Question

A cone has a height of `9 cm` and its base has a radius of `6 cm`. If the cone is horizontally cut into two segments `7 cm` from the base, what would the surface area of the bottom segment be?

Answer 1

Exact total surface area would be
`6pisqrt117 -(4pi)/3sqrt(52/9)` +`36 pi+ (16pi)/9`
(if required this can be approximated in decimal figures easily)

Explanation:

First,slant height of the cone is to be determined to calculate slant surface area using formula `pi rl`.

The slant height of the cone would be `sqrt (9^2 +6^2)= sqrt 117`

The slant surface area would then be `6pisqrt117`

Now when the cone is cut 7 cm from the base, there would be a leftover cone of height 2 cms and bottom segment would be a frustum of cone. The question is about the surface area ( slant area + plane area) of the frustum.

Slant surface area of the frustum =slant surface area o f the larger cone - slant surface area of the smaller cone.

To find the slant surface area of the smaller cone, its radius needs to be calculated. As shown in the figure below, this can be done using the property of similar triangles(ratio of corresponding sides are equal). The radius of the smaller cone works out equal to `4/3 cm`. Its slant height would be `sqrt(2^2 +(4/3)^2) =sqrt(52 /9)`

Slant surface area of the smaller cone would be=`(4pi)/3 sqrt(52/9)`

Slant surface area of the frustum would be `6pisqrt117 -(4pi)/3sqrt(52/9)`

The plane surface area of the bottom of the frustum would be `36 pi` (a circle of radius 6). Plane surface area of the top of frustum would be `(16pi)/9` ( a circle of radius `4/3`)

The total surface area of the frustum would thus be `6pisqrt117 -(4pi)/3sqrt(52/9)` +`36 pi+ (16pi)/9`