A solid right circular cone has a base radius of r and a height of h = \frac{4}{3}r. A sol… — GRE General Test

A solid right circular cone has a base radius of $r$ and a height of $h = \frac{4}{3}r$ . A solid sphere has a radius of $R$ . If the total surface area of the cone is equal to the total surface area of the sphere, which of the following statements must be true? Select all that apply.

The base radius of the cone, $r$ , is strictly greater than the radius of the sphere, $R$ .Answer
The volume of the cone is strictly less than the volume of the sphere.Answer
The ratio of the volume of the cone to the volume of the sphere is $\frac{\sqrt{6}}{4}$ .Answer
D
The volume of the cone is strictly greater than the volume of the sphere.
E
The ratio of the base radius of the cone to the radius of the sphere, $\frac{r}{R}$ , is equal to $\frac{3}{2}$ .

Answer

The statements confirming that the base radius of the cone is strictly greater than the radius of the sphere, that the volume of the cone is strictly less than the volume of the sphere, and that the ratio of the volume of the cone to the volume of the sphere is

\frac{\sqrt{6}}{4}

are all correct.

Equating the total surface area of the cone

\frac{8}{3}\pi r^2

with the surface area of the sphere

4\pi R^2

yields

\frac{r^2}{R^2} = \frac{3}{2}

, which simplifies to

\frac{r}{R} = \sqrt{\frac{3}{2}}

. Because

\sqrt{1.5} > 1

, the base radius of the cone is strictly greater than the radius of the sphere. Furthermore, evaluating the ratio of their volumes gives

\frac{V_{\text{cone}}}{V_{\text{sphere}}} = \frac{\frac{4}{9}\pi r^3}{\frac{4}{3}\pi R^3} = \frac{1}{3}\left(\frac{r}{R}\right)^3 = \frac{\sqrt{6}}{4} \approx 0.612

. Because this ratio is strictly less than

1

, the volume of the cone is strictly less than the volume of the sphere. Thus, the three true statements are those stating

r > R

, that the volume of the cone is strictly less than the sphere's volume, and that their volume ratio is

\frac{\sqrt{6}}{4}

Step-by-Step Solution

Calculate the slant height and total surface area of the cone in terms of

r

Slant height

l = \sqrt{r^2 + \left(\frac{4}{3}r\right)^2} = \sqrt{\frac{25}{9}r^2} = \frac{5}{3}r

. Total surface area

A_{\text{cone}} = \pi r^2 + \pi r l = \pi r^2 + \pi r\left(\frac{5}{3}r\right) = \frac{8}{3}\pi r^2

The total surface area of a right circular cone is the sum of its base area

\pi r^2

and lateral area

\pi r l

Equate the total surface area of the cone to the total surface area of the sphere to find the ratio

\frac{r}{R}

\frac{8}{3}\pi r^2 = 4\pi R^2 \implies 2 r^2 = 3 R^2 \implies \frac{r^2}{R^2} = \frac{3}{2} \implies \frac{r}{R} = \sqrt{\frac{3}{2}} = \frac{\sqrt{6}}{2}.

The surface area of a sphere of radius

R

4\pi R^2

Compare the linear dimensions

r

and

R

Since

\frac{r}{R} = \sqrt{1.5} \approx 1.225 > 1

, it follows that

r > R

A ratio greater than

1

implies the numerator is larger than the denominator.

Express the volumes of both solids and compute their ratio

\frac{V_{\text{cone}}}{V_{\text{sphere}}}

V_{\text{cone}} = \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi r^2 \left(\frac{4}{3}r\right) = \frac{4}{9}\pi r^3

V_{\text{sphere}} = \frac{4}{3}\pi R^3

. Therefore,

\frac{V_{\text{cone}}}{V_{\text{sphere}}} = \frac{\frac{4}{9}\pi r^3}{\frac{4}{3}\pi R^3} = \frac{1}{3}\left(\frac{r}{R}\right)^3 = \frac{1}{3}\left(\frac{3}{2}\right)^{3/2} = \frac{1}{3} \cdot \frac{3\sqrt{3}}{2\sqrt{2}} = \frac{\sqrt{6}}{4}

Using standard volume formulas for cones and spheres and substituting the known linear dimension ratio.

Determine whether the cone's volume is greater than or less than the sphere's volume.

Since

\frac{\sqrt{6}}{4} \approx 0.612 < 1

V_{\text{cone}} < V_{\text{sphere}}

A volume ratio less than

1

proves the cone has a smaller volume than the sphere.

Key Concept

Analyzing geometric scaling, volume, and total surface area relations between cones and spheres using variable constraints.

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