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?
Correct Answer : c ) 15 : 16Explanation :
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 = 900
Therefore from Angle – Angle (AA) property we can say that ▲ADC ≅ ▲AEC
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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