Mr. John Power entered his Trig class… looking happy
Mr. John Power entered his Trig class. He looked very happy. Some students immediately sensed that Mr. Power was up to something. They knew that he liked to challenge them, and his being happy meant that he had come up with a hard one to crack.
Mr. Power addressed the class: “Today, instead of a lesson, you will have an interesting problem to solve. You can work together. You are sitting in three rows, so each row can form a group. You have 20 minutes to solve the problem.”
The Problem: Determine the value of π using your knowledge of Trig.
Most students were awed. Why would Mr. Power torture them like this? They did not have a clue on how to approach the problem because he had never talked about it in the class. All they remembered were definitions of some Trig functions and a few identities that they had memorized. It never came up in any of the worksheets they did. They just looked at each other wondering if someone else was going to come up with an answer.
After 20 minutes had passed, he asked the students in row 1 for their answer. They replied, “Mr. Power, we have not yet covered this in class.”
From row 2, one of the students proudly said, “We know that the inverse cosine (also called arccos) of -1 is π radians. We used our calculators to evaluate arccos(-1) and obtained 3.14159265359. So π ≈ 3.14159265359.”
Mr. Power was frustrated because this was almost as bad as just getting the value of π from the Internet, but he said, “Good effort.”
Then he turned to the third group and asked Sara, “I hope your group did better than that.”
Sara said that they had, and he called her to the board to explain what they did.
Sara: Actually, it was all thanks to Tommy’s genius.
Tommy said that he was hungry and wished that he had some Hawaiian pizza. That gave us the big clue on how to proceed. We had a short discussion but then we settled that a circle was nothing but a polygon with infinitely many sides like a humongous pizza with lots of slices. Tommy was more worried about the pineapple and the ham bits on the pizza. We just told him to remain quiet, otherwise he would get nothing.
To solve this problem, we decided to start with a 6-slice pizza (Fig. 7.1). First, we examined one slice of the pizza. It started from the center of the pizza at A. We drew a line BC to connect the other two corners of the slice. Instead of the slice, this gave us a triangle ABC. At this time we did not worry about the small amount of the food between the line BC and the edge of the pizza. We argued but decided to worry about it later.
We drew a dotted perpendicular line AD from A to BC. AD is the apothem of the regular hexagon: the distance from its center to the midpoint of a side. We called AD = r. As the number of sides increases, this apothem approaches the radius of the pizza.
The central angle BAC = 360°/6 = 60°, since the angles at the center of all six slices add up to 360°. Angle BAD is half of this, or 30°. Because triangle ABC is isosceles, BD = DC. We called BD = L/2, half the length of the base BC.
Now (L/2)/r = tan (BAD) = tan 30° = 0.57735.
Area of triangle ABC = r × L/2 = 0.57735 r2.
But the pizza contained six such triangles. So the area of the hexagon = 6 × 0.57735 r2 = 3.4641 r2. Since the area of a circle of radius r is πr2, this gives us an initial estimate of π ≈ 3.4641. The estimate is too high because this hexagon, with apothem r, extends beyond the circle of radius r.
Double the number of pizza slices
Remember, our premise was that a circle can be thought of a polygon with infinitely many sides. As a next step towards that, we examined a 12-slice pizza the same way (Fig. 7.2). But Tommy was not happy because that would give him less pizza per slice. So we said that the pizza can be larger but it can be cut into 12 equal slices. Tommy was satisfied. We drew a triangle connecting the center of the pizza and edges of the slices the same way as for the 6-slice pizza. We also drew a similar vertical line from A.
Here BD/AD = (L/2)/r = tan (360°/(12 × 2)) = tan 15° = 0.26795.
Therefore, the area of triangle ABC
= 0.26795 r2.
The area of the dodecagon in the pizza
= 12 × 0.26795 r2 = 3.2154 r2.
This gave us a new estimate of π ≈ 3.2154, smaller than the 3.4641 obtained from the 6-slice pizza.
