We know that the area of a circle is: A=πr². But this is actually hard to prove.[br][br]So we cut the circle into wedges and place half of the wedges face-up and half face-down. [br]The hanging-out yellow pieces always "fill-up" the empty areas of the rectangle with A=πr²[br][br]As the number of wedges increases, the teal line -> radius and the hanging-out pieces start to fit inside the rectangle.[br]Isn't that cool? Showing this mathematically is called calculus!

What is the total length of the curved parts of the yellow wedges?[br]Why did we label the x-axis with units of π and the y-axis with units of numbers?

The GOLD point in the left panel is chosen at random inside the equilateral triangle. The red, blue and green segments are lines drawn from the GOLD point and perpendicular to each of the sides. Their lengths vary in size as you move the GOLD point from place to place inside the triangle. However, the sum of their lengths is constant. What is the sum equal to? Why? Can you prove it?[br] [[i]You can change the size of the equilateral triangle by dragging the BLACK dots.[/i]][br][br]Would a similar thing be true in a square? Why or why not? [br][br]What about other regular polygons with an odd number of sides? with an even number of sides?[br][br]Under what circumstances can the red, green and blue lengths can form a triangle?[br][br]What is the probability that a point chosen at random inside the triangle will determine three lengths that can form a triangle?[br][br]Can you prove your conjecture?[br][size=85][i][b][br][see further challenge below][/b][/i][/size]

[b][color=#1551b5]- GOING FURTHER[br][br]What positions of the GOLD dot determine an isosceles triangle?[br][br]Can you use the fact that the sum of the distances from any point in the interior of an equilateral triangle to its sides is constant to prove that -[br][br][i]knowing the area and the perimeter of a triangle does NOT uniquely determine the triangle[/i][/color][/b]