Question

A plane, parallel to the base of a cone intersects the cone at the midpoint between points A and B. Determine the area of the cross-section formed by the intersection of the plane and the cone. Leave your answer in terms of pi. 2o height 6 radius
A) 1.5π in2
Eliminate
B) 3π in2
C) 9π in2
D) 36π in2

Answer 1

Answer:

Area of cross-section is $9\pi in^{2}$

Step-by-step explanation:

Given a cone and a plane parallel to base intersect the cone at the mid point of points between A and B.

We have to find the area of cross-section formed by intersection of the plane and the cone which is a new circle formed.

Given,   height=20 in

             radius=6 in

By mid point theorem i.e the line segment formed by joining the mid points of two side of triangle is parallel to third equal to half of the third side.

therefore new circle formed is of radius 3 in

Hence, area of cross-section formed by intersection of the plane and the cone which is a new circle formed= $\pir^{2}$

                                                        = $3^{2}\pi$

                                                        = $9\pi$$in^{2}$

Answer 2

Given that point of intersection is at halft ot the heigth, by similarity of triangles, the radius is halft of 6 in, i.e 3 in.

The the area is Pi*r^2 = Pi*(3in)^2 = Pi*9 in^2 = 9Piin^2

This is option C)