標題:
F3 Mensuration. be quick
發問:
1. An inverted right pyramid is removed form a solid hemisphere, where the square of the hemisphere has been inscribed in the base of the hemisphere. Given that the slant edge of the pyramid is 5 cm long, fibnd the volume of the remaining solid. (Correct to 3 significant figures.) ans. 63.1 更新: * find
最佳解答:
Let radius of hemisphere = r, so diagonal of the square base of the inverted right pyramid = 2r and height of pyramid = r. Since slant height = 5, by Pythagoras thm., r^2 + r^2 = 5^2 = 25 2r^2 = 25, r = 5/sqrt 2. Therefore, volume of hemisphere = (2/3)(3.1416)(5/sqrt 2)^3 = 92.56 cm^3. Since diagonal of the square base = 2r, let side of the inverted right pyramid = L Again by Pythagoras thm., L^2 + L^2 = (2r)^2 = 4r^2 so L^2 = 2r^2 = Base area of the pyramid, therefore, volume of pyramid = (1/3)(L^2)(height 0f pyramid) = (1/3)(2r^2)(r) = (1/3)(25)(5/sqrt 2) = 29.46 cm^3. So volume remaining = 92.56 - 29.46 = 63.1 cm^3.
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