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Understanding Pappus Theorem in Coordinate Geometry

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Statement Proof and Applications of Pappus Theorem with Examples

Pappus theorem (also known as Pappus - Guldinus theorem, pappus theorem) is  named after the Greek Mathematician Pappus of Alexandria (4th century A.D.) It is a theorem that deals with the area of surface of revolution and volume of surfaces and solids of revolution. 


The Pappus Guldinus theorem helps to find the surface area and volume of surfaces and solids of revolution.


The statement of the Pappus theorem was introduced in the print for the first time in 1659, but it was known before by Kepler in 1659, and Guldin in 1640.


The First Pappus Guldinus Theorem

Let C be any curve in the plane. The area of the surface (A) obtained when C is revolved around an external axis is equivalent to the product of the arc length L of C and distance (d) travelled by the Centroid of C.

\[A = Ld\]

For example, the surface area of torus with minor and major radius r and R respectively is:  V = 2\[\pi r\]+ 2\[\pi R\]=  4\[\pi^{2}r R\]


The Second Pappus Guldinus Theorem

Let R be any region in the plane. The volume of the solid obtained when R is revolved around a central axis and is equivalent to the product of the area A of R and the distance covered (d) by the centroid of R.

\[V = Ad\]

For example, the volume of torus with minor and major radius r and R respectively is:

\[V = (\pi r^{2})(2\pi R) = 2\pi^{2}Rr^{2}\]

This special case was introduced by Johannes Kepler using infinitesimals.


Examples

Let us consider the cylinder obtained by revolving a rectangle with vertical side h and horizontal side r around one of its vertical sides (say its left side). The surface area of the cylinder, excluding the top and bottom, is calculated using the Pappus theorem as the surface is obtained by revolving its right side around its left side. The length of the arc of its right side is h and the distance travelled by its centroid is simply 2r. Therefore, the area of the circular cylinder will be 2\[\pi rh\].


The volume of the cylinder is the distance travelled by its centroid multiplied by the area rh of the rectangle. The centroid of the rectangle is considered as its centre, which is a distance \[\frac{r}{2}\] from the axis of revolution. Hence, it covers a distance of  2\[\pi\] \[\frac{r}{2}\]  as it revolves. Therefore, the volume of the cylinder is \[(\pi r)\] \[(rh)\]=\[\pi r^{2}h\].


Pappus And Guldinus Theorem Examples With Solutions

Example 1:

The surface area and volume of torus can easily be calculated using the Pappus theorem. Considering the radius of circle r, and distance from the centre of the circle to the axis of its revolution is R, then the surface area of torus can be calculated using the first Pappus theorem as:


\[A = Ld\] = \[2 \pi r\].\[2\pi R\]= \[4\pi^{2}rR\] 


(Image will be uploaded soon)


The volume inside the torus can be calculated using the second Pappus theorem as:

\[V = Ad\] = \[\pi r^{2}\].\[2\pi R\]= \[2\pi^{2}r^{2}R\]

 

Example 2: 

Find the volume of the right circular cone with radius r and height (h) = 10 

The Pappus theorem for volume is defined as\[V = Ad\]. Before calculating the volume, it is important to find the area of the triangle to which we are revolving. The right circular cone shape can be considered as a triangle rotated around an axis. The formula for the area of triangle is given as:

\[\frac{1}{2}\]\[base \times height\]

The base of the triangle will be considered as the radius (r) = 6 and the height (h) of the triangle = 10.

A = \[\frac{1}{2} \times 6 \times 10\] 

A = 30

The next step is to solve for distance (d). Distance will include the relationship of the centroid of the triangle and its rotation. 

In other words, the distance (d) = \[2\pi \bar{x}\], where \[\bar{x}\] is the coordinate of the centroid and 2refers to the fact that the object is being rotated around an axis. Therefore, the equation for \[\bar{x}\] is given as:

\[\bar{x}\]= \[\frac{1}{A}\] \[\int_{a}^{b}\] \[x f(x) dx\]

Here, we are still missing the f(x), which is the third side of the triangle. There are different ways to calculate the third side of the triangle. But, we will use the pythagorean theorem which states that

H² = h² + b²

Here, H is the hypotenuse, b is the base, and h is the height

Now, we will solve for H and substitute the values as shown below:

H = \[\sqrt{h^{2} + b^{2}}\]

H = \[\sqrt{(10)^{2} + (6)^{2}}\]

H = \[\sqrt{136}\]

Note : H =\[f(x)\]

Hence, \[f (x)\] = 0

Let us now solve for \[\bar{x}\]

\[\bar{x}\] = \[\frac{1}{30}\] \[\int_{0}^{6}\] \[x \sqrt{136} dx\]

\[\bar{x}\] = \[\frac{\sqrt{136}}{30}\] \[\int_{0}^{6}\] \[x\] \[dx\]

\[\bar{x}\] =\[\frac{\sqrt{136}}{30}\] \[(\frac{x^{2}}{x})\] \[\int_{0}^{6}\] 

\[\bar{x}\] =\[\frac{\sqrt{136}}{30}\] \[[\frac{(6)^{2}}{2} - \frac{(0)^{2}}{2}]\]

\[\bar{x}\] =\[\frac{6\sqrt{34}}{5}\]

