Fi & Lu Area Algebra Trigonometry Functions Complex Numbers Calculus
Fi was learning her multiplication tables, and noticed something interesting. The number in each entry was the number of entries above and to the left of that entry!
Lu explained that, looking at the entry for \(m\times n\), there are \(m\) rows above that entry and \(n\) columns to the left of that entry, so \(m\times n\) cells.
Lu then talked about area. “We can think of area as the number of square tiles we need to cover a given shape. Of course, the tiles have to be all the same size and they can’t overlap. You can see that, for a rectangle, the area is just the width times the height (in units of tile size). What about a right triangle? We can put two identical right triangles together along their hypotenuse to make a rectangle, so, if the smaller sides of the triangles are \(a\) and \(b\), the area of the rectangle is \(a\times b\), and so the area of each of the triangles must be half that, \(\frac{a\times b}{2}\). Of course, we’d have to cut up tiles in order to actually cover a single triangle.”
Mom had been listening, and now decided to show the kids one of the most famous theorems in all of mathematics. “Let me show you the Pythagorean Theorem. It says that if the sides of a right triangle are \(a\), \(b\), and \(c\), with \(c\) being the hypotenuse, then \(a^2 + b^2 = c^2\). There are many proofs of this theorem, but now that you know how to calculate the area of rectangles and right triangles, we need just a tiny bit of algebra: \((a+b)^2 = a^2+2ab+b^2\). Remember that \(x^2\) means \(x\times x\) and \(xy\) means \(x\times y\).”
“From the animation, you can see that the area of the big square is \((a+b)^2 = a^2+2ab+b^2\), and the big square is made up of four gray right triangles plus the inside white square, with total area of \(4\frac{ab}{2}+c^2 = 2ab+c^2\). So \(a^2+2ab+b^2 = 2ab+c^2\), and so \(a^2+b^2=c^2\). Algebra isn’t really needed; just watch the transformation of the white area from \(c^2\) to \(a^2+b^2\) and back.”
Fi asked about other kinds of triangles, and Lu pointed out that any triangle can be split into two right triangles by dropping a line from any vertex perpendicularly to the opposite side. It’s then easy to see that the area is half the product of the length of that side times the length of the perpendicular. Note that if the triangle is obtuse, and the chosen vertex isn’t the obtuse angle, the original triangle is actually the difference between the two resulting right triangles.
It was clear that once you could find the area of any triangle, you could find the area of any polygon by breaking it up into triangles. But what about circles? Fi and Lu both knew about the famous number \(\pi\) that was the ratio of the circumference of a circle to its diameter. So, if \(r\) is the radius, then the circumference is \(2\pi r\). Lu pointed out that you could break a circle into a bunch of equal wedges, each of which was almost a triangle with a height nearly equal to the circle’s radius and a base a bit smaller than the arc. The area of each of those triangles would be nearly half the radius times the arc length. By increasing the number of wedges, the sum of all those areas would get arbitrarily close to half the radius times the circumference, or \(\frac{1}{2}r\times 2\pi r = \pi r^2\). Alternatively, you could cover the circle with equal triangular wedges, each with a height equal to the radius, and a base a bit bigger than the arc. The area of each of those triangles would be a bit more than half the radius times the arc length. By increasing the number of triangles, the sum of all those areas would get arbitrarily close to half the radius times the circumference, again \(\pi r^2\).