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Using Pythagoras

Geometry · level 6, lesson 2 of 6

A question to start

Stand on flat ground on Mars, with your eyes about 1.8 m above the surface, and look out to the horizon. How far away is the edge of the world?

On Earth the answer is about 4.8 km. On Mars, which is a smaller ball, the ground curves away faster, and the horizon is only about 3.5 km away. Astronauts there will notice that the world feels closer. Both numbers come from the same right triangle, one with its corners at your eye, at the horizon point and at the centre of the planet. This lesson is about finding the right triangle hiding inside a problem, and then letting Pythagoras do the work.

The idea

The two ways to use the theorem

From lesson 1, in a right triangle with legs \(a\), \(b\) and hypotenuse \(c\):

\[a^2 + b^2 = c^2\]

To find the hypotenuse, add the squares of the legs, then take the square root:

\[c = \sqrt{a^2 + b^2}\]

To find a leg, subtract: the square of the hypotenuse minus the square of the other leg, then the square root:

\[a = \sqrt{c^2 - b^2}\]

A square root undoes squaring: \(\sqrt{49} = 7\) because \(7^2 = 49\). Lengths are positive, so we always take the positive root.

Exact answers and simplifying roots

Often the answer is not a whole number. For legs 1 and 2, \(c = \sqrt{1 + 4} = \sqrt5 \approx 2.236\). Writing \(\sqrt5\) is the exact answer; \(2.236\) is a rounded one. Exact answers are often neater and they never collect rounding errors.

Roots can often be simplified by pulling out a square factor, using \(\sqrt{p \times q} = \sqrt p \times \sqrt q\):

\[\sqrt{50} = \sqrt{25 \times 2} = \sqrt{25} \times \sqrt2 = 5\sqrt2\]

Look for the biggest square number that divides the number under the root: 4, 9, 16, 25, 36, 49, ... So \(\sqrt{12} = 2\sqrt3\), \(\sqrt{72} = 6\sqrt2\), \(\sqrt{20} = 2\sqrt5\).

Finding the right triangle

Most problems do not hand you a right triangle. You have to find one, or make one by drawing a line. The common places they hide:

  • A rectangle's diagonal cuts it into two right triangles. Diagonal \(= \sqrt{\text{length}^2 + \text{width}^2}\).
  • An isosceles triangle splits down its line of symmetry into two right triangles. The line from the top meets the base at its midpoint, at right angles (level 5, lesson 5), so each half has the slanted side as hypotenuse and half the base as a leg.
  • A ladder against a wall: wall and ground meet at a right angle, and the ladder is the hypotenuse.
  • Two moves at right angles, like east then north: the straight-line distance is the hypotenuse.
  • A radius and a tangent meet at a right angle (level 5, lesson 6).
  • Half a chord, the radius to its end, and the perpendicular from the centre.

The diagonal of a box

How long is the longest straight stick that fits in a box \(l\) long, \(w\) wide and \(h\) high? It goes from one bottom corner to the opposite top corner, through the inside: the space diagonal. Use Pythagoras twice.

lwhfd
  1. Across the floor, the diagonal \(f\) (orange) is the hypotenuse of a right triangle with legs \(l\) and \(w\): \(f^2 = l^2 + w^2\).
  2. The orange floor diagonal, the green upright edge \(h\) and the blue space diagonal \(d\) make a second right triangle, standing up inside the box (the upright edge is at right angles to the whole floor). So \(d^2 = f^2 + h^2\).
  3. Put the two together:
\[d = \sqrt{l^2 + w^2 + h^2}\]

That is Pythagoras in three dimensions: three squares instead of two.

Pythagorean triples

A Pythagorean triple is three whole numbers \(a\), \(b\), \(c\) with \(a^2 + b^2 = c^2\): 3, 4, 5 and 5, 12, 13 are the famous ones. Any multiple of a triple is a triple too (6, 8, 10 or 15, 20, 25), because multiplying every side by \(k\) multiplies every square by \(k^2\). A triple whose numbers share no common factor is called primitive.

A recipe that makes all of them. Pick two whole numbers \(m > n\). Then

\[a = m^2 - n^2, \qquad b = 2mn, \qquad c = m^2 + n^2\]

is always a triple. Why: multiply out.

\[a^2 + b^2 = (m^2 - n^2)^2 + (2mn)^2 = m^4 - 2m^2n^2 + n^4 + 4m^2n^2 = m^4 + 2m^2n^2 + n^4 = (m^2 + n^2)^2 = c^2\]
\(m\) \(n\) \(m^2 - n^2\) \(2mn\) \(m^2 + n^2\)
2 1 3 4 5
3 2 5 12 13
4 1 15 8 17
4 3 7 24 25
5 2 21 20 29
5 4 9 40 41

This recipe is in Euclid's Elements. Every primitive triple comes from it, with \(m\) and \(n\) sharing no factor and one of them even (that part takes more proof, found in Number theory).

An odd-number shortcut. For any odd number \(a\) from 3 up, the two whole numbers either side of half of \(a^2\) make a triple: \(a = 7\), \(a^2 = 49\), half is 24.5, so 7, 24, 25. Try \(a = 9\): 9, 40, 41.

The distance to the horizon

Here is the opening question. Let the planet have radius \(R\) and your eye be a height \(h\) above the ground. Your line of sight to the horizon just grazes the planet: it is a tangent, touching at the horizon point \(T\).