We noticed that the polygons in the 6-slice and 12-slice pizzas followed the same formula. The area of one triangle ABC = tan (180°/n) × r2, and the area of the regular polygon = n × tan (180°/n) × r2, where n is the number of sides and r is the apothem. As n increases, the polygon approaches a circle of radius r.
100-slice pizza
We next thought of a 100-slice pizza. Tommy didn’t like the idea again because each slice would be too small and he would remain hungry. So, we told him not to worry because this pizza would be very large about the size of our classroom.
With the same equation, the area of a 100-sided polygon in the pizza would be:
100 × tan (360°/200) × r2 = 100 × 0.03142626604 r2 = 3.142626604 r2. Again, this value was slightly smaller than the two values we had obtained before.
For a 10,000-sided polygon, the area would be approximately 10,000 × 0.000314159276 r2 = 3.14159276 r2. Here are the values we got for the areas of the different pizzas: 3.4641 r2 (6 slices), 3.2154 r2 (12 slices), 3.142626604 r2 (100 slices), 3.14159276 r2 (10000 slices). We could have kept going but the 20 minutes were up.
Since the area of a circle is πr2 and the area of our 10,000-sided polygon is approximately 3.14159276 r2, we concluded that π ≈ 3.14159276.
Mr. Power was thrilled with the answer.
He told the class that this was brilliant but not the only way. He continued saying that although they had started learning Trig using right-angled triangles, they should now know that Trig can be used for any triangles and polygons. They should also explore how Trig can also be used for any quadrilaterals and polyhedrons. He gave them several problems from the Trig book as their homework.
By now, Johnny liked the amazing things he could do with Trig because of the help of the girlfriend he loved. He wrote the final exam with confidence. Indeed, he obtained an overall grade of 99% in the course. This mark helped him keep the bike because his overall average from all the courses was well above 85%. Sara told this story to her Nana (grandma) whom she loved very much. Nana said, “I am not surprised because you are Sarasvati – the goddess of knowledge. Sara is only your nickname.”
Challenge
Challenge: Johnny enters a bike ride. Sara asks him, “How long is the ride?” Johnny replies, “I don’t know, but according to this map, we start by traveling 20 km west. Then we turn onto a road that makes an interior angle of 70° with the first road and travel 18 km. Next, we turn onto another road that makes an interior angle of 110° with the second road and travel 15 km. Finally, we take a straight road back to our starting point.” Sara says, “So the route forms a trapezoid.” How long is the bike ride?
Solution: We draw a quadrilateral ABCD (Fig. 10.6). Given: AB = 20 km, BC = 18 km, CD = 15 km, angle ABC = 70°, angle BCD = 110°. Since the sum of the angles ABC and BCD = 180°, the lines AB and CD must be parallel thus making the quadrilateral a trapezoid. We need the sum of all the four sides to get the total distance of the bike route. The lengths of AB, BC and CD are given, but we need to determine the length of DA to solve this problem.
We draw two vertical lines: CE from C to AB and AF from A to CD.
In the triangle EBC, angle EBC = angle ABC = 70°.
Therefore, CE/BC = sin 70° or CE = BC sin 70° = 18 × sin 70° ≈ 16.9145 km.
BE/BC = cos 70° or BE = BC cos 70° = 18 × cos 70° ≈ 6.1564 km.
AE = AB – BE = 20 – 6.1564 ≈ 13.8436 km.
Because AB and CD are parallel, CE and AF being perpendiculars to them must also be parallel.
Therefore AF = CE ≈ 16.9145 km.
and CF = EA ≈ 13.8436 km.
Because we are given that CD = 15 km, FD = 15 – 13.8436 ≈ 1.1564 km.
Because AF was drawn as a line perpendicular to CD, the triangle AFD is a right angle triangle. By the Pythagorean theorem,
DA2 = AF2 + FD2 ≈ 16.91452 + 1.15642, so DA ≈ 16.954 km.
Therefore, the total distance of the bike route = 20 + 18 + 15 + 16.954 ≈ 69.954 km.