Question

a reservoir is in the shape of a frustum of a right circular cone. it is 8m across at the top and 4m across at the bottom. if it is 6m deep, then its capacity is
A)176m^3
B)196m^3
C)200m^3
D)110m^3

Answer the question with the explanation....

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Answer 1

Answer:

Learner

4 years ago

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i) Volume of a frustum of cone of two end sections radii as R & r and height h units is:

V = (πh/3)*(R² + r² + Rr) cu. units

ii) Here R = 8/2 = 4m and r = 4/2 = 2 m and height h = 6 m

So its capacity = (6π/3)*(4² + 2² + 4*2) cu.m = 56π cu.m

This is nearly 176 m³ [Hence choice A]...Show more

Answer 2

The capacity of the reservoir is (A) 176 m³.

Given: A reservoir is in the shape of a frustum of a right circular cone. it is 8m across at the top and 4m across at the bottom. if it is 6m deep.

To Find: The capacity of the reservoir.

Solution:

• Frustum of a right circular cone is that portion of the right circular cone included between the base and a section parallel to the base not passing through the vertex.

• Volume of a frustum of a right circular cone can be calculated by,

        Volume (V) = 1 / 3 πh ( R² + r² + Rr )        

[where, h = Height, R = Upper radius, r = Lower radius ]

In this question, h = 6 m, R = 4 m, r =  2 m.

So, the volume (V) = 1 / 3 πh ( R² + r² + Rr )  

                                = ( 1/3 ) × ( 22/7 ) × 6 × [ 4² + 2² + 4×2 ]  m³

                                = ( 22/7 ) × 2 × [ 16 + 4 + 8]  m³

                                = ( 22/7 ) × 2 × 28 m³

                                = 176 m³

Hence, the capacity of the reservoir is (A) 176 m³.

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