Pythagoras and square roots
Tonight's idea
Take any triangle with a right angle. Call the two short sides \(a\) and \(b\) and the long side (the hypotenuse) \(c\). Then \(a^2 + b^2 = c^2\), always. A triangle with sides 3 and 4 has hypotenuse 5, because \(9 + 16 = 25\). Knot a rope into 12 equal lengths and it pulls into a 3-4-5 triangle with a perfect right angle, a trick ancient builders are said to have used.
Why is it true? Look at the two big squares below. Both have side \(a + b\), and both contain four copies of the same triangle. In the left square, what is left over is one tilted square of side \(c\). In the right square, what is left over is two squares, of sides \(a\) and \(b\). Same big square, same four triangles, so the leftovers must be equal: \(c^2 = a^2 + b^2\).
The theorem forces square roots on you. A square with side 1 has a diagonal of length \(\sqrt{1 + 1} = \sqrt{2} = 1.41421\ldots\), a number whose decimals never end and never repeat.
Watch one: find √50 without a square root button
Squeeze it. \(7^2 = 49\) and \(8^2 = 64\), so \(\sqrt{50}\) is between 7 and 8, and much closer to 7.
Try 7.1: \(7.1^2 = 50.41\). Too big. Try 7.05: \(7.05^2 = 49.7025\). Too small. So \(\sqrt{50}\) is between 7.05 and 7.1. Try 7.07: \(7.07^2 = 49.9849\). Very close and just under. So \(\sqrt{50} \approx 7.07\). (A calculator says 7.0711.)
Every few guesses gain you another decimal place. Computers do essentially this, just faster.
Your turn
Tonight's core: the example and exercises 1 to 3. The rest can wait for another visit.
- Find the hypotenuse of right triangles with short sides 5 and 12; 8 and 15; 7 and 24. All three come out as whole numbers. Sets like these are called Pythagorean triples.
- A 13-foot ladder leans against a wall with its base 5 feet from the wall. How high up the wall does it reach?
- Is a triangle with sides 20, 21, 29 right-angled? What about 7, 8, 11? (You do not need to draw them: the theorem works backwards too.)
- A square field is 100 m on each side. How much shorter is it to walk across the diagonal than around two sides? Use \(\sqrt{2} \approx 1.414\).
- Squeeze \(\sqrt{10}\) to two decimal places, showing your guesses.
- A television screen is 48 inches wide and 27 inches tall. TVs are sold by their diagonal. What size is this one sold as? Also check that 48 and 27 are in the ratio 16 to 9.
Stretch: every triple with a 20 in it
Find every Pythagorean triple (whole-number sides of a right triangle) that includes the number 20. Some have 20 as a short side and one has it as the hypotenuse. There are five in total. Two hints: scaling the 3-4-5 triangle gives triples for free; and if 20 is a short side then \(c^2 - b^2 = 400\), which factors as \((c - b)(c + b) = 400\), so you can hunt for pairs of factors of 400.
If you still have energy: explain why \(\sqrt{2}\) cannot be a fraction. Suppose \(\sqrt{2} = \frac{p}{q}\) in lowest terms, so \(p^2 = 2q^2\). Why must \(p\) be even? Write \(p = 2m\) and see what that forces on \(q\). Why is that a contradiction?
Answers
- 13 (\(25 + 144 = 169\)), 17 (\(64 + 225 = 289\)), 25 (\(49 + 576 = 625\)).
- \(\sqrt{169 - 25} = \sqrt{144} = 12\) feet.
- 20, 21, 29: \(400 + 441 = 841 = 29^2\), yes. 7, 8, 11: \(49 + 64 = 113\), but \(11^2 = 121\), no.
- Diagonal \(= 100\sqrt{2} \approx 141.4\) m. Two sides \(= 200\) m. Saving \(\approx 58.6\) m.
- \(3^2 = 9\), \(4^2 = 16\). \(3.1^2 = 9.61\), \(3.2^2 = 10.24\). \(3.15^2 = 9.9225\), \(3.17^2 = 10.0489\), \(3.16^2 = 9.9856\). So \(\sqrt{10} \approx 3.16\).
- \(\sqrt{48^2 + 27^2} = \sqrt{2304 + 729} = \sqrt{3033} \approx 55.07\), sold as a 55-inch TV. \(48 : 27 = 16 : 9\) (divide both by 3).
Stretch. The five triples: (12, 16, 20) which is 3-4-5 times 4, with 20 as hypotenuse; (15, 20, 25) which is 3-4-5 times 5; (20, 21, 29); (20, 48, 52) which is 5-12-13 times 4; (20, 99, 101). The factor hunt: \(c - b\) and \(c + b\) must multiply to 400 and be both even (so that \(b\) and \(c\) come out whole): \(2 \times 200\) gives \(c = 101, b = 99\); \(4 \times 100\) gives \(52, 48\); \(8 \times 50\) gives \(29, 21\); \(10 \times 40\) gives \(25, 15\). For \(\sqrt{2}\): \(p^2 = 2q^2\) is even, and only even numbers have even squares, so \(p\) is even. Then \(p^2 = 4m^2 = 2q^2\) gives \(q^2 = 2m^2\), so \(q\) is even too. But the fraction was in lowest terms, so \(p\) and \(q\) cannot both be even. The assumption was impossible, so \(\sqrt{2}\) is not a fraction.