Question

The surface area (not including the base) of a right circular cone of radius \(r\) and height \(h>0\) is given by $$ \pi r \sqrt{r^{2}+h^{2}} $$ Explain why the surface area is always greater than \(\pi r^{2}\) (a) In terms of the structure of the expression. (b) In terms of geometry.

Short answer

Question: Explain why the surface area of a right circular cone, not including the base, is always greater than \(\pi r^{2}\). Answer: The surface area of a right circular cone, not including the base, is given by the expression \(A = \pi r (\sqrt{r^{2}+h^{2}})\), which is always greater than \(\pi r^{2}\) by its algebraic structure. Moreover, geometrically, when the surface of the cone is flattened out, it forms a sector with a larger area than a circle of radius \(r\), as its slant height is always greater than the radius. Both explanations support the conclusion that the surface area of a right circular cone, not including the base, is always greater than \(\pi r^{2}\).

Step-by-step solution

Part (a): Expression Structure Explanation

From the given expression for the surface area (not including the base) of a right circular cone: $$ A = \pi r \sqrt{r^{2}+h^{2}} $$ We want to show that this surface area is always greater than \(\pi r^{2}\). We can do that by dividing both sides of the inequality by \(\pi r\) (since both \(r\) and \(\pi\) are positive) and comparing the remaining expressions: $$ \sqrt{r^{2}+h^{2}} > r $$ Now we can square both sides of the inequality to remove the square root: $$ r^{2}+h^{2} > r^{2} $$ Removing \(r^{2}\) from both sides leaves us with: $$ h^{2} > 0 $$ Since \(h>0\), then \(h^{2}\) is always positive. This proves that the given expression's structure supports the conclusion that the surface area of the cone is always greater than \(\pi r^{2}\).

Part (b): Geometric Explanation

If we "peel" the surface of the right circular cone, we obtain a sector of a circle with radius \(s\), where \(s\) is the slant height of the cone: \(s = \sqrt{r^{2}+h^{2}}\). The angle of the sector corresponds to a circle of radius \(r\), as shown by the lateral area of the cone: $$ A = \pi r \sqrt{r^{2}+h^{2}} = \pi r s $$ The surface area of a circle with radius \(r\) is given by: $$ A_{circle} = \pi r^{2} $$ Now, since \(s > r\) (slant height is always greater than the radius), the sector we obtain by peeling the cone will have a larger area than the circle with radius \(r\). The larger area can be seen as the sector being "stretched" when the surface of the cone is flattened; it becomes a disk with a hole in the center. Hence, the geometric explanation shows that the circular cone's surface area (not including base) is always greater than a circle with radius \(r\).

Key concepts

Inequality in Geometry

In geometry, inequalities help us compare different quantities, and they can reveal interesting insights about various shapes. In the case of the surface area of a right circular cone, there is an inequality between the cone's surface area and the area of a circle with the same radius. Understanding this inequality requires some manipulation of the expression for the surface area, excluding the base. The formula for the lateral surface area of a cone is given by \( \pi r \sqrt{r^2 + h^2} \). We want to demonstrate that this is always greater than \( \pi r^2 \), the area of a circle with radius \( r \). By manipulating the equation, we find \( \sqrt{r^2 + h^2} > r \) after dividing through by \( \pi r \). Squaring both sides gives \( r^2 + h^2 > r^2 \). Removing \( r^2 \) from both sides leaves \( h^2 > 0 \), which is always true given \( h > 0 \). This inequality confirms that the surface area of the cone, due to its height, surpasses the area of the circle, highlighting a fundamental geometric inequality.

Slant Height

The slant height of a cone is a significant feature that influences its surface area, excluding the base. It's represented by \( s \) and reflects the distance from the base of the cone up to the apex along the cone's surface.You may wonder how the slant height relates to the radius and height of the cone. It's actually calculated using the Pythagorean theorem, resulting in \( s = \sqrt{r^2 + h^2} \). This tells us the slant height is always greater than the radius \( r \) or the height \( h \) alone, unless one of them equals zero, forming a degenerate cone. In the context of our geometric inequality, the slant height \( s \) plays a crucial role. Since \( s \) always exceeds \( r \), the surface area of the cone (\( \pi r s \)), excluding its base, will indeed be larger than the area of a circle defined merely by \( \pi r^2 \). This understanding emphasizes how the slant height's value far surpasses a circle's influence when determining a cone's surface complexity.

Circle Sector

A vital concept in understanding the surface area of a cone—when 'unwrapped'—is the circle sector. Essentially, this involves transforming the lateral surface area into a two-dimensional shape when visualizing the cone's surface.When a cone's surface is "peeled," it forms a sector of a circle with radius equal to the slant height \( s \). The total lateral surface area is then represented by the sweep of this sector, covering a portion of a circle greater than one with radius \( r \). This sector's central angle correlates with the cone's dimensions and effectively shows why the cone's lateral surface area is more significant. Since the slant height \( s \) is larger than the radius \( r \), the sector covers a broader area, underscoring the inequality shown by geometry. Thus, the cone's lateral surface area is a "stretched" disk when analyzed as a sector, weaving geometry with algebra to articulate why the surface area is always beyond \( \pi r^2 \).