Question

Стороны оснований правильной четырехугольной усеченной пирамиды равны 10 и 2 дм, а высота ее 2 дм. Найти боковое ребро пирамиды.

Answer 1

Дано:

AB=2 дм; ВС=10 дм; А1К=2 дм;

Найти:

АА1-?

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Решение:

A1С1 и АС — диагонали квадратов, лежащих в основании усеченной пирамиды

 

$A_1C_1=A_1B_1*\sqrt{2}=2\sqrt{2}$ (дм)

$AC=AB*\sqrt{2}=10\sqrt{2}$ (дм)

 

A1C1HK - прямоугольник, A1K=C1H=2 (дм)

 

AA1K=CC1H(п/у тр-ки) -> AK=CH

 

$CH=AK=\frac{1}{2}(AC-A_1C_1)=\frac{1}{2}(10\sqrt{2}-\sqrt{2})=4\sqrt{2}$ (дм)

 

По теореме Пифагора:

$AA_1=\sqrt{AK^2+A_1K^2}=\sqrt{(4\sqrt{2})^2+2^2}=\sqrt{32+4}=\sqrt{36}=6$ (дм)

 

Ответ: 6 дм

Answer 2

Given Information:

We are given that the sides of the base of a right quadrilateral truncated pyramid are equal to 10 dm and 2 dm, and its height is 2 dm. We need to find the length of the lateral edge of the pyramid.

Solution:

To find the length of the lateral edge of the pyramid, we can use the Pythagorean theorem. The lateral edge of a right quadrilateral truncated pyramid is the slant height of the lateral face.

Let's denote the length of the lateral edge as l.

Using the Pythagorean theorem, we can write:

l^2 = h^2 + (a1 - a2)^2

Where: - l is the length of the lateral edge (slant height) - h is the height of the pyramid - a1 and a2 are the sides of the base of the pyramid

Substituting the given values into the equation, we have:

l^2 = 2^2 + (10 - 2)^2

Simplifying the equation:

l^2 = 4 + 8^2

l^2 = 4 + 64

l^2 = 68

Taking the square root of both sides:

l = √68

l ≈ 8.246 dm

Therefore, the length of the lateral edge of the pyramid is approximately 8.246 dm.

Answer:

The length of the lateral edge of the given quadrilateral truncated pyramid is approximately 8.246 dm.