F3 Mensuration. be quick

VINZ

2010-07-03 7:23 pm

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

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回答 (1)

用戶10012

2010-07-04 1:41 am

✔ 最佳答案

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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