A cone of height 16 cm into two parts by a line parallel to the base of the cone such that a smaller cone of height 4 cm and a frustum are obtained. What will be the ratio of the curved surface area of the frustum and the larger cone?

## Answer

Correct Answer : c ) 15 : 16

Explanation :Let us assume the height, radius and slant height of a smaller cone are h, r, and l respectively

Let us assume the height, radius and slant height of a larger cone are H, R and L respectively

Depicting the above information in the figure we get,

From the above figure,

In triangle ADE and triangle AFC

∠DAE = ∠FAC (common angle)

∠ADE = ∠AFC = 90^{0}

Therefore from Angle – Angle (AA) property we can say that ▲ADC ≅ ▲AEC

Thus,

Therefore,

As we know that

The curved surface area of the smaller cone (CSA)_{S} = πrl ….. (2)

The curved surface area of larger cone (CSA)_{L} = πRL …… (3)

Dividing equation 3 by 2 we get

As Curved surface area of Frustum (CSA)_{F} will be = (CSA)_{L} – (CSA)_{S} ….. (5)

By comparing equation 4 and 5 we get

Hence, (c) is the correct answer.

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