Question

Через точку К на ребре CD тетраэдра DABC проведено сечение параллельно грани ABC. Площадь сечения равна 48, CK:KD=1:4. Найдите площадь грани abc

Answer 1

Площади подобных треугольников относятся как отношение сторон в квадрате

Answer 2

I can answer your question in detail. Through point K on the edge CD of the tetrahedron DABC, a section parallel to the face ABC is drawn. The area of the section is 48, CK:KD=1:4. Find the area of the face abc.

Let's denote the section by KLM, where L and M are the points of intersection of the section with the edges AB and AC, respectively. Then, by the property of parallel sections of a pyramid, we have:

$$\frac{S_{KLM}}{S_{ABC}} = \left(\frac{CK}{CD}\right)^2$$

Since CK:KD=1:4, we have CK=CD/5 and CD=5CK. Substituting this into the equation, we get:

$$\frac{S_{KLM}}{S_{ABC}} = \left(\frac{1}{5}\right)^2 = \frac{1}{25}$$

Multiplying both sides by $S_{ABC}$ and using the given value of $S_{KLM}=48$, we obtain:

$$S_{ABC} = 25 \cdot S_{KLM} = 25 \cdot 48 = 1200$$

Therefore, the area of the face abc is 1200.