This GRE quant practice question is a geometry - solid geometry problem solving question. Computing the volume of a right circular cone obtained by recasting or rotating a sector of a circle.

Question 1 : A sector of a circle of radius 5 cm is recast into a right circular cone of height 4 cm. What is the volume of the resulting cone?

- 12 π cm
^{3} - 100 π cm
^{3} - 33 π cm
^{3} - 32 π cm
^{3} - 4 π cm
^{3}

From INR 2999

The sector of a circle with radius 5 cm.

When recast into a right-circular cone of height 4 cm the sector will appear as shown in the diagram.

The radius of the sector will be the slant height of the cone.

So, the slant height of the cone is 5 cm.

The base radius, slant height, and the height of the cone form a right triangle, with the slant height being the hypotenuse of the triangle.

Slant height of the cone, which is the hypotenuse of the triangle is 5 cm.

The height of the cone, which is one of the perpendicular sides of the triangle is 4 cm.

By Pythagoras theorem, we can compute the radius r =\\sqrt{(\text{Slant Height})^2 - (\text{Height})^2})

r = \\sqrt{(5)^2 -(4)^2}) = \\sqrt{25 − 16}) = \\sqrt{9}) → **r = 3 cm**

Volume of a right circular cone = \\frac{1}{3}) π r^{2} h

radius = 3 cm and height = 4 cm

So, volume = \\frac{1}{3}) π × 32 × 4 = **12 π cm ^{3}**

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