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mezya [45]
3 years ago
14

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 cr

oss-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
Mathematics
2 answers:
maksim [4K]3 years ago
8 0

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\piin^{2}


IRINA_888 [86]3 years ago
3 0
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)
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2 years ago
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Answer:

1) C. 4 - 3·i

2) D. The second graph shares the same vertex, is inverted, and opens wider than the first graph

3) C. y = (x - 2)² + 3 would shift right two units

4) B. Figure B' is congruent but not similar to figure B

5) A. m∠k' = m∠k

Step-by-step explanation:

1) Given that the real part of the complex number = 4

The imaginary of the complex number = -3

The general form of representing complex numbers is z = a + b·i, we have;

The binomial equivalent to the complex number is z = 4 - 3·i

2) The first graph equation is y = 2·x²

When x = 1, y = 2 and when x = 2, y = 8

The vertex = (h, k)

Where;

h = -b/(2.a) and b = 0, a = 2

∴ h = 0/(2 × 2) = 0

h = 0

k = f(h) = f(0) = 2 × 0² = 0

k = 0

The vertex, (h, k) = (0, 0)

The coefficient, 'a' is positive, therefore, the graph opens down

The second function, y = -(1/2)·x² also has a vertex (h, k) = (0, 0)

The coefficient, 'a' is positive, therefore, the graph opens up

When x = 1, y = -1/2 and when x = 2, y = -2

Therefore, the second function is wider

Therefore;

The second graph shares the same vertex, is inverted, and opens wider than the first graph

3) The given functions are;

First function; y = x² + 3 and second function; y = (x - 2)² + 3

First function;

When x = 1, y = x² + 3 = 1 + 3 = 4

∴ When x = 1, y = 4

Second function;

When y = 4, y = 4 = (x - 2)² + 3

√(1) = x - 2

x = 3

∴ When x = 3, y = 4

First function;

When x = 2, y = x² + 3 = 4 + 3 = 7

∴ When x = 2, y = 7

Second function;

When y = 7, y = 7 = (x - 2)² + 3

√4 = 2 = (x - 2)

x = 2 + 2 = 4

x = 4

∴ When x = 4, y = 7

Therefore, the second function, y = (x - 2)² + 3, has the x-value shifted 2 units to the right for a given value of 'y'

4) The lengths of the sides of figure B are 3 by 4, the lengths of the sides of figure B' 4.5 by 6

The ratio of the corresponding length and width of figures B and B' are;

3/4.5 = 4/6

Therefore, figure B' is similar but not congruent to figure B

5) A rotation and a reflection are rigid transformations and therefore, the dimensions and measure of the original figure and the image are the same;

∴ m∠k' = m∠k.

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