## The Dancing Triangle and Its Applications

In the figure below, lines l and m are parallel lines. What can you say about the areas of triangle ABC and triangle ADC?

The distance between two parallel lines is equal at any point, so the two triangles have the same altitude (can you see why?). Further, the two triangles have a common base, therefore, their base lengths are equal. So, the areas of the two triangles are equal. In fact, you can choose any point P on line l and the areas of the triangle ACP will always equal to the areas of triangles ABC and ADC. We like to call this triangle the dancing triangle because using an applet, you can dance it by moving P without changing the area. In the applet below, move points B and D to dance the triangle.  » Read more

## A US President’s Proof of the Pythagorean Theorem

James Garfield, the 20th president of the United States, came up with an original proof of the Pythagorean Theorem in 1876 when he was still a Congressman. His proof was published in New England Journal of Education.

Recall that the Pythagorean Theorem states that given a right triangle with sides $a$, $b$, and hypotenuse $c$, the following equation is always satisified:

$a^2 + b^2 = c^2$.

President Garfield’s proof is quite simple. We can do this in three steps:

1. Find the area of figure above using the trapezoid
2. Find the area of the same figure using the three triangles
3. Equate the results in 1 and 2. » Read more

## Area Tutorial 4 – Derivation of the Area of a Circle

We have learned that the area of a parallelogram is the product of its base and its height, and the circumference of a circle with radius $r$ is $2 \pi r$.

Figure 1 – A circle with radius r and a parallelogram with base b and altitude h.

To find the area of a circle with radius $r$, divide it into congruent sectors (blue and red divisions) then arrange them as shown below.

Figure 2 – As the number of sectors increases, the shape of the rearranged sector is becoming more and more parallelogram-like.

Observe that as the number of sectors increases, the shape of the rearranged sectors is becoming more and more like a parallelogram. In fact, if we can divide the circle into an infinite a number of sectors, it seems that the shape of the rearranged sector is a parallelogram.  Assuming that this is true, then the base of a parallelogram is $\pi r$  (Explain why.), and its altitude is $r$.

Figure 3 – The base of the parallelogram is pi*r and its height is r.

Since the area of a parallelogram is $bh$, we just have to multiply the base of the parallelogram which is $\pi r$ and its height which is $r$ to find its area. Therefore, the area of the parallelogram, which is equal to the area of a circle, is $\pi r^2$.

Another derivation

We can also derive the area of a circle by unwinding an infinite number of circular tracks. The smaller the width of our track becomes, the rearranged figure becomes more and more like a triangle. If it is indeed a triangle, then its area is the product of its height and its width.

Figure 4 – The base of the parallelogram is pi*r and its height is r.

Recall that the area of a triangle is the product of its base and height divided by 2.  Since the base of the of the triangle is equal to the circumference of the circle  ($2 \pi r$), and its height is equal to its radius ($r$), therefore, the area of the triangle, which is equal to the area of the circle, is $\displaystyle\frac{(2 \pi r)(r)}{2} = \pi r^2$.

The processes that we have done above are logical; however,  we only assumed that we can divide the circle into infinite number of sectors, or we can unwind an infinite number of tracks.  These are just assumptions, hence, we are not yet sure if the area of the circle is indeed $\pi r^2$. Of course, we know it is true but we need a proof.  The proof finding the area of a circle needs knowledge on integral calculus. We will discuss the proof of the area of the circle in the future.