ос­но­ва­ния равен Найти сто­ро­ну ос­но­ва­ния пирамиды.

Answer 1

Ответ:

Ответ: 11.

Пошаговое объяснение:

Answer 2

I can answer your question in detail. In a regular quadrilateral pyramid, the lateral edge is equal to 22, and the tangent of the angle between the lateral face and the base plane is equal to the square root of 14. To find the side of the base of the pyramid, we can use the following steps:

- Let a be the side of the base, and let h be the height of the pyramid. - Since the base is a square, we can find the diagonal of the base by using the Pythagorean theorem: $$d = \sqrt{2a^2}$$ - We can also find the slant height of the pyramid by using the Pythagorean theorem: $$s = \sqrt{h^2 + \frac{d^2}{4}}$$ - Since we know that the lateral edge is 22, we can substitute it for s and get: $$22 = \sqrt{h^2 + \frac{d^2}{4}}$$ - We can also use the given tangent value to find h in terms of d: $$\tan{\alpha} = \frac{h}{\frac{d}{2}}$$ - Substituting $\tan{\alpha} = \sqrt{14}$ and solving for h, we get: $$h = \frac{d}{2}\sqrt{14}$$ - Now we have two equations with two unknowns, d and h. We can eliminate h by substituting it in the first equation and get: $$22 = \sqrt{\left(\frac{d}{2}\sqrt{14}\right)^2 + \frac{d^2}{4}}$$ - Simplifying and squaring both sides, we get: $$484 = 7d^2 + \frac{d^2}{4}$$ - Multiplying both sides by 4 and rearranging, we get: $$1936 - 29d^2 = 0$$ - Solving for d, we get two possible values: $$d = \pm\sqrt{\frac{1936}{29}} \approx \pm 8.06$$ - Since d must be positive, we take the positive value and substitute it in the formula for a: $$a = \sqrt{\frac{d^2}{2}} = \sqrt{\frac{\left(\sqrt{\frac{1936}{29}}\right)^2}{2}} = \sqrt{\frac{1936}{58}} \approx 5.71$$

Therefore, the side of the base of the pyramid is approximately 5.71 units.

You can find more information about this problem and similar ones on these websites: [YouTube video](https://www.youtube.com/watch?v=uB6bfIwtfVY), [EGE–2023](https://ege.sdamgia.ru/problem?id=541379), [Geometry trainer](https://nsportal.ru/shkola/geometriya/library/2017/01/28/trenazher-geometriya-11-po-teme-piramida-v-zadachah-ege). I hope this helps you.