A right circular cone, with radius to height ratio as 12:5, is cut parallel to its base to get a smaller cone and a frustum. If height of smaller cone to the height of the frustum are in the ratio 3:1, by what percentage is the combined total surface area of the smaller cone and frustum more with respect to the original cone?
Answers
Given : A right circular cone, with radius to height ratio as 12:5, is cut parallel to its base to get a smaller cone and a frustum.
height of smaller cone to the height of the frustum are in the ratio 3:1,
To Find : By what percentage is the combined total surface area of the smaller cone and frustum more with respect to the original cone?
Solution:
Original cone Radius = 12k
Height = 5k
Lateral height = √(5k)² + (12k)² = 13k
Original cone total surface area = πr² + πrL
= π (12k)² + π12k(13k)
= 300πk²
height of smaller cone to the height of the frustum are in the ratio 3:1,
Height of smaller cone = (3/4)5k = (15/4)k
radius of smaller cone = r
r/12k = (15/4)k / 5k
=> r = 9k
Additional surface area added - base of new smaller cone and upper base of frustum shape.
Surface area added = 2 * π (9k)² = 162πk²
combined total surface area of the smaller cone and frustum more with respect to the original cone = 100 x 162πk² / 300πk²
= 54 %
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