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arhumsid

Radius of Sphere = (1/3)Height = \((1/3)6\sqrt{3}\) = \(2\sqrt{3}\)

Im sorry, i did not get Radius of Sphere = (1/3)Height.
Could you please explain how can we arrive at above relationship?

Thanks!
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arhumsid
GMATinsight
arhumsid

Radius of Sphere = (1/3)Height = \((1/3)6\sqrt{3}\) = \(2\sqrt{3}\)

Im sorry, i did not get Radius of Sphere = (1/3)Height.
Could you please explain how can we arrive at above relationship?

Thanks!

(Both) Slant Height = 12
Radius = 6
i.e. base Diameter = 12

i.e. In 2 Dimension the Cone is like an Equilateral Triangle in which we can derive the radius as follows
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GMATinsight, Bunuel
it seems that this math problem is beyond the scope of GMAT geometry because the problem requires outside knowledge.
Do you agree with me?
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arhumsid
What is the radius of the largest sphere that can fit inside a right circular cone of base radius \(6 m\) and slant height \(12 m\)?

A. \(2 m\)
B. \(2\sqrt{3} m\)
C. \(4\sqrt{3} m\)
D. \(6 m\)
E. \(6\sqrt{3} m\)

You can use the properties of equilateral triangle, if you remember them.

Imagine cutting a vertical cross section of the cone with a sphere inscribed in it. The front will look like the figure above (https://gmatclub.com/forum/what-is-the- ... l#p1580927)

The radius of the cone is 6 so the base of the triangle is 12. The slant height of the cone is 12 so both other sides are also 12. Hence it is an equilateral triangle.
In an equilateral triangle, the inradius is \(s*\sqrt{3}/6 = 12*\sqrt{3}/6 = 2*\sqrt{3}\)

Answer (B)
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arhumsid
What is the radius of the largest sphere that can fit inside a right circular cone of base radius \(6 m\) and slant height \(12 m\)?

A. \(2 m\)
B. \(2\sqrt{3} m\)
C. \(4\sqrt{3} m\)
D. \(6 m\)
E. \(6\sqrt{3} m\)

You can use the properties of equilateral triangle, if you remember them.

Imagine cutting a vertical cross section of the cone with a sphere inscribed in it. The front will look like the figure above (https://gmatclub.com/forum/what-is-the- ... l#p1580927)

The radius of the cone is 6 so the base of the triangle is 12. The slant height of the cone is 12 so both other sides are also 12. Hence it is an equilateral triangle.
In an equilateral triangle, the inradius is \(s*\sqrt{3}/6 = 12*\sqrt{3}/6 = 2*\sqrt{3}\)

Answer (B)

sure, even if the ath problem gives a very special scenario, I believe this question emphasizes too much on the 3D geometry.
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GMATinsight, Bunuel
it seems that this math problem is beyond the scope of GMAT geometry because the problem requires outside knowledge.
Do you agree with me?

Hi chesstitans

This question may be solved by the properties that GMAT expects from tests takers hence I wouldn't discount the possibility of such a question to appear in gMAT however I will assign a very low probability for such question to appear in exam as it's too complex and also uses a solid rarely used by GMAT I.e. Cone

Posted from my mobile device
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arhumsid
What is the radius of the largest sphere that can fit inside a right circular cone of base radius \(6 m\) and slant height \(12 m\)?

A. \(2 m\)
B. \(2\sqrt{3} m\)
C. \(4\sqrt{3} m\)
D. \(6 m\)
E. \(6\sqrt{3} m\)

Radius of the circle = tan 30 * radius of the cone

height of the cone = 12^2 -6^2 =144 -36 =108 =6*rt3

angle inside the triangle = height of the cone / radius of the cone = 60

Since the triangles are equally divided therefore we can use similar triangles and divide the triangle

THerefore IMO 6/rt 3 = 2* rt3 B
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