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2 years ago

If a right circular cone intersects a plane that runs parallel to the edge of the cone, the resulting curve will be a

2 answers:

5 0

If a right circular cone intersects a plane that runs parallel to the edge of the cone, the resulting curve will be a <u>parabola</u>.

A parabola is curved from one end and open from other end, therefore it would fit the best.

Hope I could help! :)

slega [8]2 years ago

4 0

<h2>Answer:</h2>

If a right circular cone intersects a plane that runs parallel to the edge of the cone, the resulting curve will be a: Parabola.

<h2>Step-by-step explanation:</h2>

If a plane is perpendicular to the axis of rotation of the cone then the resulting cross section will be a <em>Circle.</em>

If the plane is placed slightly tilted across the axis of rotation of the cone then the resulting cross section will be a<em> Ellipse.</em>

If a plane is placed such that it is parallel to the side of the cone i.e. the edge of a cone then the resulting cross section is <em>Parabola.</em>

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Answer: $\text{Length of AG=}\frac{2\sqrt{63}}{3}$

Explanation:

Please follow the diagram in attachment.

As we know median from vertex C to hypotenuse is CM

$\therefore CM=\frac{1}{2}AB$

We are given length of CG=4

Median divide by centroid 2:1

CG:GM=2:1

Where, CG=4

$\therefore GM=2$ ft

Length of CM=4+2= 6 ft

$\therefore CM=\frac{1}{2}AB\Rightarrow AB=12$

In $\triangle ABC, \angle C=90^0$

Using trigonometry ratio identities

$AC=AB\sin 30^0\Rightarrow AC=6$ ft

$BC=AB\cos 30^0\Rightarrow BC=6\sqrt{3}$ ft

$CN=\frac{1}{2}BC\Rightarrow CN=3\sqrt{3}$ ft

In $\triangle CAN, \angle C=90^0$

Using pythagoreous theorem

$AN=\sqrt{6^2+(3\sqrt{3})^2\Rightarrow \sqrt{63}$

Length of AG=2/3 AN

$\text{Length of AG=}\frac{2\sqrt{63}}{3}$ ft

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