Answer on Question #58514 – Math – Geometry
Question
1. The frustum of a right circular cone has a slant height of 9 ft. and the radii of the bases are 5 ft. and 7 ft.
a. Find the lateral area and total area of the frustum.
b. What is the altitude of this frustum?
c. Find the altitude of the cone that was removed to leave this frustum.
d. What is the volume of the entire cone?
Solution
a. The lateral area is
𝑀 = 𝜋 𝑠 (𝑅 + 𝑟) = 𝜋 ∙ 9 ∙ (7 + 5) = 126𝜋 𝑓𝑡 2
The total area is
𝐴 = 𝑀 + 𝜋𝑅 2 + 𝜋𝑟 2 = 𝜋(126 + 49 + 25) = 200𝜋 𝑓𝑡 2
b. The altitude of this frustum is
ℎ = √𝑠 2 − (𝑅 − 𝑟)2 = √92 − (7 − 5)2 = √77 𝑓𝑡.
c. The altitude 𝐻 of the cone that was removed to leave this frustum
ℎ + ℎ𝑟 ℎ𝑟
=
7
5
5ℎ + 5ℎ𝑟 = 7ℎ𝑟
5ℎ = 2ℎ𝑟
5
5√77
ℎ𝑟 = ℎ =
2
2
d. The volume of the entire cone is
1
1
1
5
𝑉 = 𝜋ℎ𝑡𝑜𝑡𝑎𝑙 (𝑅2 ) = 𝜋(ℎ + ℎ𝑟 )(𝑅 2 ) = 𝜋 (√77 + √77) (72 ) =
3
3
3
2
343𝜋
√77 𝑓𝑡 3 .
6
Question
2. Consider the right pentagon pyramid. The sides of the upper and lower bases of the frustum are 4 and 10
inches, respectively, and the altitude of a lateral face is 6 inches. Find:
a. Lateral area of the frustum
b. Total area of the frustum
c. Volume of the frustum
d. Volume of the entire pyramid.
Solution
a. Lateral area of the frustum
𝐴𝐿 =
1
1
∙ 𝑛(𝑎 + 𝑏)𝑠 = ∙ 5(4 + 10)6 = 210 𝑖𝑛2
2
2
b. Total area of the frustum
1
180 1 2
180
1
180
𝐴 = 𝐴𝐿 + 𝑛𝑎2 cot
+ 𝑛𝑏 cot
= 210 + 5(42 + 102 ) cot
= 410 𝑖𝑛2
4
𝑛
4
𝑛
4
5
c. Volume of the frustum
1
𝑉 = ℎ(𝐴 + 𝐴′ + √𝐴𝐴′ ),
3
where
𝐴=
𝐴′ =
1
180
∙ 5 ∙ 42 ∙ cot
= 28
4
5
1
180
∙ 5 ∙ 102 ∙ cot
= 172
4
5
ℎ = √62 − (
10 − 4 2
) = √27
2
1
𝑉 = √27(28 + 172 + √28 ∙ 172) = 424 𝑖𝑛3
3
d. Volume of the entire pyramid.
1
𝑉𝑡𝑜𝑡𝑎𝑙 = 𝐴′ 𝐻
3
𝐻
𝐻−ℎ
5
=
→ 𝐻 = ℎ = 5√3
10
4
3
1
𝑉𝑡𝑜𝑡𝑎𝑙 = 172 ∙ 5√3 = 497 𝑖𝑛2
3
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