PTdh RRdown to O

The radius \(OT\) is perpendicular to the tangent, so triangle \(OTP\) has a right angle at \(T\). Its hypotenuse is \(OP = R + h\) and its legs are \(R\) and the distance \(d\). So

\[d^2 + R^2 = (R + h)^2 = R^2 + 2Rh + h^2, \qquad d = \sqrt{2Rh + h^2}\]

For a person, \(h\) is tiny compared with \(R\), and \(h^2\) hardly matters, so \(d \approx \sqrt{2Rh}\).

Worked examples

1. A ladder. A 13 m ladder leans against a wall with its foot 5 m out. How high does it reach?

  1. The ladder is the hypotenuse: \(c = 13\). One leg is 5.
  2. Height \(= \sqrt{13^2 - 5^2} = \sqrt{169 - 25} = \sqrt{144} = 12\) m.

2. The sliding ladder. A 10 m ladder reaches 8 m up a wall, so its foot is \(\sqrt{100 - 64} = 6\) m out. The top slips down 1 m, to 7 m. How far does the foot slide out?

  1. New distance out: \(\sqrt{10^2 - 7^2} = \sqrt{51} \approx 7.14\) m.
  2. The foot moves \(7.14 - 6 = 1.14\) m, a bit more than the 1 m the top dropped. The ladder's length stays the same, but the two legs do not change by the same amount.

3. An equilateral triangle's height. Side 2. Cut it down the middle: the hypotenuse is 2 and one leg is half the base, 1. Height \(= \sqrt{4 - 1} = \sqrt3 \approx 1.732\). (For side \(s\) the height is \(\tfrac{\sqrt3}{2}s\).)

4. Will the stick fit? A box is 60 cm by 50 cm by 40 cm. Will a 1 m stick fit inside?

\(d = \sqrt{60^2 + 50^2 + 40^2} = \sqrt{3600 + 2500 + 1600} = \sqrt{7700} \approx 87.7\) cm. No: the longest stick that fits is under 88 cm.

5. The horizon. Mars has radius about 3389.5 km. An eye height of 1.8 m is 0.0018 km.

  1. \(d = \sqrt{2 \times 3389.5 \times 0.0018 + 0.0018^2} \approx \sqrt{12.2} \approx 3.49\) km.
  2. On Earth, radius about 6371 km: \(d \approx \sqrt{2 \times 6371 \times 0.0018} \approx \sqrt{22.9} \approx 4.79\) km.

Try it

  1. A right triangle has legs 9 cm and 12 cm. Find the hypotenuse.
  2. The hypotenuse is 26 and one leg is 10. Find the other leg.
  3. Find the diagonal of a 7 m by 24 m rectangle.
  4. Find the exact hypotenuse when the legs are 2 and 4, in simplest form, and then to two decimal places.
  5. An equilateral triangle has sides of 10 cm. Find its height, exactly and to one decimal place.
  6. A box measures 2 by 3 by 6. How long is its space diagonal?
  7. Use \(m^2 - n^2\), \(2mn\), \(m^2 + n^2\) with \(m = 4, n = 1\) and with \(m = 5, n = 2\). Check one of them.
  8. Harder: a rover's mast camera on Mars is lifted on a tower so that it is 100 m above the ground. Roughly how far away is its horizon? (Use \(R = 3389.5\) km.)
Answers
  1. \(\sqrt{81 + 144} = \sqrt{225} = 15\) cm.
  2. \(\sqrt{26^2 - 10^2} = \sqrt{676 - 100} = \sqrt{576} = 24\).
  3. \(\sqrt{49 + 576} = \sqrt{625} = 25\) m.
  4. \(\sqrt{4 + 16} = \sqrt{20} = \sqrt{4 \times 5} = 2\sqrt5 \approx 4.47\).
  5. Halves: hypotenuse 10, leg 5. Height \(= \sqrt{100 - 25} = \sqrt{75} = \sqrt{25 \times 3} = 5\sqrt3 \approx 8.7\) cm.
  6. \(\sqrt{4 + 9 + 36} = \sqrt{49} = 7\).
  7. \(m = 4, n = 1\): \(16 - 1 = 15\), \(2 \times 4 \times 1 = 8\), \(16 + 1 = 17\). Check: \(225 + 64 = 289 = 17^2\). \(m = 5, n = 2\): \(25 - 4 = 21\), \(20\), \(29\). Check: \(441 + 400 = 841 = 29^2\).
  8. \(h = 0.1\) km. \(d = \sqrt{2 \times 3389.5 \times 0.1 + 0.01} = \sqrt{677.91} \approx 26.0\) km. Raising the camera from 1.8 m to 100 m pushes the horizon from about 3.5 km to about 26 km.

Common mistake

Adding when you should subtract. If the hypotenuse is one of the two lengths you know, you must subtract squares. Check your answer: the hypotenuse is the longest side, so a leg can never come out longer than it.

Square-rooting each term. \(\sqrt{a^2 + b^2}\) is not \(a + b\). For 3 and 4, \(\sqrt{9 + 16} = 5\), but \(3 + 4 = 7\). The square root of a sum is not the sum of the square roots. (The triangle inequality says the hypotenuse is always shorter than \(a + b\).)

Where it leads