Let be a positive integer.
If the radius of a sphere inscribed in a -gonal truncated pyramid is , and the slant height of the -gonal truncated pyramid is , find the volume of the -gonal truncated pyramid.
Calculate and express the result to five decimal places.
Bonus: Given the radius of a sphere inscribed in a truncated cone is , and the slant height of the truncated cone is , find the volume of the truncated cone and show that
This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try
refreshing the page, (b) enabling javascript if it is disabled on your browser and,
finally, (c)
loading the
non-javascript version of this page
. We're sorry about the hassle.
Cut the truncated 4 n -gonal pyramid and inscribed sphere with a vertical plane passing through the center of the sphere and perpendicular to the two bases.
Let x be the lower base of the 4 n -gonal truncated pyramid and y be the upper base of the 4 n -gonal truncated pyramid, where x > y .
From the diagram above the altitudes of the upper and lower bases are h ∗ = 2 x cot ( 4 n π ) and h ∗ ∗ = 2 y cot ( 4 n π ) respectively.
h ∗ = 1 0 − h ∗ ∗ = 2 x cot ( 4 n π ) and 1 0 − 2 h ∗ ∗ = 1 0 − y cot ( 4 n π ) ⟹ 6 4 + ( 1 0 − y cot ( 4 n π ) ) 2 = 1 0 0 ⟹ 6 4 + 1 0 0 − 2 0 cot ( 4 n π ) y + cot 2 ( 4 n π ) y 2 = 1 0 0 ⟹ cot 2 ( 4 n π ) y 2 − 2 0 cot ( 4 n π ) y + 6 4 = 0 ⟹ y = tan ( 4 n π ) ( 1 0 ± 6 )
⟹ y 1 = 1 6 tan ( 4 n π ) and y 2 = 4 tan ( 4 n π ) .
Letting the smaller of the two bases y 2 = 4 tan ( 4 n π ) ⟹ h ∗ ∗ = 2 tan ( 4 n π ) cot ( 4 n π ) = 2 ⟹
h ∗ = 1 0 − h ∗ ∗ = 8 = 2 x cot ( 4 n π ) ⟹ x = 1 6 tan ( 4 n π ) .
The volume of a 4 n -gonal truncated pyramid is V n = 1 2 4 n h ( x 2 + x y + y 2 ) cot ( 4 n π ) = 3 8 n ( 2 5 6 + 6 4 + 1 6 ) tan 2 ( 4 n π ) cot ( 4 n π ) = 8 9 6 tan ( 4 n π ) n
V 1 0 0 = 8 9 6 0 0 tan ( 4 0 0 π ) ≈ 7 0 3 . 7 3 1 2 2
To find the volume V c of the truncated cone, given the radius of the sphere inscribed in the truncated cone is 4 , and the slant height of the truncated cone is 1 0
Let x be the lower base of the truncated cone and y be the upper base of the truncated cone, where x > y .
Similar to the truncated 4 n -gonal pyramid above we obtain:
6 4 + 1 0 0 − 4 0 y + 4 y 2 = 1 0 0 ⟹ 4 y 2 − 4 0 y + 6 4 = 0 ⟹ 4 ( y 2 − 1 0 y + 1 6 ) = 0 ⟹ ( y − 8 ) ( y − 2 ) = 0 ⟹ y 1 = 8 and y 2 = 2
Letting the smaller of the two bases y 2 = 2 ⟹ x = 1 0 − 2 = 8 ⟹ the volume of the truncated cone V c = 3 π h ( x 2 + x y + y 2 ) = 3 8 π ( 8 4 ) = 2 2 4 π
To show lim n → ∞ V n = V c .
Let V n = 8 9 6 tan ( 4 n π ) n .
Using the inequality cos ( x ) < x sin ( x ) < 1 ⟹ 4 π < n tan ( 4 π ) < 4 π sec ( 4 π ) ⟹ lim n → ∞ n tan ( 4 π ) = 4 π ⟹ lim n → ∞ V n = 8 9 6 ( 4 π ) = 2 2 4 π = V c