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Khối đa diện đều loại 5 ; 3 là khối mười hai mặt đều
gồm 12 mặt là các ngũ giác đều nên tổng các góc bằng 12 . 3 π = 36 π (mỗi mặt chia thành 5 tam giác để tổng góc)
Chọn đáp án B
* Gọi a là số cạnh, b là số mặt của khối đa diện.
Nếu khối đa diện có các mặt là tam giác thì mỗi mặt có ba cạnh. Trong ba cạnh đó mỗi cạnh lần lượt là cạnh chung của hai mặt.
Ta có 3b = 2a. Nghĩa là b chẵn.
Mà 2a chia hết cho 2 nên 3b cũng chia hết cho 2
⇒ b chia hết cho 2 hay b là số chẵn.
* Ví dụ: hình tứ diện đều có 4 mặt
Giả sử đa diện (H) có m mặt. Vì mỗi mặt của (H) có 3 cạnh, nên m mặt có 3m cạnh. Nhưng mỗi cạnh của (H) là cạnh chung của đúng hai mặt nên số cạnh của (H) bằng c = \(\dfrac{3m}{2}\). Do c là số nguyên dương nên m phải là số chẵn. Ví dụ : Số cạnh của tứ diện bằng sáu.
Xét hai đường sinh SA , SB tùy ý của hình nón. Vẽ đường kính AC của đường tròn đáy. Ta có góc ASC là góc ở đỉnh của hình nón. Hai tam giác ASC và ASB có hai cặp cạnh bằng nhau vì chúng cùng là đường sinh của hình nón.
Ta có cạnh AC ≥ AB nên ∠ ASC ≥ ∠ ASB. Đó là điều cần chứng minh.