Now, let us solve for distance d = 2\[\lambda\] \[\bar{x}\] 

\[d = 2 \pi \frac{6\sqrt{34}}{5}\]

\[d = \frac{12\pi\sqrt{34}}{5}\]

At last, we will solve for volume using V = Ad

V = (20) \[(\frac{12\pi\sqrt{34}}{5})\]

V = 72\[\pi \sqrt{34}\]

Therefore, the volume of right circular cone with base radius 6 and height 10 is V = 72\[\pi \sqrt{34}\]

FAQs on Understanding Pappus Theorem in Coordinate Geometry

1. What is Pappus’s Theorem?

**Pappus’s Theorem** states that if two sets of three collinear points are taken on two distinct lines, then the intersection points of opposite sides of the hexagon formed are collinear. In other words, given points A, B, C on one line and D, E, F on another line, the intersection points of AD with BE, AE with BD, and AF with CD lie on a single straight line called the Pappus line.

  • It is a fundamental result in projective geometry.
  • It shows a deep relationship between alignment (collinearity) and intersections.
  • It works even when the lines are extended beyond the segments.

2. What is the statement of Pappus’s Hexagon Theorem?

The statement of Pappus’s Hexagon Theorem is that for two lines with points A, B, C on one line and D, E, F on the other, the three intersection points of pairs (AD ∩ BE), (AE ∩ BD), and (AF ∩ CD) are collinear.

  • Start with two distinct straight lines.
  • Select three distinct points on each line.
  • Join crosswise pairs to form a hexagon.
  • The three new intersection points lie on one straight line.
This result is central in classical projective geometry.

3. What is the Pappus line?

The Pappus line is the straight line on which the three intersection points in Pappus’s Theorem lie. Specifically:

  • Let P = AD ∩ BE
  • Let Q = AE ∩ BD
  • Let R = AF ∩ CD
Then points P, Q, and R are collinear, and that common line is called the Pappus line. It is the main geometric conclusion of the theorem.

4. How do you construct Pappus’s Theorem step by step?

To construct Pappus’s Theorem, you draw two lines, mark three points on each, connect crosswise pairs, and locate their intersections.

  • Draw two distinct lines ℓ and m.
  • Mark A, B, C on ℓ and D, E, F on m.
  • Draw lines AD and BE, mark their intersection P.
  • Draw lines AE and BD, mark their intersection Q.
  • Draw lines AF and CD, mark their intersection R.
  • Check that P, Q, and R lie on one straight line.
This confirms the geometric property described by Pappus’s Theorem.

5. Why is Pappus’s Theorem important in projective geometry?

Pappus’s Theorem is important because it characterizes projective planes that arise from fields and forms a foundation of projective geometry.

  • It connects algebra and geometry.
  • It is used to define coordinate systems in projective planes.
  • It helps distinguish between different types of geometric structures.
In advanced mathematics, a projective plane satisfying Pappus’s Theorem is closely related to one defined over a field.

6. What is an example of Pappus’s Theorem?

An example of Pappus’s Theorem is when two horizontal lines are drawn and three points are chosen on each, forming a hexagon whose opposite sides intersect in collinear points. For instance:

  • Let line ℓ contain points A, B, C.
  • Let parallel line m contain D, E, F.
  • Join AD, BE; AE, BD; AF, CD.
The three resulting intersection points always lie on a straight line, demonstrating the collinearity property of Pappus’s Theorem.

7. What is the difference between Pappus’s Theorem and Pascal’s Theorem?

The main difference is that Pappus’s Theorem applies to two straight lines, while Pascal’s Theorem applies to six points on a conic section.

  • Pappus: Points lie on two straight lines.
  • Pascal: Points lie on a conic (circle, ellipse, parabola, hyperbola).
  • Both conclude that three intersection points are collinear.
Pascal’s Theorem can be seen as a generalization of Pappus’s Theorem.

8. Does Pappus’s Theorem work for parallel lines?

Yes, Pappus’s Theorem also works when the two given lines are parallel. In projective geometry:

  • Parallel lines meet at a point at infinity.
  • The theorem still guarantees three intersection points are collinear.
Thus, even if ℓ ∥ m in Euclidean geometry, the Pappus line still exists in the projective plane.

9. What are the conditions required for Pappus’s Theorem?

The conditions for Pappus’s Theorem are that six points must lie three on each of two distinct lines. Specifically:

  • Two distinct straight lines ℓ and m.
  • Three distinct collinear points A, B, C on ℓ.
  • Three distinct collinear points D, E, F on m.
  • All required intersection points must be well-defined.
Under these conditions, the three constructed intersection points are guaranteed to be collinear.

10. Is Pappus’s Theorem Euclidean or projective?

Pappus’s Theorem is fundamentally a projective geometry theorem, though it can be visualized in Euclidean geometry.

  • In Euclidean geometry, it can be drawn with straight lines and intersections.
  • In projective geometry, it holds universally, including cases with parallel lines.
  • It plays a structural role in defining projective planes over fields.
Therefore, it is mainly studied within projective geometry.