In a regular triangular pyramid PABC with base ABC, the APB angle is 36 °. Point N is taken

| September 21, 2021 | education

In a regular triangular pyramid PABC with base ABC, the APB angle is 36 °. Point N is taken on the edge PC so that AN is the bisector of the angle PAC. The cross-sectional area of the pyramid passing through A, N and B is 25√3. Find the side of the base?

Since the pyramid is correct, the flat angles at the top of the pyramid are equal. Angle APB = APC = BPC = 36.

Consider an isosceles triangle RAS, in which the angle PAC = OCA = (180 – 36) / 2 = 72. Then the angle CAH = 72/2 = 36, since AH is a bisector, and the angle AHC = (180 – 72 – 36) = 72 In triangle ACH the angles at the side CH are equal, then triangle ACH is isosceles, AH = AC.

Similarly BH = BC.

Since the triangle ABC is equilateral, AB = AC = BC, then the triangle ABN is also equilateral, the area of ​​which is 25 * √3 cm2.

Then Savn = AB ^ 2 * √3 / 4 = 25 * √3 cm2.

AB ^ 2 = 4 * 25 * √3 / √3 = 100.

AB = 10 cm.

Answer: The length of the side of the base is 10 cm.



One of the components of a person's success in our time is receiving modern high-quality education, mastering the knowledge, skills and abilities necessary for life in society. A person today needs to study almost all his life, mastering everything new and new, acquiring the necessary professional qualities.